difference of mean hypothesis testing

Null Hypothesis

H : μ= 191         (no change)

Research Hypothesis

H : μ> 191         (investigator's belief)

Upper-tailed, Lower-tailed, Two-tailed Tests

The research or alternative hypothesis can take one of three forms. An investigator might believe that the parameter has increased, decreased or changed. For example, an investigator might hypothesize:  

The exact form of the research hypothesis depends on the investigator's belief about the parameter of interest and whether it has possibly increased, decreased or is different from the null value. The research hypothesis is set up by the investigator before any data are collected.

Rejection Region for Upper-Tailed Z Test (H : μ > μ ) with α=0.05

The decision rule is: Reject H if Z 1.645.

Rejection Region for Lower-Tailed Z Test (H 1 : μ < μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < 1.645.

Rejection Region for Two-Tailed Z Test (H 1 : μ ≠ μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < -1.960 or if Z > 1.960.

The complete table of critical values of Z for upper, lower and two-tailed tests can be found in the table of Z values to the right in "Other Resources."

Critical values of t for upper, lower and two-tailed tests can be found in the table of t values in "Other Resources."

• Step 4. Compute the test statistic.  

Here we compute the test statistic by substituting the observed sample data into the test statistic identified in Step 2.

• Step 5. Conclusion.  

The final conclusion is made by comparing the test statistic (which is a summary of the information observed in the sample) to the decision rule. The final conclusion will be either to reject the null hypothesis (because the sample data are very unlikely if the null hypothesis is true) or not to reject the null hypothesis (because the sample data are not very unlikely).  

If the null hypothesis is rejected, then an exact significance level is computed to describe the likelihood of observing the sample data assuming that the null hypothesis is true. The exact level of significance is called the p-value and it will be less than the chosen level of significance if we reject H 0 .

Statistical computing packages provide exact p-values as part of their standard output for hypothesis tests. In fact, when using a statistical computing package, the steps outlined about can be abbreviated. The hypotheses (step 1) should always be set up in advance of any analysis and the significance criterion should also be determined (e.g., α =0.05). Statistical computing packages will produce the test statistic (usually reporting the test statistic as t) and a p-value. The investigator can then determine statistical significance using the following: If p < α then reject H 0 .  

• Step 1. Set up hypotheses and determine level of significance

H 0 : μ = 191 H 1 : μ > 191                 α =0.05

The research hypothesis is that weights have increased, and therefore an upper tailed test is used.

• Step 2. Select the appropriate test statistic.

Because the sample size is large (n > 30) the appropriate test statistic is

• Step 3. Set up decision rule.  

In this example, we are performing an upper tailed test (H 1 : μ> 191), with a Z test statistic and selected α =0.05.   Reject H 0 if Z > 1.645.

We now substitute the sample data into the formula for the test statistic identified in Step 2.  

We reject H 0 because 2.38 > 1.645. We have statistically significant evidence at a =0.05, to show that the mean weight in men in 2006 is more than 191 pounds. Because we rejected the null hypothesis, we now approximate the p-value which is the likelihood of observing the sample data if the null hypothesis is true. An alternative definition of the p-value is the smallest level of significance where we can still reject H 0 . In this example, we observed Z=2.38 and for α=0.05, the critical value was 1.645. Because 2.38 exceeded 1.645 we rejected H 0 . In our conclusion we reported a statistically significant increase in mean weight at a 5% level of significance. Using the table of critical values for upper tailed tests, we can approximate the p-value. If we select α=0.025, the critical value is 1.96, and we still reject H 0 because 2.38 > 1.960. If we select α=0.010 the critical value is 2.326, and we still reject H 0 because 2.38 > 2.326. However, if we select α=0.005, the critical value is 2.576, and we cannot reject H 0 because 2.38 < 2.576. Therefore, the smallest α where we still reject H 0 is 0.010. This is the p-value. A statistical computing package would produce a more precise p-value which would be in between 0.005 and 0.010. Here we are approximating the p-value and would report p < 0.010.                  

Type I and Type II Errors

In all tests of hypothesis, there are two types of errors that can be committed. The first is called a Type I error and refers to the situation where we incorrectly reject H 0 when in fact it is true. This is also called a false positive result (as we incorrectly conclude that the research hypothesis is true when in fact it is not). When we run a test of hypothesis and decide to reject H 0 (e.g., because the test statistic exceeds the critical value in an upper tailed test) then either we make a correct decision because the research hypothesis is true or we commit a Type I error. The different conclusions are summarized in the table below. Note that we will never know whether the null hypothesis is really true or false (i.e., we will never know which row of the following table reflects reality).

Table - Conclusions in Test of Hypothesis

In the first step of the hypothesis test, we select a level of significance, α, and α= P(Type I error). Because we purposely select a small value for α, we control the probability of committing a Type I error. For example, if we select α=0.05, and our test tells us to reject H 0 , then there is a 5% probability that we commit a Type I error. Most investigators are very comfortable with this and are confident when rejecting H 0 that the research hypothesis is true (as it is the more likely scenario when we reject H 0 ).

When we run a test of hypothesis and decide not to reject H 0 (e.g., because the test statistic is below the critical value in an upper tailed test) then either we make a correct decision because the null hypothesis is true or we commit a Type II error. Beta (β) represents the probability of a Type II error and is defined as follows: β=P(Type II error) = P(Do not Reject H 0 | H 0 is false). Unfortunately, we cannot choose β to be small (e.g., 0.05) to control the probability of committing a Type II error because β depends on several factors including the sample size, α, and the research hypothesis. When we do not reject H 0 , it may be very likely that we are committing a Type II error (i.e., failing to reject H 0 when in fact it is false). Therefore, when tests are run and the null hypothesis is not rejected we often make a weak concluding statement allowing for the possibility that we might be committing a Type II error. If we do not reject H 0 , we conclude that we do not have significant evidence to show that H 1 is true. We do not conclude that H 0 is true.

 The most common reason for a Type II error is a small sample size.

Tests with One Sample, Continuous Outcome

Hypothesis testing applications with a continuous outcome variable in a single population are performed according to the five-step procedure outlined above. A key component is setting up the null and research hypotheses. The objective is to compare the mean in a single population to known mean (μ 0 ). The known value is generally derived from another study or report, for example a study in a similar, but not identical, population or a study performed some years ago. The latter is called a historical control. It is important in setting up the hypotheses in a one sample test that the mean specified in the null hypothesis is a fair and reasonable comparator. This will be discussed in the examples that follow.

Test Statistics for Testing H 0 : μ= μ 0

• if n > 30
• if n < 30

Note that statistical computing packages will use the t statistic exclusively and make the necessary adjustments for comparing the test statistic to appropriate values from probability tables to produce a p-value. 

The National Center for Health Statistics (NCHS) published a report in 2005 entitled Health, United States, containing extensive information on major trends in the health of Americans. Data are provided for the US population as a whole and for specific ages, sexes and races.  The NCHS report indicated that in 2002 Americans paid an average of $3,302 per year on health care and prescription drugs. An investigator hypothesizes that in 2005 expenditures have decreased primarily due to the availability of generic drugs. To test the hypothesis, a sample of 100 Americans are selected and their expenditures on health care and prescription drugs in 2005 are measured.   The sample data are summarized as follows: n=100, x̄

=$3,190 and s=$890. Is there statistical evidence of a reduction in expenditures on health care and prescription drugs in 2005? Is the sample mean of $3,190 evidence of a true reduction in the mean or is it within chance fluctuation? We will run the test using the five-step approach. 

• Step 1.  Set up hypotheses and determine level of significance

H 0 : μ = 3,302 H 1 : μ < 3,302           α =0.05

The research hypothesis is that expenditures have decreased, and therefore a lower-tailed test is used.

This is a lower tailed test, using a Z statistic and a 5% level of significance.   Reject H 0 if Z < -1.645.

•   Step 4. Compute the test statistic.  

We do not reject H 0 because -1.26 > -1.645. We do not have statistically significant evidence at α=0.05 to show that the mean expenditures on health care and prescription drugs are lower in 2005 than the mean of $3,302 reported in 2002.  

Recall that when we fail to reject H 0 in a test of hypothesis that either the null hypothesis is true (here the mean expenditures in 2005 are the same as those in 2002 and equal to $3,302) or we committed a Type II error (i.e., we failed to reject H 0 when in fact it is false). In summarizing this test, we conclude that we do not have sufficient evidence to reject H 0 . We do not conclude that H 0 is true, because there may be a moderate to high probability that we committed a Type II error. It is possible that the sample size is not large enough to detect a difference in mean expenditures.      

The NCHS reported that the mean total cholesterol level in 2002 for all adults was 203. Total cholesterol levels in participants who attended the seventh examination of the Offspring in the Framingham Heart Study are summarized as follows: n=3,310, x̄ =200.3, and s=36.8. Is there statistical evidence of a difference in mean cholesterol levels in the Framingham Offspring?

Here we want to assess whether the sample mean of 200.3 in the Framingham sample is statistically significantly different from 203 (i.e., beyond what we would expect by chance). We will run the test using the five-step approach.

H 0 : μ= 203 H 1 : μ≠ 203                       α=0.05

The research hypothesis is that cholesterol levels are different in the Framingham Offspring, and therefore a two-tailed test is used.

•   Step 3. Set up decision rule.  

This is a two-tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.960 or is Z > 1.960.

We reject H 0 because -4.22 ≤ -1. .960. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level in the Framingham Offspring is different from the national average of 203 reported in 2002.   Because we reject H 0 , we also approximate a p-value. Using the two-sided significance levels, p < 0.0001.  

Statistical Significance versus Clinical (Practical) Significance

This example raises an important concept of statistical versus clinical or practical significance. From a statistical standpoint, the total cholesterol levels in the Framingham sample are highly statistically significantly different from the national average with p < 0.0001 (i.e., there is less than a 0.01% chance that we are incorrectly rejecting the null hypothesis). However, the sample mean in the Framingham Offspring study is 200.3, less than 3 units different from the national mean of 203. The reason that the data are so highly statistically significant is due to the very large sample size. It is always important to assess both statistical and clinical significance of data. This is particularly relevant when the sample size is large. Is a 3 unit difference in total cholesterol a meaningful difference?  

Consider again the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. Suppose a new drug is proposed to lower total cholesterol. A study is designed to evaluate the efficacy of the drug in lowering cholesterol.   Fifteen patients are enrolled in the study and asked to take the new drug for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows:   n=15, x̄ =195.9 and s=28.7. Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new drug for 6 weeks? We will run the test using the five-step approach. 

H 0 : μ= 203 H 1 : μ< 203                   α=0.05

•  Step 2. Select the appropriate test statistic.  

Because the sample size is small (n<30) the appropriate test statistic is

This is a lower tailed test, using a t statistic and a 5% level of significance. In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. In this example df=15-1=14. The critical value for a lower tailed test with df=14 and a =0.05 is -2.145 and the decision rule is as follows:   Reject H 0 if t < -2.145.

We do not reject H 0 because -0.96 > -2.145. We do not have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower than the national mean in patients taking the new drug for 6 weeks. Again, because we failed to reject the null hypothesis we make a weaker concluding statement allowing for the possibility that we may have committed a Type II error (i.e., failed to reject H 0 when in fact the drug is efficacious).

This example raises an important issue in terms of study design. In this example we assume in the null hypothesis that the mean cholesterol level is 203. This is taken to be the mean cholesterol level in patients without treatment. Is this an appropriate comparator? Alternative and potentially more efficient study designs to evaluate the effect of the new drug could involve two treatment groups, where one group receives the new drug and the other does not, or we could measure each patient's baseline or pre-treatment cholesterol level and then assess changes from baseline to 6 weeks post-treatment. These designs are also discussed here.

Video - Comparing a Sample Mean to Known Population Mean (8:20)

Link to transcript of the video

Tests with One Sample, Dichotomous Outcome

Hypothesis testing applications with a dichotomous outcome variable in a single population are also performed according to the five-step procedure. Similar to tests for means, a key component is setting up the null and research hypotheses. The objective is to compare the proportion of successes in a single population to a known proportion (p 0 ). That known proportion is generally derived from another study or report and is sometimes called a historical control. It is important in setting up the hypotheses in a one sample test that the proportion specified in the null hypothesis is a fair and reasonable comparator.    

In one sample tests for a dichotomous outcome, we set up our hypotheses against an appropriate comparator. We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size (n) and the sample proportion which is computed by taking the ratio of the number of successes to the sample size,

We then determine the appropriate test statistic (Step 2) for the hypothesis test. The formula for the test statistic is given below.

Test Statistic for Testing H 0 : p = p 0

if min(np 0 , n(1-p 0 )) > 5

The formula above is appropriate for large samples, defined when the smaller of np 0 and n(1-p 0 ) is at least 5. This is similar, but not identical, to the condition required for appropriate use of the confidence interval formula for a population proportion, i.e.,

Here we use the proportion specified in the null hypothesis as the true proportion of successes rather than the sample proportion. If we fail to satisfy the condition, then alternative procedures, called exact methods must be used to test the hypothesis about the population proportion.

Example:  

The NCHS report indicated that in 2002 the prevalence of cigarette smoking among American adults was 21.1%.  Data on prevalent smoking in n=3,536 participants who attended the seventh examination of the Offspring in the Framingham Heart Study indicated that 482/3,536 = 13.6% of the respondents were currently smoking at the time of the exam. Suppose we want to assess whether the prevalence of smoking is lower in the Framingham Offspring sample given the focus on cardiovascular health in that community. Is there evidence of a statistically lower prevalence of smoking in the Framingham Offspring study as compared to the prevalence among all Americans?

H 0 : p = 0.211 H 1 : p < 0.211                     α=0.05

We must first check that the sample size is adequate.   Specifically, we need to check min(np 0 , n(1-p 0 )) = min( 3,536(0.211), 3,536(1-0.211))=min(746, 2790)=746. The sample size is more than adequate so the following formula can be used:

This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.645.

