99% Upper | |||||
N | Mean | SE Mean | Bound | Z | P |
227 | 19.600 | 0.485 | 20.727 | -7.02 | 0.000 |
Excel does not offer 1-sample hypothesis testing.
Frequently, the population standard deviation (σ) is not known. We can estimate the population standard deviation (σ) with the sample standard deviation (s). However, the test statistic will no longer follow the standard normal distribution. We must rely on the student’s t-distribution with n-1 degrees of freedom. Because we use the sample standard deviation (s), the test statistic will change from a Z-score to a t-score.
Steps for a hypothesis test are the same that we covered in Section 2.
Just as with the hypothesis test from the previous section, the data for this test must be from a random sample and requires either that the population from which the sample was drawn be normal or that the sample size is sufficiently large (n≥30). A t-test is robust, so small departures from normality will not adversely affect the results of the test. That being said, if the sample size is smaller than 30, it is always good to verify the assumption of normality through a normal probability plot.
We will still have the same three pairs of null and alternative hypotheses and we can still use either the classical approach or the p-value approach.
Selecting the correct critical value from the student’s t-distribution table depends on three factors: the type of test (one-sided or two-sided alternative hypothesis), the sample size, and the level of significance.
For a two-sided test (“not equal” alternative hypothesis), the critical value (t α /2 ), is determined by alpha ( α ), the level of significance, divided by two, to deal with the possibility that the result could be less than OR greater than the known value.
For a one-sided test (“a less than” or “greater than” alternative hypothesis), the critical value (t α ) , is determined by alpha ( α ), the level of significance, being all in the one side.
Find the critical value you would use to test the claim that μ ≠ 112 with a sample size of 18 and a 5% level of significance.
In this case, the critical value (t α /2 ) would be 2.110. This is a two-sided question (≠) so you would divide alpha by 2 (0.05/2 = 0.025) and go down the 0.025 column to 17 degrees of freedom.
What would the critical value be if you wanted to test that μ < 112 for the same data?
In this case, the critical value would be 1.740. This is a one-sided question (<) so alpha would be divided by 1 (0.05/1 = 0.05). You would go down the 0.05 column with 17 degrees of freedom to get the correct critical value.
In 2005, the mean pH level of rain in a county in northern New York was 5.41. A biologist believes that the rain acidity has changed. He takes a random sample of 11 rain dates in 2010 and obtains the following data. Use a 1% level of significance to test his claim.
4.70, 5.63, 5.02, 5.78, 4.99, 5.91, 5.76, 5.54, 5.25, 5.18, 5.01
The sample size is small and we don’t know anything about the distribution of the population, so we examine a normal probability plot. The distribution looks normal so we will continue with our test.
The sample mean is 5.343 with a sample standard deviation of 0.397.
We will fail to reject the null hypothesis. We do not have enough evidence to support the claim that the mean rain pH has changed.
Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. The government has set safety limits for cadmium in dry vegetables at 0.5 ppm. Biologists believe that the mean level of cadmium in mushrooms growing near strip mines is greater than the recommended limit of 0.5 ppm, negatively impacting the animals that live in this ecosystem. A random sample of 51 mushrooms gave a sample mean of 0.59 ppm with a sample standard deviation of 0.29 ppm. Use a 5% level of significance to test the claim that the mean cadmium level is greater than the acceptable limit of 0.5 ppm.
The sample size is greater than 30 so we are assured of a normal distribution of the means.
Step 4) State a Conclusion.
The test statistic falls in the rejection zone. We will reject the null hypothesis. We have enough evidence to support the claim that the mean cadmium level is greater than the acceptable safe limit.
BUT, what happens if the significance level changes to 1%?
The critical value is now found by going down the 0.01 column with 50 degrees of freedom. The critical value is 2.403. The test statistic is now LESS THAN the critical value. The test statistic does not fall in the rejection zone. The conclusion will change. We do NOT have enough evidence to support the claim that the mean cadmium level is greater than the acceptable safe limit of 0.5 ppm.
The level of significance is the probability that you, as the researcher, set to decide if there is enough statistical evidence to support the alternative claim. It should be set before the experiment begins.
We can also use the p-value approach for a hypothesis test about the mean when the population standard deviation ( σ ) is unknown. However, when using a student’s t-table, we can only estimate the range of the p-value, not a specific value as when using the standard normal table. The student’s t-table has area (probability) across the top row in the table, with t-scores in the body of the table.
Estimating P-value from a Student’s T-table
If your test statistic is 3.789 with 3 degrees of freedom, you would go across the 3 df row. The value 3.789 falls between the values 3.482 and 4.541 in that row. Therefore, the p-value is between 0.02 and 0.01. The p-value will be greater than 0.01 but less than 0.02 (0.01<p<0.02).