We reject H 0 because -10.93 < -1.645. We have statistically significant evidence at α=0.05 to show that the prevalence of smoking in the Framingham Offspring is lower than the prevalence nationally (21.1%). Here, p < 0.0001.  

The NCHS report indicated that in 2002, 75% of children aged 2 to 17 saw a dentist in the past year. An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of 125 children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data?

Calculate this on your own before checking the answer.

Video - Hypothesis Test for One Sample and a Dichotomous Outcome (3:55)

Tests with Two Independent Samples, Continuous Outcome

There are many applications where it is of interest to compare two independent groups with respect to their mean scores on a continuous outcome. Here we compare means between groups, but rather than generating an estimate of the difference, we will test whether the observed difference (increase, decrease or difference) is statistically significant or not. Remember, that hypothesis testing gives an assessment of statistical significance, whereas estimation gives an estimate of effect and both are important.

Here we discuss the comparison of means when the two comparison groups are independent or physically separate. The two groups might be determined by a particular attribute (e.g., sex, diagnosis of cardiovascular disease) or might be set up by the investigator (e.g., participants assigned to receive an experimental treatment or placebo). The first step in the analysis involves computing descriptive statistics on each of the two samples. Specifically, we compute the sample size, mean and standard deviation in each sample and we denote these summary statistics as follows:

for sample 1:

for sample 2:

The designation of sample 1 and sample 2 is arbitrary. In a clinical trial setting the convention is to call the treatment group 1 and the control group 2. However, when comparing men and women, for example, either group can be 1 or 2.  

In the two independent samples application with a continuous outcome, the parameter of interest in the test of hypothesis is the difference in population means, μ 1 -μ 2 . The null hypothesis is always that there is no difference between groups with respect to means, i.e.,

The null hypothesis can also be written as follows: H 0 : μ 1 = μ 2 . In the research hypothesis, an investigator can hypothesize that the first mean is larger than the second (H 1 : μ 1 > μ 2 ), that the first mean is smaller than the second (H 1 : μ 1 < μ 2 ), or that the means are different (H 1 : μ 1 ≠ μ 2 ). The three different alternatives represent upper-, lower-, and two-tailed tests, respectively. The following test statistics are used to test these hypotheses.

Test Statistics for Testing H 0 : μ 1 = μ 2

• if n 1 > 30 and n 2 > 30
• if n 1 < 30 or n 2 < 30

NOTE: The formulas above assume equal variability in the two populations (i.e., the population variances are equal, or s 1 2 = s 2 2 ). This means that the outcome is equally variable in each of the comparison populations. For analysis, we have samples from each of the comparison populations. If the sample variances are similar, then the assumption about variability in the populations is probably reasonable. As a guideline, if the ratio of the sample variances, s 1 2 /s 2 2 is between 0.5 and 2 (i.e., if one variance is no more than double the other), then the formulas above are appropriate. If the ratio of the sample variances is greater than 2 or less than 0.5 then alternative formulas must be used to account for the heterogeneity in variances.    

The test statistics include Sp, which is the pooled estimate of the common standard deviation (again assuming that the variances in the populations are similar) computed as the weighted average of the standard deviations in the samples as follows:

Because we are assuming equal variances between groups, we pool the information on variability (sample variances) to generate an estimate of the variability in the population. Note: Because Sp is a weighted average of the standard deviations in the sample, Sp will always be in between s 1 and s 2 .)

Data measured on n=3,539 participants who attended the seventh examination of the Offspring in the Framingham Heart Study are shown below.  

Suppose we now wish to assess whether there is a statistically significant difference in mean systolic blood pressures between men and women using a 5% level of significance.  

H 0 : μ 1 = μ 2

H 1 : μ 1 ≠ μ 2                       α=0.05

Because both samples are large ( > 30), we can use the Z test statistic as opposed to t. Note that statistical computing packages use t throughout. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The guideline suggests investigating the ratio of the sample variances, s 1 2 /s 2 2 . Suppose we call the men group 1 and the women group 2. Again, this is arbitrary; it only needs to be noted when interpreting the results. The ratio of the sample variances is 17.5 2 /20.1 2 = 0.76, which falls between 0.5 and 2 suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is

We now substitute the sample data into the formula for the test statistic identified in Step 2. Before substituting, we will first compute Sp, the pooled estimate of the common standard deviation.

Notice that the pooled estimate of the common standard deviation, Sp, falls in between the standard deviations in the comparison groups (i.e., 17.5 and 20.1). Sp is slightly closer in value to the standard deviation in the women (20.1) as there were slightly more women in the sample.   Recall, Sp is a weight average of the standard deviations in the comparison groups, weighted by the respective sample sizes.  

Now the test statistic:

We reject H 0 because 2.66 > 1.960. We have statistically significant evidence at α=0.05 to show that there is a difference in mean systolic blood pressures between men and women. The p-value is p < 0.010.  

Here again we find that there is a statistically significant difference in mean systolic blood pressures between men and women at p < 0.010. Notice that there is a very small difference in the sample means (128.2-126.5 = 1.7 units), but this difference is beyond what would be expected by chance. Is this a clinically meaningful difference? The large sample size in this example is driving the statistical significance. A 95% confidence interval for the difference in mean systolic blood pressures is: 1.7 + 1.26 or (0.44, 2.96). The confidence interval provides an assessment of the magnitude of the difference between means whereas the test of hypothesis and p-value provide an assessment of the statistical significance of the difference.  

Above we performed a study to evaluate a new drug designed to lower total cholesterol. The study involved one sample of patients, each patient took the new drug for 6 weeks and had their cholesterol measured. As a means of evaluating the efficacy of the new drug, the mean total cholesterol following 6 weeks of treatment was compared to the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. At the end of the example, we discussed the appropriateness of the fixed comparator as well as an alternative study design to evaluate the effect of the new drug involving two treatment groups, where one group receives the new drug and the other does not. Here, we revisit the example with a concurrent or parallel control group, which is very typical in randomized controlled trials or clinical trials (refer to the EP713 module on Clinical Trials).  

A new drug is proposed to lower total cholesterol. A randomized controlled trial is designed to evaluate the efficacy of the medication in lowering cholesterol. Thirty participants are enrolled in the trial and are randomly assigned to receive either the new drug or a placebo. The participants do not know which treatment they are assigned. Each participant is asked to take the assigned treatment for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows.

Is there statistical evidence of a reduction in mean total cholesterol in patients taking the new drug for 6 weeks as compared to participants taking placebo? We will run the test using the five-step approach.

H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2                         α=0.05

Because both samples are small (< 30), we use the t test statistic. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The ratio of the sample variances, s 1 2 /s 2 2 =28.7 2 /30.3 2 = 0.90, which falls between 0.5 and 2, suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is:

This is a lower-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table (in More Resources to the right). In order to determine the critical value of t we need degrees of freedom, df, defined as df=n 1 +n 2 -2 = 15+15-2=28. The critical value for a lower tailed test with df=28 and α=0.05 is -1.701 and the decision rule is: Reject H 0 if t < -1.701.

Now the test statistic,

We reject H 0 because -2.92 < -1.701. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower in patients taking the new drug for 6 weeks as compared to patients taking placebo, p < 0.005.

The clinical trial in this example finds a statistically significant reduction in total cholesterol, whereas in the previous example where we had a historical control (as opposed to a parallel control group) we did not demonstrate efficacy of the new drug. Notice that the mean total cholesterol level in patients taking placebo is 217.4 which is very different from the mean cholesterol reported among all Americans in 2002 of 203 and used as the comparator in the prior example. The historical control value may not have been the most appropriate comparator as cholesterol levels have been increasing over time. In the next section, we present another design that can be used to assess the efficacy of the new drug.

Video - Comparison of Two Independent Samples With a Continuous Outcome (8:02)

Tests with Matched Samples, Continuous Outcome

In the previous section we compared two groups with respect to their mean scores on a continuous outcome. An alternative study design is to compare matched or paired samples. The two comparison groups are said to be dependent, and the data can arise from a single sample of participants where each participant is measured twice (possibly before and after an intervention) or from two samples that are matched on specific characteristics (e.g., siblings). When the samples are dependent, we focus on difference scores in each participant or between members of a pair and the test of hypothesis is based on the mean difference, μ d . The null hypothesis again reflects "no difference" and is stated as H 0 : μ d =0 . Note that there are some instances where it is of interest to test whether there is a difference of a particular magnitude (e.g., μ d =5) but in most instances the null hypothesis reflects no difference (i.e., μ d =0).  

The appropriate formula for the test of hypothesis depends on the sample size. The formulas are shown below and are identical to those we presented for estimating the mean of a single sample presented (e.g., when comparing against an external or historical control), except here we focus on difference scores.

Test Statistics for Testing H 0 : μ d =0

A new drug is proposed to lower total cholesterol and a study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients agree to participate in the study and each is asked to take the new drug for 6 weeks. However, before starting the treatment, each patient's total cholesterol level is measured. The initial measurement is a pre-treatment or baseline value. After taking the drug for 6 weeks, each patient's total cholesterol level is measured again and the data are shown below. The rightmost column contains difference scores for each patient, computed by subtracting the 6 week cholesterol level from the baseline level. The differences represent the reduction in total cholesterol over 4 weeks. (The differences could have been computed by subtracting the baseline total cholesterol level from the level measured at 6 weeks. The way in which the differences are computed does not affect the outcome of the analysis only the interpretation.)

Because the differences are computed by subtracting the cholesterols measured at 6 weeks from the baseline values, positive differences indicate reductions and negative differences indicate increases (e.g., participant 12 increases by 2 units over 6 weeks). The goal here is to test whether there is a statistically significant reduction in cholesterol. Because of the way in which we computed the differences, we want to look for an increase in the mean difference (i.e., a positive reduction). In order to conduct the test, we need to summarize the differences. In this sample, we have

The calculations are shown below.  

Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new medication for 6 weeks? We will run the test using the five-step approach.

H 0 : μ d = 0 H 1 : μ d > 0                 α=0.05

NOTE: If we had computed differences by subtracting the baseline level from the level measured at 6 weeks then negative differences would have reflected reductions and the research hypothesis would have been H 1 : μ d < 0. 

• Step 2 . Select the appropriate test statistic.

This is an upper-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table at the right, with df=15-1=14. The critical value for an upper-tailed test with df=14 and α=0.05 is 2.145 and the decision rule is Reject H 0 if t > 2.145.

We now substitute the sample data into the formula for the test statistic identified in Step 2.

We reject H 0 because 4.61 > 2.145. We have statistically significant evidence at α=0.05 to show that there is a reduction in cholesterol levels over 6 weeks.  

Here we illustrate the use of a matched design to test the efficacy of a new drug to lower total cholesterol. We also considered a parallel design (randomized clinical trial) and a study using a historical comparator. It is extremely important to design studies that are best suited to detect a meaningful difference when one exists. There are often several alternatives and investigators work with biostatisticians to determine the best design for each application. It is worth noting that the matched design used here can be problematic in that observed differences may only reflect a "placebo" effect. All participants took the assigned medication, but is the observed reduction attributable to the medication or a result of these participation in a study.

Video - Hypothesis Testing With a Matched Sample and a Continuous Outcome (3:11)

Tests with Two Independent Samples, Dichotomous Outcome

There are several approaches that can be used to test hypotheses concerning two independent proportions. Here we present one approach - the chi-square test of independence is an alternative, equivalent, and perhaps more popular approach to the same analysis. Hypothesis testing with the chi-square test is addressed in the third module in this series: BS704_HypothesisTesting-ChiSquare.

In tests of hypothesis comparing proportions between two independent groups, one test is performed and results can be interpreted to apply to a risk difference, relative risk or odds ratio. As a reminder, the risk difference is computed by taking the difference in proportions between comparison groups, the risk ratio is computed by taking the ratio of proportions, and the odds ratio is computed by taking the ratio of the odds of success in the comparison groups. Because the null values for the risk difference, the risk ratio and the odds ratio are different, the hypotheses in tests of hypothesis look slightly different depending on which measure is used. When performing tests of hypothesis for the risk difference, relative risk or odds ratio, the convention is to label the exposed or treated group 1 and the unexposed or control group 2.      

For example, suppose a study is designed to assess whether there is a significant difference in proportions in two independent comparison groups. The test of interest is as follows:

H 0 : p 1 = p 2 versus H 1 : p 1 ≠ p 2 .  

The following are the hypothesis for testing for a difference in proportions using the risk difference, the risk ratio and the odds ratio. First, the hypotheses above are equivalent to the following:

• For the risk difference, H 0 : p 1 - p 2 = 0 versus H 1 : p 1 - p 2 ≠ 0 which are, by definition, equal to H 0 : RD = 0 versus H 1 : RD ≠ 0.
• If an investigator wants to focus on the risk ratio, the equivalent hypotheses are H 0 : RR = 1 versus H 1 : RR ≠ 1.
• If the investigator wants to focus on the odds ratio, the equivalent hypotheses are H 0 : OR = 1 versus H 1 : OR ≠ 1.  

Suppose a test is performed to test H 0 : RD = 0 versus H 1 : RD ≠ 0 and the test rejects H 0 at α=0.05. Based on this test we can conclude that there is significant evidence, α=0.05, of a difference in proportions, significant evidence that the risk difference is not zero, significant evidence that the risk ratio and odds ratio are not one. The risk difference is analogous to the difference in means when the outcome is continuous. Here the parameter of interest is the difference in proportions in the population, RD = p 1 -p 2 and the null value for the risk difference is zero. In a test of hypothesis for the risk difference, the null hypothesis is always H 0 : RD = 0. This is equivalent to H 0 : RR = 1 and H 0 : OR = 1. In the research hypothesis, an investigator can hypothesize that the first proportion is larger than the second (H 1 : p 1 > p 2 , which is equivalent to H 1 : RD > 0, H 1 : RR > 1 and H 1 : OR > 1), that the first proportion is smaller than the second (H 1 : p 1 < p 2 , which is equivalent to H 1 : RD < 0, H 1 : RR < 1 and H 1 : OR < 1), or that the proportions are different (H 1 : p 1 ≠ p 2 , which is equivalent to H 1 : RD ≠ 0, H 1 : RR ≠ 1 and H 1 : OR ≠

1). The three different alternatives represent upper-, lower- and two-tailed tests, respectively.  

The formula for the test of hypothesis for the difference in proportions is given below.

Test Statistics for Testing H 0 : p 1 = p

The formula above is appropriate for large samples, defined as at least 5 successes (np > 5) and at least 5 failures (n(1-p > 5)) in each of the two samples. If there are fewer than 5 successes or failures in either comparison group, then alternative procedures, called exact methods must be used to estimate the difference in population proportions.

The following table summarizes data from n=3,799 participants who attended the fifth examination of the Offspring in the Framingham Heart Study. The outcome of interest is prevalent CVD and we want to test whether the prevalence of CVD is significantly higher in smokers as compared to non-smokers.

The prevalence of CVD (or proportion of participants with prevalent CVD) among non-smokers is 298/3,055 = 0.0975 and the prevalence of CVD among current smokers is 81/744 = 0.1089. Here smoking status defines the comparison groups and we will call the current smokers group 1 (exposed) and the non-smokers (unexposed) group 2. The test of hypothesis is conducted below using the five step approach.

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2                 α=0.05

• Step 2.  Select the appropriate test statistic.  

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group. In this example, we have more than enough successes (cases of prevalent CVD) and failures (persons free of CVD) in each comparison group. The sample size is more than adequate so the following formula can be used:

Reject H 0 if Z < -1.960 or if Z > 1.960.

We now substitute the sample data into the formula for the test statistic identified in Step 2. We first compute the overall proportion of successes:

We now substitute to compute the test statistic.

• Step 5. Conclusion.

We do not reject H 0 because -1.960 < 0.927 < 1.960. We do not have statistically significant evidence at α=0.05 to show that there is a difference in prevalent CVD between smokers and non-smokers.  

A 95% confidence interval for the difference in prevalent CVD (or risk difference) between smokers and non-smokers as 0.0114 + 0.0247, or between -0.0133 and 0.0361. Because the 95% confidence interval for the risk difference includes zero we again conclude that there is no statistically significant difference in prevalent CVD between smokers and non-smokers.    

Smoking has been shown over and over to be a risk factor for cardiovascular disease. What might explain the fact that we did not observe a statistically significant difference using data from the Framingham Heart Study? HINT: Here we consider prevalent CVD, would the results have been different if we considered incident CVD?

A randomized trial is designed to evaluate the effectiveness of a newly developed pain reliever designed to reduce pain in patients following joint replacement surgery. The trial compares the new pain reliever to the pain reliever currently in use (called the standard of care). A total of 100 patients undergoing joint replacement surgery agreed to participate in the trial. Patients were randomly assigned to receive either the new pain reliever or the standard pain reliever following surgery and were blind to the treatment assignment. Before receiving the assigned treatment, patients were asked to rate their pain on a scale of 0-10 with higher scores indicative of more pain. Each patient was then given the assigned treatment and after 30 minutes was again asked to rate their pain on the same scale. The primary outcome was a reduction in pain of 3 or more scale points (defined by clinicians as a clinically meaningful reduction). The following data were observed in the trial.

We now test whether there is a statistically significant difference in the proportions of patients reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) using the five step approach.  

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2              α=0.05

Here the new or experimental pain reliever is group 1 and the standard pain reliever is group 2.

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group, i.e.,

In this example, we have min(50(0.46), 50(1-0.46), 50(0.22), 50(1-0.22)) = min(23, 27, 11, 39) = 11. The sample size is adequate so the following formula can be used

We reject H 0 because 2.526 > 1960. We have statistically significant evidence at a =0.05 to show that there is a difference in the proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever.

A 95% confidence interval for the difference in proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever is 0.24 + 0.18 or between 0.06 and 0.42. Because the 95% confidence interval does not include zero we concluded that there was a statistically significant difference in proportions which is consistent with the test of hypothesis result. 

Again, the procedures discussed here apply to applications where there are two independent comparison groups and a dichotomous outcome. There are other applications in which it is of interest to compare a dichotomous outcome in matched or paired samples. For example, in a clinical trial we might wish to test the effectiveness of a new antibiotic eye drop for the treatment of bacterial conjunctivitis. Participants use the new antibiotic eye drop in one eye and a comparator (placebo or active control treatment) in the other. The success of the treatment (yes/no) is recorded for each participant for each eye. Because the two assessments (success or failure) are paired, we cannot use the procedures discussed here. The appropriate test is called McNemar's test (sometimes called McNemar's test for dependent proportions).  

Vide0 - Hypothesis Testing With Two Independent Samples and a Dichotomous Outcome (2:55)

Here we presented hypothesis testing techniques for means and proportions in one and two sample situations. Tests of hypothesis involve several steps, including specifying the null and alternative or research hypothesis, selecting and computing an appropriate test statistic, setting up a decision rule and drawing a conclusion. There are many details to consider in hypothesis testing. The first is to determine the appropriate test. We discussed Z and t tests here for different applications. The appropriate test depends on the distribution of the outcome variable (continuous or dichotomous), the number of comparison groups (one, two) and whether the comparison groups are independent or dependent. The following table summarizes the different tests of hypothesis discussed here.

• Continuous Outcome, One Sample: H0: μ = μ0
• Continuous Outcome, Two Independent Samples: H0: μ1 = μ2
• Continuous Outcome, Two Matched Samples: H0: μd = 0
• Dichotomous Outcome, One Sample: H0: p = p 0
• Dichotomous Outcome, Two Independent Samples: H0: p1 = p2, RD=0, RR=1, OR=1

Once the type of test is determined, the details of the test must be specified. Specifically, the null and alternative hypotheses must be clearly stated. The null hypothesis always reflects the "no change" or "no difference" situation. The alternative or research hypothesis reflects the investigator's belief. The investigator might hypothesize that a parameter (e.g., a mean, proportion, difference in means or proportions) will increase, will decrease or will be different under specific conditions (sometimes the conditions are different experimental conditions and other times the conditions are simply different groups of participants). Once the hypotheses are specified, data are collected and summarized. The appropriate test is then conducted according to the five step approach. If the test leads to rejection of the null hypothesis, an approximate p-value is computed to summarize the significance of the findings. When tests of hypothesis are conducted using statistical computing packages, exact p-values are computed. Because the statistical tables in this textbook are limited, we can only approximate p-values. If the test fails to reject the null hypothesis, then a weaker concluding statement is made for the following reason.

In hypothesis testing, there are two types of errors that can be committed. A Type I error occurs when a test incorrectly rejects the null hypothesis. This is referred to as a false positive result, and the probability that this occurs is equal to the level of significance, α. The investigator chooses the level of significance in Step 1, and purposely chooses a small value such as α=0.05 to control the probability of committing a Type I error. A Type II error occurs when a test fails to reject the null hypothesis when in fact it is false. The probability that this occurs is equal to β. Unfortunately, the investigator cannot specify β at the outset because it depends on several factors including the sample size (smaller samples have higher b), the level of significance (β decreases as a increases), and the difference in the parameter under the null and alternative hypothesis.    

We noted in several examples in this chapter, the relationship between confidence intervals and tests of hypothesis. The approaches are different, yet related. It is possible to draw a conclusion about statistical significance by examining a confidence interval. For example, if a 95% confidence interval does not contain the null value (e.g., zero when analyzing a mean difference or risk difference, one when analyzing relative risks or odds ratios), then one can conclude that a two-sided test of hypothesis would reject the null at α=0.05. It is important to note that the correspondence between a confidence interval and test of hypothesis relates to a two-sided test and that the confidence level corresponds to a specific level of significance (e.g., 95% to α=0.05, 90% to α=0.10 and so on). The exact significance of the test, the p-value, can only be determined using the hypothesis testing approach and the p-value provides an assessment of the strength of the evidence and not an estimate of the effect.

Answers to Selected Problems

Dental services problem - bottom of page 5.

• Step 1: Set up hypotheses and determine the level of significance.

α=0.05

• Step 2: Select the appropriate test statistic.

First, determine whether the sample size is adequate.

Therefore the sample size is adequate, and we can use the following formula:

• Step 3: Set up the decision rule.

Reject H0 if Z is less than or equal to -1.96 or if Z is greater than or equal to 1.96.

• Step 4: Compute the test statistic
• Step 5: Conclusion.

We reject the null hypothesis because -6.15<-1.96. Therefore there is a statistically significant difference in the proportion of children in Boston using dental services compated to the national proportion.

Module 10: Inference for Means

Hypothesis test for a difference in two population means (1 of 2), learning outcomes.

• Under appropriate conditions, conduct a hypothesis test about a difference between two population means. State a conclusion in context.

Using the Hypothesis Test for a Difference in Two Population Means

The general steps of this hypothesis test are the same as always. As expected, the details of the conditions for use of the test and the test statistic are unique to this test (but similar in many ways to what we have seen before.)

Step 1: Determine the hypotheses.

The hypotheses for a difference in two population means are similar to those for a difference in two population proportions. The null hypothesis, H 0 , is again a statement of “no effect” or “no difference.”

• H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2

The alternative hypothesis, H a , can be any one of the following.

• H a : μ 1 – μ 2 < 0, which is the same as H a : μ 1 < μ 2
• H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2
• H a : μ 1 – μ 2 ≠ 0, which is the same as H a : μ 1 ≠ μ 2

Step 2: Collect the data.

As usual, how we collect the data determines whether we can use it in the inference procedure. We have our usual two requirements for data collection.

• Samples must be random to remove or minimize bias.
• Samples must be representative of the populations in question.

We use this hypothesis test when the data meets the following conditions.

• The two random samples are independent .
• The variable is normally distributed in both populations . If this variable is not known, samples of more than 30 will have a difference in sample means that can be modeled adequately by the t-distribution. As we discussed in “Hypothesis Test for a Population Mean,” t-procedures are robust even when the variable is not normally distributed in the population. If checking normality in the populations is impossible, then we look at the distribution in the samples. If a histogram or dotplot of the data does not show extreme skew or outliers, we take it as a sign that the variable is not heavily skewed in the populations, and we use the inference procedure. (Note: This is the same condition we used for the one-sample t-test in “Hypothesis Test for a Population Mean.”)

Step 3: Assess the evidence.

If the conditions are met, then we calculate the t-test statistic. The t-test statistic has a familiar form.

[latex]T\text{}=\text{}\frac{(\mathrm{Observed}\text{}\mathrm{difference}\text{}\mathrm{in}\text{}\mathrm{sample}\text{}\mathrm{means})-(\mathrm{Hypothesized}\text{}\mathrm{difference}\text{}\mathrm{in}\text{}\mathrm{population}\text{}\mathrm{means})}{\mathrm{Standard}\text{}\mathrm{error}}[/latex]

[latex]T\text{}=\text{}\frac{({\stackrel{¯}{x}}_{1}-{\stackrel{¯}{x}}_{2})-({μ}_{1}-{μ}_{2})}{\sqrt{\frac{{{s}_{1}}^{2}}{{n}_{1}}+\frac{{{s}_{2}}^{2}}{{n}_{2}}}}[/latex]

Since the null hypothesis assumes there is no difference in the population means, the expression (μ 1 – μ 2 ) is always zero.

As we learned in “Estimating a Population Mean,” the t-distribution depends on the degrees of freedom (df) . In the one-sample and matched-pair cases df = n – 1. For the two-sample t-test, determining the correct df is based on a complicated formula that we do not cover in this course. We will either give the df or use technology to find the df . With the t-test statistic and the degrees of freedom, we can use the appropriate t-model to find the P-value, just as we did in “Hypothesis Test for a Population Mean.” We can even use the same simulation.

Step 4: State a conclusion.

To state a conclusion, we follow what we have done with other hypothesis tests. We compare our P-value to a stated level of significance.

• If the P-value ≤ α, we reject the null hypothesis in favor of the alternative hypothesis.
• If the P-value > α, we fail to reject the null hypothesis. We do not have enough evidence to support the alternative hypothesis.

As always, we state our conclusion in context, usually by referring to the alternative hypothesis.

“Context and Calories”

Does the company you keep impact what you eat? This example comes from an article titled “Impact of Group Settings and Gender on Meals Purchased by College Students” (Allen-O’Donnell, M., T. C. Nowak, K. A. Snyder, and M. D. Cottingham, Journal of Applied Social Psychology 49(9), 2011, onlinelibrary.wiley.com/doi/10.1111/j.1559-1816.2011.00804.x/full) . In this study, researchers examined this issue in the context of gender-related theories in their field. For our purposes, we look at this research more narrowly.

Step 1: Stating the hypotheses.

In the article, the authors make the following hypothesis. “The attempt to appear feminine will be empirically demonstrated by the purchase of fewer calories by women in mixed-gender groups than by women in same-gender groups.” We translate this into a simpler and narrower research question: Do women purchase fewer calories when they eat with men compared to when they eat with women?

Here the two populations are “women eating with women” (population 1) and “women eating with men” (population 2). The variable is the calories in the meal. We test the following hypotheses at the 5% level of significance.

The null hypothesis is always H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2 .

The alternative hypothesis H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2 .

Here μ 1 represents the mean number of calories ordered by women when they were eating with other women, and μ 2 represents the mean number of calories ordered by women when they were eating with men.

Note: It does not matter which population we label as 1 or 2, but once we decide, we have to stay consistent throughout the hypothesis test. Since we expect the number of calories to be greater for the women eating with other women, the difference is positive if “women eating with women” is population 1. If you prefer to work with positive numbers, choose the group with the larger expected mean as population 1. This is a good general tip.

Step 2: Collect Data.

As usual, there are two major things to keep in mind when considering the collection of data.

• Samples need to be representative of the population in question.
• Samples need to be random in order to remove or minimize bias.

Representative Samples?

The researchers state their hypothesis in terms of “women.” We did the same. But the researchers gathered data by watching people eat at the HUB Rock Café II on the campus of Indiana University of Pennsylvania during the Spring semester of 2006. Almost all of the women in the data set were white undergraduates between the ages of 18 and 24, so there are some definite limitations on the scope of this study. These limitations will affect our conclusion (and the specific definition of the population means in our hypotheses.)

Random Samples?

The observations were collected on February 13, 2006, through February 22, 2006, between 11 a.m. and 7 p.m. We can see that the researchers included both lunch and dinner. They also made observations on all days of the week to ensure that weekly customer patterns did not confound their findings. The authors state that “since the time period for observations and the place where [they] observed students were limited, the sample was a convenience sample.” Despite these limitations, the researchers conducted inference procedures with the data, and the results were published in a reputable journal. We will also conduct inference with this data, but we also include a discussion of the limitations of the study with our conclusion. The authors did this, also.

Do the data met the conditions for use of a t-test?

The researchers reported the following sample statistics.

• In a sample of 45 women dining with other women, the average number of calories ordered was 850, and the standard deviation was 252.
• In a sample of 27 women dining with men, the average number of calories ordered was 719, and the standard deviation was 322.

One of the samples has fewer than 30 women. We need to make sure the distribution of calories in this sample is not heavily skewed and has no outliers, but we do not have access to a spreadsheet of the actual data. Since the researchers conducted a t-test with this data, we will assume that the conditions are met. This includes the assumption that the samples are independent.

As noted previously, the researchers reported the following sample statistics.

To compute the t-test statistic, make sure sample 1 corresponds to population 1. Here our population 1 is “women eating with other women.” So x 1 = 850, s 1 = 252, n 1 =45, and so on.

[latex]T\text{}=\text{}\frac{{\stackrel{¯}{x}}_{1}\text{}\text{−}\text{}{\stackrel{¯}{x}}_{2}}{\sqrt{\frac{{{s}_{1}}^{2}}{{n}_{1}}+\frac{{{s}_{2}}^{2}}{{n}_{2}}}}\text{}=\text{}\frac{850\text{}\text{−}\text{}719}{\sqrt{\frac{{252}^{2}}{45}+\frac{{322}^{2}}{27}}}\text{}\approx \text{}\frac{131}{72.47}\text{}\approx \text{}1.81[/latex]

Using technology, we determined that the degrees of freedom are about 45 for this data. To find the P-value, we use our familiar simulation of the t-distribution. Since the alternative hypothesis is a “greater than” statement, we look for the area to the right of T = 1.81. The P-value is 0.0385.

Generic Conclusion

The hypotheses for this test are H 0 : μ 1 – μ 2 = 0 and H a : μ 1 – μ 2 > 0. Since the P-value is less than the significance level (0.0385 < 0.05), we reject H 0 and accept H a .

Conclusion in context

At Indiana University of Pennsylvania, the mean number of calories ordered by undergraduate women eating with other women is greater than the mean number of calories ordered by undergraduate women eating with men (P-value = 0.0385).

Comment about Conclusions

In the conclusion above, we did not generalize the findings to all women. Since the samples included only undergraduate women at one university, we included this information in our conclusion. But our conclusion is a cautious statement of the findings. The authors see the results more broadly in the context of theories in the field of social psychology. In the context of these theories, they write, “Our findings support the assertion that meal size is a tool for influencing the impressions of others. For traditional-age, predominantly White college women, diminished meal size appears to be an attempt to assert femininity in groups that include men.” This viewpoint is echoed in the following summary of the study for the general public on National Public Radio (npr.org).

• Both men and women appear to choose larger portions when they eat with women, and both men and women choose smaller portions when they eat in the company of men, according to new research published in the Journal of Applied Social Psychology . The study, conducted among a sample of 127 college students, suggests that both men and women are influenced by unconscious scripts about how to behave in each other’s company. And these scripts change the way men and women eat when they eat together and when they eat apart.

Should we be concerned that the findings of this study are generalized in this way? Perhaps. But the authors of the article address this concern by including the following disclaimer with their findings: “While the results of our research are suggestive, they should be replicated with larger, representative samples. Studies should be done not only with primarily White, middle-class college students, but also with students who differ in terms of race/ethnicity, social class, age, sexual orientation, and so forth.” This is an example of good statistical practice. It is often very difficult to select truly random samples from the populations of interest. Researchers therefore discuss the limitations of their sampling design when they discuss their conclusions.

In the following activities, you will have the opportunity to practice parts of the hypothesis test for a difference in two population means. On the next page, the activities focus on the entire process and also incorporate technology.

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Statistics By Jim

Making statistics intuitive

Statistical Hypothesis Testing Overview

By Jim Frost 59 Comments

In this blog post, I explain why you need to use statistical hypothesis testing and help you navigate the essential terminology. Hypothesis testing is a crucial procedure to perform when you want to make inferences about a population using a random sample. These inferences include estimating population properties such as the mean, differences between means, proportions, and the relationships between variables.

This post provides an overview of statistical hypothesis testing. If you need to perform hypothesis tests, consider getting my book, Hypothesis Testing: An Intuitive Guide .

Why You Should Perform Statistical Hypothesis Testing

Hypothesis testing is a form of inferential statistics that allows us to draw conclusions about an entire population based on a representative sample. You gain tremendous benefits by working with a sample. In most cases, it is simply impossible to observe the entire population to understand its properties. The only alternative is to collect a random sample and then use statistics to analyze it.

While samples are much more practical and less expensive to work with, there are trade-offs. When you estimate the properties of a population from a sample, the sample statistics are unlikely to equal the actual population value exactly.  For instance, your sample mean is unlikely to equal the population mean. The difference between the sample statistic and the population value is the sample error.

Differences that researchers observe in samples might be due to sampling error rather than representing a true effect at the population level. If sampling error causes the observed difference, the next time someone performs the same experiment the results might be different. Hypothesis testing incorporates estimates of the sampling error to help you make the correct decision. Learn more about Sampling Error .

For example, if you are studying the proportion of defects produced by two manufacturing methods, any difference you observe between the two sample proportions might be sample error rather than a true difference. If the difference does not exist at the population level, you won’t obtain the benefits that you expect based on the sample statistics. That can be a costly mistake!

Let’s cover some basic hypothesis testing terms that you need to know.

Background information : Difference between Descriptive and Inferential Statistics and Populations, Parameters, and Samples in Inferential Statistics

Hypothesis Testing

Hypothesis testing is a statistical analysis that uses sample data to assess two mutually exclusive theories about the properties of a population. Statisticians call these theories the null hypothesis and the alternative hypothesis. A hypothesis test assesses your sample statistic and factors in an estimate of the sample error to determine which hypothesis the data support.

When you can reject the null hypothesis, the results are statistically significant, and your data support the theory that an effect exists at the population level.

The effect is the difference between the population value and the null hypothesis value. The effect is also known as population effect or the difference. For example, the mean difference between the health outcome for a treatment group and a control group is the effect.

Typically, you do not know the size of the actual effect. However, you can use a hypothesis test to help you determine whether an effect exists and to estimate its size. Hypothesis tests convert your sample effect into a test statistic, which it evaluates for statistical significance. Learn more about Test Statistics .

An effect can be statistically significant, but that doesn’t necessarily indicate that it is important in a real-world, practical sense. For more information, read my post about Statistical vs. Practical Significance .

Null Hypothesis

The null hypothesis is one of two mutually exclusive theories about the properties of the population in hypothesis testing. Typically, the null hypothesis states that there is no effect (i.e., the effect size equals zero). The null is often signified by H 0 .

In all hypothesis testing, the researchers are testing an effect of some sort. The effect can be the effectiveness of a new vaccination, the durability of a new product, the proportion of defect in a manufacturing process, and so on. There is some benefit or difference that the researchers hope to identify.

However, it’s possible that there is no effect or no difference between the experimental groups. In statistics, we call this lack of an effect the null hypothesis. Therefore, if you can reject the null, you can favor the alternative hypothesis, which states that the effect exists (doesn’t equal zero) at the population level.

You can think of the null as the default theory that requires sufficiently strong evidence against in order to reject it.

For example, in a 2-sample t-test, the null often states that the difference between the two means equals zero.

When you can reject the null hypothesis, your results are statistically significant. Learn more about Statistical Significance: Definition & Meaning .

Related post : Understanding the Null Hypothesis in More Detail

Alternative Hypothesis

The alternative hypothesis is the other theory about the properties of the population in hypothesis testing. Typically, the alternative hypothesis states that a population parameter does not equal the null hypothesis value. In other words, there is a non-zero effect. If your sample contains sufficient evidence, you can reject the null and favor the alternative hypothesis. The alternative is often identified with H 1 or H A .

For example, in a 2-sample t-test, the alternative often states that the difference between the two means does not equal zero.

You can specify either a one- or two-tailed alternative hypothesis:

If you perform a two-tailed hypothesis test, the alternative states that the population parameter does not equal the null value. For example, when the alternative hypothesis is H A : μ ≠ 0, the test can detect differences both greater than and less than the null value.

A one-tailed alternative has more power to detect an effect but it can test for a difference in only one direction. For example, H A : μ > 0 can only test for differences that are greater than zero.

Related posts : Understanding T-tests and One-Tailed and Two-Tailed Hypothesis Tests Explained

P-values are the probability that you would obtain the effect observed in your sample, or larger, if the null hypothesis is correct. In simpler terms, p-values tell you how strongly your sample data contradict the null. Lower p-values represent stronger evidence against the null. You use P-values in conjunction with the significance level to determine whether your data favor the null or alternative hypothesis.

Related post : Interpreting P-values Correctly

Significance Level (Alpha)

For instance, a significance level of 0.05 signifies a 5% risk of deciding that an effect exists when it does not exist.

Use p-values and significance levels together to help you determine which hypothesis the data support. If the p-value is less than your significance level, you can reject the null and conclude that the effect is statistically significant. In other words, the evidence in your sample is strong enough to be able to reject the null hypothesis at the population level.

Related posts : Graphical Approach to Significance Levels and P-values and Conceptual Approach to Understanding Significance Levels

Types of Errors in Hypothesis Testing

Statistical hypothesis tests are not 100% accurate because they use a random sample to draw conclusions about entire populations. There are two types of errors related to drawing an incorrect conclusion.

• False positives: You reject a null that is true. Statisticians call this a Type I error . The Type I error rate equals your significance level or alpha (α).
• False negatives: You fail to reject a null that is false. Statisticians call this a Type II error. Generally, you do not know the Type II error rate. However, it is a larger risk when you have a small sample size , noisy data, or a small effect size. The type II error rate is also known as beta (β).

Statistical power is the probability that a hypothesis test correctly infers that a sample effect exists in the population. In other words, the test correctly rejects a false null hypothesis. Consequently, power is inversely related to a Type II error. Power = 1 – β. Learn more about Power in Statistics .

Related posts : Types of Errors in Hypothesis Testing and Estimating a Good Sample Size for Your Study Using Power Analysis

Which Type of Hypothesis Test is Right for You?

There are many different types of procedures you can use. The correct choice depends on your research goals and the data you collect. Do you need to understand the mean or the differences between means? Or, perhaps you need to assess proportions. You can even use hypothesis testing to determine whether the relationships between variables are statistically significant.

To choose the proper statistical procedure, you’ll need to assess your study objectives and collect the correct type of data . This background research is necessary before you begin a study.

Related Post : Hypothesis Tests for Continuous, Binary, and Count Data

Statistical tests are crucial when you want to use sample data to make conclusions about a population because these tests account for sample error. Using significance levels and p-values to determine when to reject the null hypothesis improves the probability that you will draw the correct conclusion.

To see an alternative approach to these traditional hypothesis testing methods, learn about bootstrapping in statistics !

If you want to see examples of hypothesis testing in action, I recommend the following posts that I have written:

• How Effective Are Flu Shots? This example shows how you can use statistics to test proportions.
• Fatality Rates in Star Trek . This example shows how to use hypothesis testing with categorical data.
• Busting Myths About the Battle of the Sexes . A fun example based on a Mythbusters episode that assess continuous data using several different tests.
• Are Yawns Contagious? Another fun example inspired by a Mythbusters episode.

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Reader Interactions

January 14, 2024 at 8:43 am

Hello professor Jim, how are you doing! Pls. What are the properties of a population and their examples? Thanks for your time and understanding.

January 14, 2024 at 12:57 pm

Please read my post about Populations vs. Samples for more information and examples.

Also, please note there is a search bar in the upper-right margin of my website. Use that to search for topics.

July 5, 2023 at 7:05 am

Hello, I have a question as I read your post. You say in p-values section

“P-values are the probability that you would obtain the effect observed in your sample, or larger, if the null hypothesis is correct. In simpler terms, p-values tell you how strongly your sample data contradict the null. Lower p-values represent stronger evidence against the null.”

But according to your definition of effect, the null states that an effect does not exist, correct? So what I assume you want to say is that “P-values are the probability that you would obtain the effect observed in your sample, or larger, if the null hypothesis is **incorrect**.”

July 6, 2023 at 5:18 am

Hi Shrinivas,

The correct definition of p-value is that it is a probability that exists in the context of a true null hypothesis. So, the quotation is correct in stating “if the null hypothesis is correct.”

Essentially, the p-value tells you the likelihood of your observed results (or more extreme) if the null hypothesis is true. It gives you an idea of whether your results are surprising or unusual if there is no effect.

Hence, with sufficiently low p-values, you reject the null hypothesis because it’s telling you that your sample results were unlikely to have occurred if there was no effect in the population.

I hope that helps make it more clear. If not, let me know I’ll attempt to clarify!

May 8, 2023 at 12:47 am

Thanks a lot Ny best regards

May 7, 2023 at 11:15 pm

Hi Jim Can you tell me something about size effect? Thanks

May 8, 2023 at 12:29 am

Here’s a post that I’ve written about Effect Sizes that will hopefully tell you what you need to know. Please read that. Then, if you have any more specific questions about effect sizes, please post them there. Thanks!

January 7, 2023 at 4:19 pm

Hi Jim, I have only read two pages so far but I am really amazed because in few paragraphs you made me clearly understand the concepts of months of courses I received in biostatistics! Thanks so much for this work you have done it helps a lot!

January 10, 2023 at 3:25 pm

Thanks so much!

June 17, 2021 at 1:45 pm

Can you help in the following question: Rocinante36 is priced at ₹7 lakh and has been designed to deliver a mileage of 22 km/litre and a top speed of 140 km/hr. Formulate the null and alternative hypotheses for mileage and top speed to check whether the new models are performing as per the desired design specifications.

April 19, 2021 at 1:51 pm

Its indeed great to read your work statistics.

I have a doubt regarding the one sample t-test. So as per your book on hypothesis testing with reference to page no 45, you have mentioned the difference between “the sample mean and the hypothesised mean is statistically significant”. So as per my understanding it should be quoted like “the difference between the population mean and the hypothesised mean is statistically significant”. The catch here is the hypothesised mean represents the sample mean.

Please help me understand this.

Regards Rajat

April 19, 2021 at 3:46 pm

Thanks for buying my book. I’m so glad it’s been helpful!

The test is performed on the sample but the results apply to the population. Hence, if the difference between the sample mean (observed in your study) and the hypothesized mean is statistically significant, that suggests that population does not equal the hypothesized mean.

For one sample tests, the hypothesized mean is not the sample mean. It is a mean that you want to use for the test value. It usually represents a value that is important to your research. In other words, it’s a value that you pick for some theoretical/practical reasons. You pick it because you want to determine whether the population mean is different from that particular value.

I hope that helps!

November 5, 2020 at 6:24 am

Jim, you are such a magnificent statistician/economist/econometrician/data scientist etc whatever profession. Your work inspires and simplifies the lives of so many researchers around the world. I truly admire you and your work. I will buy a copy of each book you have on statistics or econometrics. Keep doing the good work. Remain ever blessed

November 6, 2020 at 9:47 pm

Hi Renatus,

Thanks so much for you very kind comments. You made my day!! I’m so glad that my website has been helpful. And, thanks so much for supporting my books! 🙂

November 2, 2020 at 9:32 pm

Hi Jim, I hope you are aware of 2019 American Statistical Association’s official statement on Statistical Significance: https://www.tandfonline.com/doi/full/10.1080/00031305.2019.1583913 In case you do not bother reading the full article, may I quote you the core message here: “We conclude, based on our review of the articles in this special issue and the broader literature, that it is time to stop using the term “statistically significant” entirely. Nor should variants such as “significantly different,” “p < 0.05,” and “nonsignificant” survive, whether expressed in words, by asterisks in a table, or in some other way."

With best wishes,

November 3, 2020 at 2:09 am

I’m definitely aware of the debate surrounding how to use p-values most effectively. However, I need to correct you on one point. The link you provide is NOT a statement by the American Statistical Association. It is an editorial by several authors.

There is considerable debate over this issue. There are problems with p-values. However, as the authors state themselves, much of the problem is over people’s mindsets about how to use p-values and their incorrect interpretations about what statistical significance does and does not mean.

If you were to read my website more thoroughly, you’d be aware that I share many of their concerns and I address them in multiple posts. One of the authors’ key points is the need to be thoughtful and conduct thoughtful research and analysis. I emphasize this aspect in multiple posts on this topic. I’ll ask you to read the following three because they all address some of the authors’ concerns and suggestions. But you might run across others to read as well.

Five Tips for Using P-values to Avoid Being Misled How to Interpret P-values Correctly P-values and the Reproducibility of Experimental Results

September 24, 2020 at 11:52 pm

HI Jim, i just want you to know that you made explanation for Statistics so simple! I should say lesser and fewer words that reduce the complexity. All the best! 🙂

September 25, 2020 at 1:03 am

Thanks, Rene! Your kind words mean a lot to me! I’m so glad it has been helpful!

September 23, 2020 at 2:21 am

Honestly, I never understood stats during my entire M.Ed course and was another nightmare for me. But how easily you have explained each concept, I have understood stats way beyond my imagination. Thank you so much for helping ignorant research scholars like us. Looking forward to get hardcopy of your book. Kindly tell is it available through flipkart?

September 24, 2020 at 11:14 pm

I’m so happy to hear that my website has been helpful!

I checked on flipkart and it appears like my books are not available there. I’m never exactly sure where they’re available due to the vagaries of different distribution channels. They are available on Amazon in India.

Introduction to Statistics: An Intuitive Guide (Amazon IN) Hypothesis Testing: An Intuitive Guide (Amazon IN)

July 26, 2020 at 11:57 am

Dear Jim I am a teacher from India . I don’t have any background in statistics, and still I should tell that in a single read I can follow your explanations . I take my entire biostatistics class for botany graduates with your explanations. Thanks a lot. May I know how I can avail your books in India

July 28, 2020 at 12:31 am

Right now my books are only available as ebooks from my website. However, soon I’ll have some exciting news about other ways to obtain it. Stay tuned! I’ll announce it on my email list. If you’re not already on it, you can sign up using the form that is in the right margin of my website.

June 22, 2020 at 2:02 pm

Also can you please let me if this book covers topics like EDA and principal component analysis?

June 22, 2020 at 2:07 pm

This book doesn’t cover principal components analysis. Although, I wouldn’t really classify that as a hypothesis test. In the future, I might write a multivariate analysis book that would cover this and others. But, that’s well down the road.

My Introduction to Statistics covers EDA. That’s the largely graphical look at your data that you often do prior to hypothesis testing. The Introduction book perfectly leads right into the Hypothesis Testing book.

June 22, 2020 at 1:45 pm

Thanks for the detailed explanation. It does clear my doubts. I saw that your book related to hypothesis testing has the topics that I am studying currently. I am looking forward to purchasing it.

Regards, Take Care

June 19, 2020 at 1:03 pm

For this particular article I did not understand a couple of statements and it would great if you could help: 1)”If sample error causes the observed difference, the next time someone performs the same experiment the results might be different.” 2)”If the difference does not exist at the population level, you won’t obtain the benefits that you expect based on the sample statistics.”

I discovered your articles by chance and now I keep coming back to read & understand statistical concepts. These articles are very informative & easy to digest. Thanks for the simplifying things.

June 20, 2020 at 9:53 pm

I’m so happy to hear that you’ve found my website to be helpful!

To answer your questions, keep in mind that a central tenant of inferential statistics is that the random sample that a study drew was only one of an infinite number of possible it could’ve drawn. Each random sample produces different results. Most results will cluster around the population value assuming they used good methodology. However, random sampling error always exists and makes it so that population estimates from a sample almost never exactly equal the correct population value.

So, imagine that we’re studying a medication and comparing the treatment and control groups. Suppose that the medicine is truly not effect and that the population difference between the treatment and control group is zero (i.e., no difference.) Despite the true difference being zero, most sample estimates will show some degree of either a positive or negative effect thanks to random sampling error. So, just because a study has an observed difference does not mean that a difference exists at the population level. So, on to your questions:

1. If the observed difference is just random error, then it makes sense that if you collected another random sample, the difference could change. It could change from negative to positive, positive to negative, more extreme, less extreme, etc. However, if the difference exists at the population level, most random samples drawn from the population will reflect that difference. If the medicine has an effect, most random samples will reflect that fact and not bounce around on both sides of zero as much.

2. This is closely related to the previous answer. If there is no difference at the population level, but say you approve the medicine because of the observed effects in a sample. Even though your random sample showed an effect (which was really random error), that effect doesn’t exist. So, when you start using it on a larger scale, people won’t benefit from the medicine. That’s why it’s important to separate out what is easily explained by random error versus what is not easily explained by it.

I think reading my post about how hypothesis tests work will help clarify this process. Also, in about 24 hours (as I write this), I’ll be releasing my new ebook about Hypothesis Testing!

May 29, 2020 at 5:23 am

Hi Jim, I really enjoy your blog. Can you please link me on your blog where you discuss about Subgroup analysis and how it is done? I need to use non parametric and parametric statistical methods for my work and also do subgroup analysis in order to identify potential groups of patients that may benefit more from using a treatment than other groups.

May 29, 2020 at 2:12 pm

Hi, I don’t have a specific article about subgroup analysis. However, subgroup analysis is just the dividing up of a larger sample into subgroups and then analyzing those subgroups separately. You can use the various analyses I write about on the subgroups.

Alternatively, you can include the subgroups in regression analysis as an indicator variable and include that variable as a main effect and an interaction effect to see how the relationships vary by subgroup without needing to subdivide your data. I write about that approach in my article about comparing regression lines . This approach is my preferred approach when possible.

April 19, 2020 at 7:58 am

sir is confidence interval is a part of estimation?

April 17, 2020 at 3:36 pm

Sir can u plz briefly explain alternatives of hypothesis testing? I m unable to find the answer

April 18, 2020 at 1:22 am

Assuming you want to draw conclusions about populations by using samples (i.e., inferential statistics ), you can use confidence intervals and bootstrap methods as alternatives to the traditional hypothesis testing methods.

March 9, 2020 at 10:01 pm

Hi JIm, could you please help with activities that can best teach concepts of hypothesis testing through simulation, Also, do you have any question set that would enhance students intuition why learning hypothesis testing as a topic in introductory statistics. Thanks.

March 5, 2020 at 3:48 pm

Hi Jim, I’m studying multiple hypothesis testing & was wondering if you had any material that would be relevant. I’m more trying to understand how testing multiple samples simultaneously affects your results & more on the Bonferroni Correction

March 5, 2020 at 4:05 pm

I write about multiple comparisons (aka post hoc tests) in the ANOVA context . I don’t talk about Bonferroni Corrections specifically but I cover related types of corrections. I’m not sure if that exactly addresses what you want to know but is probably the closest I have already written. I hope it helps!

January 14, 2020 at 9:03 pm

Thank you! Have a great day/evening.

January 13, 2020 at 7:10 pm

Any help would be greatly appreciated. What is the difference between The Hypothesis Test and The Statistical Test of Hypothesis?

January 14, 2020 at 11:02 am

They sound like the same thing to me. Unless this is specialized terminology for a particular field or the author was intending something specific, I’d guess they’re one and the same.

April 1, 2019 at 10:00 am

so these are the only two forms of Hypothesis used in statistical testing?

April 1, 2019 at 10:02 am

Are you referring to the null and alternative hypothesis? If so, yes, that’s those are the standard hypotheses in a statistical hypothesis test.

April 1, 2019 at 9:57 am

year very insightful post, thanks for the write up

October 27, 2018 at 11:09 pm

hi there, am upcoming statistician, out of all blogs that i have read, i have found this one more useful as long as my problem is concerned. thanks so much

October 27, 2018 at 11:14 pm

Hi Stano, you’re very welcome! Thanks for your kind words. They mean a lot! I’m happy to hear that my posts were able to help you. I’m sure you will be a fantastic statistician. Best of luck with your studies!

October 26, 2018 at 11:39 am

Dear Jim, thank you very much for your explanations! I have a question. Can I use t-test to compare two samples in case each of them have right bias?

October 26, 2018 at 12:00 pm

Hi Tetyana,

You’re very welcome!

The term “right bias” is not a standard term. Do you by chance mean right skewed distributions? In other words, if you plot the distribution for each group on a histogram they have longer right tails? These are not the symmetrical bell-shape curves of the normal distribution.

If that’s the case, yes you can as long as you exceed a specific sample size within each group. I include a table that contains these sample size requirements in my post about nonparametric vs parametric analyses .

Bias in statistics refers to cases where an estimate of a value is systematically higher or lower than the true value. If this is the case, you might be able to use t-tests, but you’d need to be sure to understand the nature of the bias so you would understand what the results are really indicating.

I hope this helps!

April 2, 2018 at 7:28 am

Simple and upto the point 👍 Thank you so much.

April 2, 2018 at 11:11 am

Hi Kalpana, thanks! And I’m glad it was helpful!

March 26, 2018 at 8:41 am

Am I correct if I say: Alpha – Probability of wrongly rejection of null hypothesis P-value – Probability of wrongly acceptance of null hypothesis

March 28, 2018 at 3:14 pm

You’re correct about alpha. Alpha is the probability of rejecting the null hypothesis when the null is true.

Unfortunately, your definition of the p-value is a bit off. The p-value has a fairly convoluted definition. It is the probability of obtaining the effect observed in a sample, or more extreme, if the null hypothesis is true. The p-value does NOT indicate the probability that either the null or alternative is true or false. Although, those are very common misinterpretations. To learn more, read my post about how to interpret p-values correctly .

March 2, 2018 at 6:10 pm

I recently started reading your blog and it is very helpful to understand each concept of statistical tests in easy way with some good examples. Also, I recommend to other people go through all these blogs which you posted. Specially for those people who have not statistical background and they are facing to many problems while studying statistical analysis.

Thank you for your such good blogs.

March 3, 2018 at 10:12 pm

Hi Amit, I’m so glad that my blog posts have been helpful for you! It means a lot to me that you took the time to write such a nice comment! Also, thanks for recommending by blog to others! I try really hard to write posts about statistics that are easy to understand.

January 17, 2018 at 7:03 am

I recently started reading your blog and I find it very interesting. I am learning statistics by my own, and I generally do many google search to understand the concepts. So this blog is quite helpful for me, as it have most of the content which I am looking for.

January 17, 2018 at 3:56 pm

Hi Shashank, thank you! And, I’m very glad to hear that my blog is helpful!

January 2, 2018 at 2:28 pm

thank u very much sir.

January 2, 2018 at 2:36 pm

You’re very welcome, Hiral!

November 21, 2017 at 12:43 pm

Thank u so much sir….your posts always helps me to be a #statistician

November 21, 2017 at 2:40 pm

Hi Sachin, you’re very welcome! I’m happy that you find my posts to be helpful!

November 19, 2017 at 8:22 pm

great post as usual, but it would be nice to see an example.

November 19, 2017 at 8:27 pm

Thank you! At the end of this post, I have links to four other posts that show examples of hypothesis tests in action. You’ll find what you’re looking for in those posts!

Comments and Questions Cancel reply

Module 10: Inference for Means

Hypothesis Test for a Difference in Two Population Means (1 of 2)

Learning outcomes.

• Under appropriate conditions, conduct a hypothesis test about a difference between two population means. State a conclusion in context.

Using the Hypothesis Test for a Difference in Two Population Means

The general steps of this hypothesis test are the same as always. As expected, the details of the conditions for use of the test and the test statistic are unique to this test (but similar in many ways to what we have seen before.)

Step 1: Determine the hypotheses.

The hypotheses for a difference in two population means are similar to those for a difference in two population proportions. The null hypothesis, H 0 , is again a statement of “no effect” or “no difference.”

• H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2

The alternative hypothesis, H a , can be any one of the following.

• H a : μ 1 – μ 2 < 0, which is the same as H a : μ 1 < μ 2
• H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2
• H a : μ 1 – μ 2 ≠ 0, which is the same as H a : μ 1 ≠ μ 2

Step 2: Collect the data.

As usual, how we collect the data determines whether we can use it in the inference procedure. We have our usual two requirements for data collection.

• Samples must be random to remove or minimize bias.
• Samples must be representative of the populations in question.

We use this hypothesis test when the data meets the following conditions.

• The two random samples are independent .
• The variable is normally distributed in both populations . If this variable is not known, samples of more than 30 will have a difference in sample means that can be modeled adequately by the t-distribution. As we discussed in “Hypothesis Test for a Population Mean,” t-procedures are robust even when the variable is not normally distributed in the population. If checking normality in the populations is impossible, then we look at the distribution in the samples. If a histogram or dotplot of the data does not show extreme skew or outliers, we take it as a sign that the variable is not heavily skewed in the populations, and we use the inference procedure. (Note: This is the same condition we used for the one-sample t-test in “Hypothesis Test for a Population Mean.”)

Step 3: Assess the evidence.

If the conditions are met, then we calculate the t-test statistic. The t-test statistic has a familiar form.

[latex]T=\frac{Observeddifferenceinsamplemeans-Hypothesizeddiferenceinpopulationmeans}{ standarderror}[/latex]

[latex]T=\frac{(\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}}+\frac{s_{2}^{2}}{n_{2}}}[/latex]

Since the null hypothesis assumes there is no difference in the population means, the expression (μ 1 – μ 2 ) is always zero.

As we learned in “Estimating a Population Mean,” the t-distribution depends on the degrees of freedom (df) . In the one-sample and matched-pair cases df = n – 1. For the two-sample t-test, determining the correct df is based on a complicated formula that we do not cover in this course. We will either give the df or use technology to find the df . With the t-test statistic and the degrees of freedom, we can use the appropriate t-model to find the P-value, just as we did in “Hypothesis Test for a Population Mean.” We can even use the same simulation.

Step 4: State a conclusion.

To state a conclusion, we follow what we have done with other hypothesis tests. We compare our P-value to a stated level of significance.

• If the P-value ≤ α, we reject the null hypothesis in favor of the alternative hypothesis.
• If the P-value > α, we fail to reject the null hypothesis. We do not have enough evidence to support the alternative hypothesis.

As always, we state our conclusion in context, usually by referring to the alternative hypothesis.

“Context and Calories”

Does the company you keep impact what you eat? This example comes from an article titled “Impact of Group Settings and Gender on Meals Purchased by College Students” (Allen-O’Donnell, M., T. C. Nowak, K. A. Snyder, and M. D. Cottingham, Journal of Applied Social Psychology 49(9), 2011, onlinelibrary.wiley.com/doi/10.1111/j.1559-1816.2011.00804.x/full) . In this study, researchers examined this issue in the context of gender-related theories in their field. For our purposes, we look at this research more narrowly.

Step 1: Stating the hypotheses.

In the article, the authors make the following hypothesis. “The attempt to appear feminine will be empirically demonstrated by the purchase of fewer calories by women in mixed-gender groups than by women in same-gender groups.” We translate this into a simpler and narrower research question: Do women purchase fewer calories when they eat with men compared to when they eat with women?

Here the two populations are “women eating with women” (population 1) and “women eating with men” (population 2). The variable is the calories in the meal. We test the following hypotheses at the 5% level of significance.

The null hypothesis is always H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2 .

The alternative hypothesis H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2 .

Here μ 1 represents the mean number of calories ordered by women when they were eating with other women, and μ 2 represents the mean number of calories ordered by women when they were eating with men.

Note: It does not matter which population we label as 1 or 2, but once we decide, we have to stay consistent throughout the hypothesis test. Since we expect the number of calories to be greater for the women eating with other women, the difference is positive if “women eating with women” is population 1. If you prefer to work with positive numbers, choose the group with the larger expected mean as population 1. This is a good general tip.

Step 2: Collect Data.

As usual, there are two major things to keep in mind when considering the collection of data.

• Samples need to be representative of the population in question.
• Samples need to be random in order to remove or minimize bias.

Representative Samples?

The researchers state their hypothesis in terms of “women.” We did the same. But the researchers gathered data by watching people eat at the HUB Rock Café II on the campus of Indiana University of Pennsylvania during the Spring semester of 2006. Almost all of the women in the data set were white undergraduates between the ages of 18 and 24, so there are some definite limitations on the scope of this study. These limitations will affect our conclusion (and the specific definition of the population means in our hypotheses.)

Random Samples?

The observations were collected on February 13, 2006, through February 22, 2006, between 11 a.m. and 7 p.m. We can see that the researchers included both lunch and dinner. They also made observations on all days of the week to ensure that weekly customer patterns did not confound their findings. The authors state that “since the time period for observations and the place where [they] observed students were limited, the sample was a convenience sample.” Despite these limitations, the researchers conducted inference procedures with the data, and the results were published in a reputable journal. We will also conduct inference with this data, but we also include a discussion of the limitations of the study with our conclusion. The authors did this, also.

Do the data meet the conditions for use of a t-test?

The researchers reported the following sample statistics.

• In a sample of 45 women dining with other women, the average number of calories ordered was 850, and the standard deviation was 252.
• In a sample of 27 women dining with men, the average number of calories ordered was 719, and the standard deviation was 322.

One of the samples has fewer than 30 women. We need to make sure the distribution of calories in this sample is not heavily skewed and has no outliers, but we do not have access to a spreadsheet of the actual data. Since the researchers conducted a t-test with this data, we will assume that the conditions are met. This includes the assumption that the samples are independent.

As noted previously, the researchers reported the following sample statistics.

To compute the t-test statistic, make sure sample 1 corresponds to population 1. Here our population 1 is “women eating with other women.” So x 1 = 850, s 1 = 252, n 1 =45, and so on.

[latex]T=\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}}+\frac{s_{2}^{2}}{n_{2}}}= \frac{850-719}{\sqrt{\frac{252^{2}}{45}+\frac{322^{2}}{27}}}\approx \frac{131}{72.47}\approx 1.81[/latex]

Using technology, we determined that the degrees of freedom are about 45 for this data. To find the P-value, we use our familiar simulation of the t-distribution. Since the alternative hypothesis is a “greater than” statement, we look for the area to the right of T = 1.81. The P-value is 0.0385.

Generic Conclusion

The hypotheses for this test are H 0 : μ 1 – μ 2 = 0 and H a : μ 1 – μ 2 > 0. Since the P-value is less than the significance level (0.0385 < 0.05), we reject H 0 and accept H a .

Conclusion in context

At Indiana University of Pennsylvania, the mean number of calories ordered by undergraduate women eating with other women is greater than the mean number of calories ordered by undergraduate women eating with men (P-value = 0.0385).

Comment about Conclusions

In the conclusion above, we did not generalize the findings to all women. Since the samples included only undergraduate women at one university, we included this information in our conclusion. But our conclusion is a cautious statement of the findings. The authors see the results more broadly in the context of theories in the field of social psychology. In the context of these theories, they write, “Our findings support the assertion that meal size is a tool for influencing the impressions of others. For traditional-age, predominantly White college women, diminished meal size appears to be an attempt to assert femininity in groups that include men.” This viewpoint is echoed in the following summary of the study for the general public on National Public Radio (npr.org).

• Both men and women appear to choose larger portions when they eat with women, and both men and women choose smaller portions when they eat in the company of men, according to new research published in the Journal of Applied Social Psychology . The study, conducted among a sample of 127 college students, suggests that both men and women are influenced by unconscious scripts about how to behave in each other’s company. And these scripts change the way men and women eat when they eat together and when they eat apart.

Should we be concerned that the findings of this study are generalized in this way? Perhaps. But the authors of the article address this concern by including the following disclaimer with their findings: “While the results of our research are suggestive, they should be replicated with larger, representative samples. Studies should be done not only with primarily White, middle-class college students, but also with students who differ in terms of race/ethnicity, social class, age, sexual orientation, and so forth.” This is an example of good statistical practice. It is often very difficult to select truly random samples from the populations of interest. Researchers therefore discuss the limitations of their sampling design when they discuss their conclusions.

In the following activities, you will have the opportunity to practice parts of the hypothesis test for a difference in two population means. On the next page, the activities focus on the entire process and also incorporate technology.

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• Indian J Crit Care Med
• v.23(Suppl 3); 2019 Sep

An Introduction to Statistics: Understanding Hypothesis Testing and Statistical Errors

Priya ranganathan.

1 Department of Anesthesiology, Critical Care and Pain, Tata Memorial Hospital, Mumbai, Maharashtra, India

2 Department of Surgical Oncology, Tata Memorial Centre, Mumbai, Maharashtra, India

The second article in this series on biostatistics covers the concepts of sample, population, research hypotheses and statistical errors.

How to cite this article

Ranganathan P, Pramesh CS. An Introduction to Statistics: Understanding Hypothesis Testing and Statistical Errors. Indian J Crit Care Med 2019;23(Suppl 3):S230–S231.

Two papers quoted in this issue of the Indian Journal of Critical Care Medicine report. The results of studies aim to prove that a new intervention is better than (superior to) an existing treatment. In the ABLE study, the investigators wanted to show that transfusion of fresh red blood cells would be superior to standard-issue red cells in reducing 90-day mortality in ICU patients. 1 The PROPPR study was designed to prove that transfusion of a lower ratio of plasma and platelets to red cells would be superior to a higher ratio in decreasing 24-hour and 30-day mortality in critically ill patients. 2 These studies are known as superiority studies (as opposed to noninferiority or equivalence studies which will be discussed in a subsequent article).

SAMPLE VERSUS POPULATION

A sample represents a group of participants selected from the entire population. Since studies cannot be carried out on entire populations, researchers choose samples, which are representative of the population. This is similar to walking into a grocery store and examining a few grains of rice or wheat before purchasing an entire bag; we assume that the few grains that we select (the sample) are representative of the entire sack of grains (the population).

The results of the study are then extrapolated to generate inferences about the population. We do this using a process known as hypothesis testing. This means that the results of the study may not always be identical to the results we would expect to find in the population; i.e., there is the possibility that the study results may be erroneous.

HYPOTHESIS TESTING

A clinical trial begins with an assumption or belief, and then proceeds to either prove or disprove this assumption. In statistical terms, this belief or assumption is known as a hypothesis. Counterintuitively, what the researcher believes in (or is trying to prove) is called the “alternate” hypothesis, and the opposite is called the “null” hypothesis; every study has a null hypothesis and an alternate hypothesis. For superiority studies, the alternate hypothesis states that one treatment (usually the new or experimental treatment) is superior to the other; the null hypothesis states that there is no difference between the treatments (the treatments are equal). For example, in the ABLE study, we start by stating the null hypothesis—there is no difference in mortality between groups receiving fresh RBCs and standard-issue RBCs. We then state the alternate hypothesis—There is a difference between groups receiving fresh RBCs and standard-issue RBCs. It is important to note that we have stated that the groups are different, without specifying which group will be better than the other. This is known as a two-tailed hypothesis and it allows us to test for superiority on either side (using a two-sided test). This is because, when we start a study, we are not 100% certain that the new treatment can only be better than the standard treatment—it could be worse, and if it is so, the study should pick it up as well. One tailed hypothesis and one-sided statistical testing is done for non-inferiority studies, which will be discussed in a subsequent paper in this series.

STATISTICAL ERRORS

There are two possibilities to consider when interpreting the results of a superiority study. The first possibility is that there is truly no difference between the treatments but the study finds that they are different. This is called a Type-1 error or false-positive error or alpha error. This means falsely rejecting the null hypothesis.

The second possibility is that there is a difference between the treatments and the study does not pick up this difference. This is called a Type 2 error or false-negative error or beta error. This means falsely accepting the null hypothesis.

The power of the study is the ability to detect a difference between groups and is the converse of the beta error; i.e., power = 1-beta error. Alpha and beta errors are finalized when the protocol is written and form the basis for sample size calculation for the study. In an ideal world, we would not like any error in the results of our study; however, we would need to do the study in the entire population (infinite sample size) to be able to get a 0% alpha and beta error. These two errors enable us to do studies with realistic sample sizes, with the compromise that there is a small possibility that the results may not always reflect the truth. The basis for this will be discussed in a subsequent paper in this series dealing with sample size calculation.

Conventionally, type 1 or alpha error is set at 5%. This means, that at the end of the study, if there is a difference between groups, we want to be 95% certain that this is a true difference and allow only a 5% probability that this difference has occurred by chance (false positive). Type 2 or beta error is usually set between 10% and 20%; therefore, the power of the study is 90% or 80%. This means that if there is a difference between groups, we want to be 80% (or 90%) certain that the study will detect that difference. For example, in the ABLE study, sample size was calculated with a type 1 error of 5% (two-sided) and power of 90% (type 2 error of 10%) (1).

Table 1 gives a summary of the two types of statistical errors with an example

Statistical errors

In the next article in this series, we will look at the meaning and interpretation of ‘ p ’ value and confidence intervals for hypothesis testing.

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6.6 - confidence intervals & hypothesis testing.

Confidence intervals and hypothesis tests are similar in that they are both inferential methods that rely on an approximated sampling distribution. Confidence intervals use data from a sample to estimate a population parameter. Hypothesis tests use data from a sample to test a specified hypothesis. Hypothesis testing requires that we have a hypothesized parameter. 

The simulation methods used to construct bootstrap distributions and randomization distributions are similar. One primary difference is a bootstrap distribution is centered on the observed sample statistic while a randomization distribution is centered on the value in the null hypothesis. 

In Lesson 4, we learned confidence intervals contain a range of reasonable estimates of the population parameter. All of the confidence intervals we constructed in this course were two-tailed. These two-tailed confidence intervals go hand-in-hand with the two-tailed hypothesis tests we learned in Lesson 5. The conclusion drawn from a two-tailed confidence interval is usually the same as the conclusion drawn from a two-tailed hypothesis test. In other words, if the the 95% confidence interval contains the hypothesized parameter, then a hypothesis test at the 0.05 \(\alpha\) level will almost always fail to reject the null hypothesis. If the 95% confidence interval does not contain the hypothesize parameter, then a hypothesis test at the 0.05 \(\alpha\) level will almost always reject the null hypothesis.

Example: Mean Section  

This example uses the Body Temperature dataset built in to StatKey for constructing a  bootstrap confidence interval and conducting a randomization test . 

Let's start by constructing a 95% confidence interval using the percentile method in StatKey:

The 95% confidence interval for the mean body temperature in the population is [98.044, 98.474].

Now, what if we want to know if there is enough evidence that the mean body temperature is different from 98.6 degrees? We can conduct a hypothesis test. Because 98.6 is not contained within the 95% confidence interval, it is not a reasonable estimate of the population mean. We should expect to have a p value less than 0.05 and to reject the null hypothesis.

\(H_0: \mu=98.6\)

\(H_a: \mu \ne 98.6\)

\(p = 2*0.00080=0.00160\)

\(p \leq 0.05\), reject the null hypothesis

There is evidence that the population mean is different from 98.6 degrees. 

Selecting the Appropriate Procedure Section  

The decision of whether to use a confidence interval or a hypothesis test depends on the research question. If we want to estimate a population parameter, we use a confidence interval. If we are given a specific population parameter (i.e., hypothesized value), and want to determine the likelihood that a population with that parameter would produce a sample as different as our sample, we use a hypothesis test. Below are a few examples of selecting the appropriate procedure. 

Example: Cheese Consumption Section  

Research question: How much cheese (in pounds) does an average American adult consume annually? 

What is the appropriate inferential procedure? 

Cheese consumption, in pounds, is a quantitative variable. We have one group: American adults. We are not given a specific value to test, so the appropriate procedure here is a  confidence interval for a single mean .

Example: Age Section  

Research question:  Is the average age in the population of all STAT 200 students greater than 30 years?

There is one group: STAT 200 students. The variable of interest is age in years, which is quantitative. The research question includes a specific population parameter to test: 30 years. The appropriate procedure is a  hypothesis test for a single mean .

Try it! Section  

For each research question, identify the variables, the parameter of interest and decide on the the appropriate inferential procedure.

Research question:  How strong is the correlation between height (in inches) and weight (in pounds) in American teenagers?

There are two variables of interest: (1) height in inches and (2) weight in pounds. Both are quantitative variables. The parameter of interest is the correlation between these two variables.

We are not given a specific correlation to test. We are being asked to estimate the strength of the correlation. The appropriate procedure here is a  confidence interval for a correlation . 

Research question:  Are the majority of registered voters planning to vote in the next presidential election?

The parameter that is being tested here is a single proportion. We have one group: registered voters. "The majority" would be more than 50%, or p>0.50. This is a specific parameter that we are testing. The appropriate procedure here is a  hypothesis test for a single proportion .

Research question:  On average, are STAT 200 students younger than STAT 500 students?

We have two independent groups: STAT 200 students and STAT 500 students. We are comparing them in terms of average (i.e., mean) age.

If STAT 200 students are younger than STAT 500 students, that translates to \(\mu_{200}<\mu_{500}\) which is an alternative hypothesis. This could also be written as \(\mu_{200}-\mu_{500}<0\), where 0 is a specific population parameter that we are testing. 

The appropriate procedure here is a  hypothesis test for the difference in two means .

Research question:  On average, how much taller are adult male giraffes compared to adult female giraffes?

There are two groups: males and females. The response variable is height, which is quantitative. We are not given a specific parameter to test, instead we are asked to estimate "how much" taller males are than females. The appropriate procedure is a  confidence interval for the difference in two means .

Research question:  Are STAT 500 students more likely than STAT 200 students to be employed full-time?

There are two independent groups: STAT 500 students and STAT 200 students. The response variable is full-time employment status which is categorical with two levels: yes/no.

If STAT 500 students are more likely than STAT 200 students to be employed full-time, that translates to \(p_{500}>p_{200}\) which is an alternative hypothesis. This could also be written as \(p_{500}-p_{200}>0\), where 0 is a specific parameter that we are testing. The appropriate procedure is a  hypothesis test for the difference in two proportions.

Research question:  Is there is a relationship between outdoor temperature (in Fahrenheit) and coffee sales (in cups per day)?

There are two variables here: (1) temperature in Fahrenheit and (2) cups of coffee sold in a day. Both variables are quantitative. The parameter of interest is the correlation between these two variables.

If there is a relationship between the variables, that means that the correlation is different from zero. This is a specific parameter that we are testing. The appropriate procedure is a  hypothesis test for a correlation . 

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• Knowledge Base
• Choosing the Right Statistical Test | Types & Examples

Choosing the Right Statistical Test | Types & Examples

Published on January 28, 2020 by Rebecca Bevans . Revised on June 22, 2023.

Statistical tests are used in hypothesis testing . They can be used to:

• determine whether a predictor variable has a statistically significant relationship with an outcome variable.
• estimate the difference between two or more groups.

Statistical tests assume a null hypothesis of no relationship or no difference between groups. Then they determine whether the observed data fall outside of the range of values predicted by the null hypothesis.

If you already know what types of variables you’re dealing with, you can use the flowchart to choose the right statistical test for your data.

Statistical tests flowchart

Table of contents

What does a statistical test do, when to perform a statistical test, choosing a parametric test: regression, comparison, or correlation, choosing a nonparametric test, flowchart: choosing a statistical test, other interesting articles, frequently asked questions about statistical tests.

Statistical tests work by calculating a test statistic – a number that describes how much the relationship between variables in your test differs from the null hypothesis of no relationship.

It then calculates a p value (probability value). The p -value estimates how likely it is that you would see the difference described by the test statistic if the null hypothesis of no relationship were true.

If the value of the test statistic is more extreme than the statistic calculated from the null hypothesis, then you can infer a statistically significant relationship between the predictor and outcome variables.

If the value of the test statistic is less extreme than the one calculated from the null hypothesis, then you can infer no statistically significant relationship between the predictor and outcome variables.

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You can perform statistical tests on data that have been collected in a statistically valid manner – either through an experiment , or through observations made using probability sampling methods .

For a statistical test to be valid , your sample size needs to be large enough to approximate the true distribution of the population being studied.

To determine which statistical test to use, you need to know:

• whether your data meets certain assumptions.
• the types of variables that you’re dealing with.

Statistical assumptions

Statistical tests make some common assumptions about the data they are testing:

• Independence of observations (a.k.a. no autocorrelation): The observations/variables you include in your test are not related (for example, multiple measurements of a single test subject are not independent, while measurements of multiple different test subjects are independent).
• Homogeneity of variance : the variance within each group being compared is similar among all groups. If one group has much more variation than others, it will limit the test’s effectiveness.
• Normality of data : the data follows a normal distribution (a.k.a. a bell curve). This assumption applies only to quantitative data .

If your data do not meet the assumptions of normality or homogeneity of variance, you may be able to perform a nonparametric statistical test , which allows you to make comparisons without any assumptions about the data distribution.

If your data do not meet the assumption of independence of observations, you may be able to use a test that accounts for structure in your data (repeated-measures tests or tests that include blocking variables).

Types of variables

The types of variables you have usually determine what type of statistical test you can use.

Quantitative variables represent amounts of things (e.g. the number of trees in a forest). Types of quantitative variables include:

• Continuous (aka ratio variables): represent measures and can usually be divided into units smaller than one (e.g. 0.75 grams).
• Discrete (aka integer variables): represent counts and usually can’t be divided into units smaller than one (e.g. 1 tree).

Categorical variables represent groupings of things (e.g. the different tree species in a forest). Types of categorical variables include:

• Ordinal : represent data with an order (e.g. rankings).
• Nominal : represent group names (e.g. brands or species names).
• Binary : represent data with a yes/no or 1/0 outcome (e.g. win or lose).

Choose the test that fits the types of predictor and outcome variables you have collected (if you are doing an experiment , these are the independent and dependent variables ). Consult the tables below to see which test best matches your variables.

Parametric tests usually have stricter requirements than nonparametric tests, and are able to make stronger inferences from the data. They can only be conducted with data that adheres to the common assumptions of statistical tests.

The most common types of parametric test include regression tests, comparison tests, and correlation tests.

Regression tests

Regression tests look for cause-and-effect relationships . They can be used to estimate the effect of one or more continuous variables on another variable.

Comparison tests

Comparison tests look for differences among group means . They can be used to test the effect of a categorical variable on the mean value of some other characteristic.

T-tests are used when comparing the means of precisely two groups (e.g., the average heights of men and women). ANOVA and MANOVA tests are used when comparing the means of more than two groups (e.g., the average heights of children, teenagers, and adults).

Correlation tests

Correlation tests check whether variables are related without hypothesizing a cause-and-effect relationship.

These can be used to test whether two variables you want to use in (for example) a multiple regression test are autocorrelated.

Non-parametric tests don’t make as many assumptions about the data, and are useful when one or more of the common statistical assumptions are violated. However, the inferences they make aren’t as strong as with parametric tests.

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This flowchart helps you choose among parametric tests. For nonparametric alternatives, check the table above.

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient
• Null hypothesis

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Statistical tests commonly assume that:

• the data are normally distributed
• the groups that are being compared have similar variance
• the data are independent

If your data does not meet these assumptions you might still be able to use a nonparametric statistical test , which have fewer requirements but also make weaker inferences.

A test statistic is a number calculated by a  statistical test . It describes how far your observed data is from the  null hypothesis  of no relationship between  variables or no difference among sample groups.

The test statistic tells you how different two or more groups are from the overall population mean , or how different a linear slope is from the slope predicted by a null hypothesis . Different test statistics are used in different statistical tests.

Statistical significance is a term used by researchers to state that it is unlikely their observations could have occurred under the null hypothesis of a statistical test . Significance is usually denoted by a p -value , or probability value.

Statistical significance is arbitrary – it depends on the threshold, or alpha value, chosen by the researcher. The most common threshold is p < 0.05, which means that the data is likely to occur less than 5% of the time under the null hypothesis .

When the p -value falls below the chosen alpha value, then we say the result of the test is statistically significant.

Quantitative variables are any variables where the data represent amounts (e.g. height, weight, or age).

Categorical variables are any variables where the data represent groups. This includes rankings (e.g. finishing places in a race), classifications (e.g. brands of cereal), and binary outcomes (e.g. coin flips).

You need to know what type of variables you are working with to choose the right statistical test for your data and interpret your results .

Discrete and continuous variables are two types of quantitative variables :

• Discrete variables represent counts (e.g. the number of objects in a collection).
• Continuous variables represent measurable amounts (e.g. water volume or weight).

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JMP Introductory Lab Activities (JILA)

• Exploring and Describing Data
• Probability and Sampling Distributions
• Confidence Intervals

Hypothesis Testing for Means

• Hypothesis Testing for Proportions - Categorical Data

Hypothesis Testing, the z-Test (Activity 10)

Use a z-test to determine whether the students in a school have above-average intelligence.

• View activity (PDF)

Hypothesis Testing, the t-Test (Activity 11)

Examine whether the differences between measurements of the Earth´s rotation angle reveal a real change or whether they can be attributed to chance variability.

Hypothesis Testing, Paired t-Test (Activity 12)

Compare readings from a new eardrum prode thermometer to readings from a standard oral mercury thermometer.

Hypothesis Testing, Two-Sample t-Test (Activity 13)

Examine if the mean heights for 12-year-old and 15-year-old adolescent males are greater than the mean heights for similarly aged females.

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Difference in Means Hypothesis Test Calculator

Use the calculator below to analyze the results of a difference in sample means hypothesis test. Enter your sample means, sample standard deviations, sample sizes, hypothesized difference in means, test type, and significance level to calculate your results.

You will find a description of how to conduct a two sample t-test below the calculator.

Define the Two Sample t-test

The Difference Between the Sample Means Under the Null Distribution

Conducting a hypothesis test for the difference in means.

When two populations are related, you can compare them by analyzing the difference between their means.

A hypothesis test for the difference in samples means can help you make inferences about the relationships between two population means.

Testing for a Difference in Means

For the results of a hypothesis test to be valid, you should follow these steps:

Check Your Conditions

State your hypothesis, determine your analysis plan, analyze your sample, interpret your results.

To use the testing procedure described below, you should check the following conditions:

• Independence of Samples - Your samples should be collected independently of one another.
• Simple Random Sampling - You should collect your samples with simple random sampling. This type of sampling requires that every occurrence of a value in a population has an equal chance of being selected when taking a sample.
• Normality of Sample Distributions - The sampling distributions for both samples should follow the Normal or a nearly Normal distribution. A sampling distribution will be nearly Normal when the samples are collected independently and when the population distribution is nearly Normal. Generally, the larger the sample size, the more normally distributed the sampling distribution. Additionally, outlier data points can make a distribution less Normal, so if your data contains many outliers, exercise caution when verifying this condition.

You must state a null hypothesis and an alternative hypothesis to conduct an hypothesis test of the difference in means.

The null hypothesis is a skeptical claim that you would like to test.

The alternative hypothesis represents the alternative claim to the null hypothesis.

Your null hypothesis and alternative hypothesis should be stated in one of three mutually exclusive ways listed in the table below.

D is the hypothesized difference between the populations' means that you would like to test.

Before conducting a hypothesis test, you must determine a reasonable significance level, α, or the probability of rejecting the null hypothesis assuming it is true. The lower your significance level, the more confident you can be of the conclusion of your hypothesis test. Common significance levels are 10%, 5%, and 1%.

To evaluate your hypothesis test at the significance level that you set, consider if you are conducting a one or two tail test:

• Two-tail tests divide the rejection region, or critical region, evenly above and below the null distribution, i.e. to the tails of the null sampling distribution. For example, in a two-tail test with a 5% significance level, your rejection region would be the upper and lower 2.5% of the null distribution. An alternative hypothesis of μ 1 - μ 2 ≠ D requires a two tail test.
• One-tail tests place the rejection region entirely on one side of the distribution i.e. to the right or left tail of the null distribution. For example, in a one-tail test evaluating if the actual difference in means, D, is above the null distribution with a 5% significance level, your rejection region would be the upper 5% of the null distribution. μ 1 - μ 2 > D and μ 1 - μ 2 < D alternative hypotheses require one-tail tests.

The graphical results section of the calculator above shades rejection regions blue.

After checking your conditions, stating your hypothesis, determining your significance level, and collecting your sample, you are ready to analyze your hypothesis.

Sample means follow the Normal distribution with the following parameters:

• The Difference in the Population Means, D - The true difference in the population means is unknown, but we use the hypothesized difference in the means, D, from the null hypothesis in the calculations.
• The Standard Error, SE - The standard error of the difference in the sample means can be computed as follows:      SE = (s 1 2 /n 1 + s 2 2 /n 2 ) (1/2) with s 1 being the standard deviation of sample one, n 1 being the sample size of sample one, s 2 being the standard deviation of sample one, and n 2 being the sample size of sample two. The standard error defines how differences in sample means are expected to vary around the null difference in means sampling distribution given the sample sizes and under the assumption that the null hypothesis is true.
• The Degrees of Freedom, DF - The degrees of freedom calculation can be estimated as the smaller of n 1 - 1 or n 2 - 1. For more accurate results, use the following formula for the degrees of freedom (DF):      DF = (s 1 2 /n 1 + s 2 2 /n 2 ) 2 / ((s 1 2 /n 1 ) 2 / (n 1 - 1) + (s 2 2 /n 2 ) 2 / (n 2 - 1))

In a difference in means hypothesis test, we calculate the probability that we would observe the difference in sample means (x̄ 1 - x̄ 2 ), assuming the null hypothesis is true, also known as the p-value . If the p-value is less than the significance level, then we can reject the null hypothesis.

You can determine a precise p-value using the calculator above, but we can find an estimate of the p-value manually by calculating the t-score, or t-statistic, as follows: t = (x̄ 1 - x̄ 2 - D) / SE

The t-score is a test statistic that tells you how far our observation is from the null hypothesis's difference in means under the null distribution. Using any t-score table, you can look up the probability of observing the results under the null distribution. You will need to look up the t-score for the type of test you are conducting, i.e. one or two tail. A hypothesis test for the difference in means is sometimes known as a two sample mean t-test because of the use of a t-score in analyzing results.

The conclusion of a hypothesis test for the difference in means is always either:

• Reject the null hypothesis
• Do not reject the null hypothesis

If you reject the null hypothesis, you cannot say that your sample difference in means is the true difference between the means. If you do not reject the null hypothesis, you cannot say that the hypothesized difference in means is true.

A hypothesis test is simply a way to look at evidence and conclude if it provides sufficient evidence to reject the null hypothesis.

Example: Hypothesis Test for the Difference in Two Means

Let’s say you are a manager at a company that designs batteries for smartphones. One of your engineers believes that she has developed a battery that will last more than two hours longer than your standard battery.

Before you can consider if you should replace your standard battery with the new one, you need to test the engineer’s claim. So, you decided to run a difference in means hypothesis test to see if her claim that the new battery will last two hours longer than the standard one is reasonable.

You direct your team to run a study. They will take a sample of 100 of the new batteries and compare their performance to 1,000 of the old standard batteries.

• Check the conditions - Your test consists of independent samples . Your team collects your samples using simple random sampling , and you have reason to believe that all your batteries' performances are always close to normally distributed . So, the conditions are met to conduct a two sample t-test.
• State Your Hypothesis - Your null hypothesis is that the charge of the new battery lasts at most two hours longer than your standard battery (i.e. μ 1 - μ 2 ≤ 2). Your alternative hypothesis is that the new battery lasts more than two hours longer than the standard battery (i.e. μ 1 - μ 2 > 2).
• Determine Your Analysis Plan - You believe that a 1% significance level is reasonable. As your test is a one-tail test, you will evaluate if the difference in mean charge between the samples would occur at the upper 1% of the null distribution.
• Analyze Your Sample - After collecting your samples (which you do after steps 1-3), you find the new battery sample had a mean charge of 10.4 hours, x̄ 1 , with a 0.8 hour standard deviation, s 1 . Your standard battery sample had a mean charge of 8.2 hours, x̄ 2 , with a standard deviation of 0.2 hours, s 2 . Using the calculator above, you find that a difference in sample means of 2.2 hours [2 = 10.4 – 8.2] would results in a t-score of 2.49 under the null distribution, which translates to a p-value of 0.72%.
• Interpret Your Results - Since your p-value of 0.72% is less than the significance level of 1%, you have sufficient evidence to reject the null hypothesis.

In this example, you found that you can reject your null hypothesis that the new battery design does not result in more than 2 hours of extra battery life. The test does not guarantee that your engineer’s new battery lasts two hours longer than your standard battery, but it does give you strong reason to believe her claim.

Statistics Tutorial

Descriptive statistics, inferential statistics, stat reference, statistics - hypothesis testing a mean.

A population mean is an average of value a population.

Hypothesis tests are used to check a claim about the size of that population mean.

Hypothesis Testing a Mean

The following steps are used for a hypothesis test:

• Check the conditions
• Define the claims
• Decide the significance level
• Calculate the test statistic

For example:

• Population : Nobel Prize winners
• Category : Age when they received the prize.

And we want to check the claim:

"The average age of Nobel Prize winners when they received the prize is more than 55"

By taking a sample of 30 randomly selected Nobel Prize winners we could find that:

The mean age in the sample (\(\bar{x}\)) is 62.1

The standard deviation of age in the sample (\(s\)) is 13.46

From this sample data we check the claim with the steps below.

1. Checking the Conditions

The conditions for calculating a confidence interval for a proportion are:

• The sample is randomly selected
• The population data is normally distributed
• Sample size is large enough

A moderately large sample size, like 30, is typically large enough.

In the example, the sample size was 30 and it was randomly selected, so the conditions are fulfilled.

Note: Checking if the data is normally distributed can be done with specialized statistical tests.

2. Defining the Claims

We need to define a null hypothesis (\(H_{0}\)) and an alternative hypothesis (\(H_{1}\)) based on the claim we are checking.

The claim was:

In this case, the parameter is the mean age of Nobel Prize winners when they received the prize (\(\mu\)).

The null and alternative hypothesis are then:

Null hypothesis : The average age was 55.

Alternative hypothesis : The average age was more than 55.

Which can be expressed with symbols as:

\(H_{0}\): \(\mu = 55 \)

\(H_{1}\): \(\mu > 55 \)

This is a ' right tailed' test, because the alternative hypothesis claims that the proportion is more than in the null hypothesis.

If the data supports the alternative hypothesis, we reject the null hypothesis and accept the alternative hypothesis.

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3. Deciding the Significance Level

The significance level (\(\alpha\)) is the uncertainty we accept when rejecting the null hypothesis in a hypothesis test.

The significance level is a percentage probability of accidentally making the wrong conclusion.

Typical significance levels are:

• \(\alpha = 0.1\) (10%)
• \(\alpha = 0.05\) (5%)
• \(\alpha = 0.01\) (1%)

A lower significance level means that the evidence in the data needs to be stronger to reject the null hypothesis.

There is no "correct" significance level - it only states the uncertainty of the conclusion.

Note: A 5% significance level means that when we reject a null hypothesis:

We expect to reject a true null hypothesis 5 out of 100 times.

4. Calculating the Test Statistic

The test statistic is used to decide the outcome of the hypothesis test.

The test statistic is a standardized value calculated from the sample.

The formula for the test statistic (TS) of a population mean is:

\(\displaystyle \frac{\bar{x} - \mu}{s} \cdot \sqrt{n} \)

\(\bar{x}-\mu\) is the difference between the sample mean (\(\bar{x}\)) and the claimed population mean (\(\mu\)).

\(s\) is the sample standard deviation .

\(n\) is the sample size.

In our example:

The claimed (\(H_{0}\)) population mean (\(\mu\)) was \( 55 \)

The sample mean (\(\bar{x}\)) was \(62.1\)

The sample standard deviation (\(s\)) was \(13.46\)

The sample size (\(n\)) was \(30\)

So the test statistic (TS) is then:

\(\displaystyle \frac{62.1-55}{13.46} \cdot \sqrt{30} = \frac{7.1}{13.46} \cdot \sqrt{30} \approx 0.528 \cdot 5.477 = \underline{2.889}\)

You can also calculate the test statistic using programming language functions:

With Python use the scipy and math libraries to calculate the test statistic.

With R use built-in math and statistics functions to calculate the test statistic.

5. Concluding

There are two main approaches for making the conclusion of a hypothesis test:

• The critical value approach compares the test statistic with the critical value of the significance level.
• The P-value approach compares the P-value of the test statistic and with the significance level.

Note: The two approaches are only different in how they present the conclusion.

The Critical Value Approach

For the critical value approach we need to find the critical value (CV) of the significance level (\(\alpha\)).

For a population mean test, the critical value (CV) is a T-value from a student's t-distribution .

This critical T-value (CV) defines the rejection region for the test.

The rejection region is an area of probability in the tails of the standard normal distribution.

Because the claim is that the population mean is more than 55, the rejection region is in the right tail:

The student's t-distribution is adjusted for the uncertainty from smaller samples.

This adjustment is called degrees of freedom (df), which is the sample size \((n) - 1\)

In this case the degrees of freedom (df) is: \(30 - 1 = \underline{29} \)

Choosing a significance level (\(\alpha\)) of 0.01, or 1%, we can find the critical T-value from a T-table , or with a programming language function:

With Python use the Scipy Stats library t.ppf() function find the T-Value for an \(\alpha\) = 0.01 at 29 degrees of freedom (df).

With R use the built-in qt() function to find the t-value for an \(\alpha\) = 0.01 at 29 degrees of freedom (df).

Using either method we can find that the critical T-Value is \(\approx \underline{2.462}\)

For a right tailed test we need to check if the test statistic (TS) is bigger than the critical value (CV).

If the test statistic is bigger than the critical value, the test statistic is in the rejection region .

When the test statistic is in the rejection region, we reject the null hypothesis (\(H_{0}\)).

Here, the test statistic (TS) was \(\approx \underline{2.889}\) and the critical value was \(\approx \underline{2.462}\)

Here is an illustration of this test in a graph:

Since the test statistic was bigger than the critical value we reject the null hypothesis.

This means that the sample data supports the alternative hypothesis.

And we can summarize the conclusion stating:

The sample data supports the claim that "The average age of Nobel Prize winners when they received the prize is more than 55" at a 1% significance level .

The P-Value Approach

For the P-value approach we need to find the P-value of the test statistic (TS).

If the P-value is smaller than the significance level (\(\alpha\)), we reject the null hypothesis (\(H_{0}\)).

The test statistic was found to be \( \approx \underline{2.889} \)

For a population proportion test, the test statistic is a T-Value from a student's t-distribution .

Because this is a right tailed test, we need to find the P-value of a t-value bigger than 2.889.

The student's t-distribution is adjusted according to degrees of freedom (df), which is the sample size \((30) - 1 = \underline{29}\)

We can find the P-value using a T-table , or with a programming language function:

With Python use the Scipy Stats library t.cdf() function find the P-value of a T-value bigger than 2.889 at 29 degrees of freedom (df):

With R use the built-in pt() function find the P-value of a T-Value bigger than 2.889 at 29 degrees of freedom (df):

Using either method we can find that the P-value is \(\approx \underline{0.0036}\)

This tells us that the significance level (\(\alpha\)) would need to be bigger than 0.0036, or 0.36%, to reject the null hypothesis.

This P-value is smaller than any of the common significance levels (10%, 5%, 1%).

So the null hypothesis is rejected at all of these significance levels.

The sample data supports the claim that "The average age of Nobel Prize winners when they received the prize is more than 55" at a 10%, 5%, or 1% significance level .

Note: An outcome of an hypothesis test that rejects the null hypothesis with a p-value of 0.36% means:

For this p-value, we only expect to reject a true null hypothesis 36 out of 10000 times.

Calculating a P-Value for a Hypothesis Test with Programming

Many programming languages can calculate the P-value to decide outcome of a hypothesis test.

Using software and programming to calculate statistics is more common for bigger sets of data, as calculating manually becomes difficult.

The P-value calculated here will tell us the lowest possible significance level where the null-hypothesis can be rejected.

With Python use the scipy and math libraries to calculate the P-value for a right tailed hypothesis test for a mean.

Here, the sample size is 30, the sample mean is 62.1, the sample standard deviation is 13.46, and the test is for a mean bigger than 55.

With R use built-in math and statistics functions find the P-value for a right tailed hypothesis test for a mean.

Left-Tailed and Two-Tailed Tests

This was an example of a right tailed test, where the alternative hypothesis claimed that parameter is bigger than the null hypothesis claim.

You can check out an equivalent step-by-step guide for other types here:

• Left-Tailed Test
• Two-Tailed Test

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