If your level of significance is 5%, you would reject the null hypothesis as the p-value (0.01-0.02) is less than alpha ( α ) of 0.05.
If your level of significance is 1%, you would fail to reject the null hypothesis as the p-value (0.01-0.02) is greater than alpha ( α ) of 0.01.
Software packages typically output p-values. It is easy to use the Decision Rule to answer your research question by the p-value method.
(referring to Ex. 12)
Test of mu = 0.5 vs. > 0.5
95% Lower | ||||||
N | Mean | StDev | SE Mean | Bound | T | P |
51 | 0.5900 | 0.2900 | 0.0406 | 0.5219 | 2.22 | 0.016 |
Additional example: www.youtube.com/watch?v=WwdSjO4VUsg .
Frequently, the parameter we are testing is the population proportion.
Recall that the best point estimate of p , the population proportion, is given by
when np (1 – p )≥10. We can use both the classical approach and the p-value approach for testing.
The steps for a hypothesis test are the same that we covered in Section 2.
The test statistic follows the standard normal distribution. Notice that the standard error (the denominator) uses p instead of p̂ , which was used when constructing a confidence interval about the population proportion. In a hypothesis test, the null hypothesis is assumed to be true, so the known proportion is used.
A botanist has produced a new variety of hybrid soy plant that is better able to withstand drought than other varieties. The botanist knows the seed germination for the parent plants is 75%, but does not know the seed germination for the new hybrid. He tests the claim that it is different from the parent plants. To test this claim, 450 seeds from the hybrid plant are tested and 321 have germinated. Use a 5% level of significance to test this claim that the germination rate is different from 75%.
This is a two-sided question so alpha is divided by 2.
The test statistic does not fall in the rejection zone. We fail to reject the null hypothesis. We do not have enough evidence to support the claim that the germination rate of the hybrid plant is different from the parent plants.
Let’s answer this question using the p-value approach. Remember, for a two-sided alternative hypothesis (“not equal”), the p-value is two times the area of the test statistic. The test statistic is -1.81 and we want to find the area to the left of -1.81 from the standard normal table.
Now compare the p-value to alpha. The Decision Rule states that if the p-value is less than alpha, reject the H 0 . In this case, the p-value (0.0702) is greater than alpha (0.05) so we will fail to reject H 0 . We do not have enough evidence to support the claim that the germination rate of the hybrid plant is different from the parent plants.
You are a biologist studying the wildlife habitat in the Monongahela National Forest. Cavities in older trees provide excellent habitat for a variety of birds and small mammals. A study five years ago stated that 32% of the trees in this forest had suitable cavities for this type of wildlife. You believe that the proportion of cavity trees has increased. You sample 196 trees and find that 79 trees have cavities. Does this evidence support your claim that there has been an increase in the proportion of cavity trees?
Use a 10% level of significance to test this claim.
This is a one-sided question so alpha is divided by 1.
The test statistic is larger than the critical value (it falls in the rejection zone). We will reject the null hypothesis. We have enough evidence to support the claim that there has been an increase in the proportion of cavity trees.
Now use the p-value approach to answer the question. This is a right-sided question (“greater than”), so the p-value is equal to the area to the right of the test statistic. Go to the positive side of the standard normal table and find the area associated with the Z-score of 2.49. The area is 0.9936. Remember that this table is cumulative from the left. To find the area to the right of 2.49, we subtract from one.
p-value = (1 – 0.9936) = 0.0064
The p-value is less than the level of significance (0.10), so we reject the null hypothesis. We have enough evidence to support the claim that the proportion of cavity trees has increased.
(referring to Ex. 15)
Test of p = 0.32 vs. p > 0.32
90% Lower | ||||||
Sample | X | N | Sample p | Bound | Z-Value | p-Value |
1 | 79 | 196 | 0.403061 | 0.358160 | 2.49 | 0.006 |
Using the normal approximation. |
When people think of statistical inference, they usually think of inferences involving population means or proportions. However, the particular population parameter needed to answer an experimenter’s practical questions varies from one situation to another, and sometimes a population’s variability is more important than its mean. Thus, product quality is often defined in terms of low variability.
Sample variance S 2 can be used for inferences concerning a population variance σ 2 . For a random sample of n measurements drawn from a normal population with mean μ and variance σ 2 , the value S 2 provides a point estimate for σ 2 . In addition, the quantity ( n – 1) S 2 / σ 2 follows a Chi-square ( χ 2 ) distribution, with df = n – 1.
The properties of Chi-square ( χ 2 ) distribution are:
Alternative hypothesis:
where the χ 2 critical value in the rejection region is based on degrees of freedom df = n – 1 and a specified significance level of α .
As with previous sections, if the test statistic falls in the rejection zone set by the critical value, you will reject the null hypothesis.
A forester wants to control a dense understory of striped maple that is interfering with desirable hardwood regeneration using a mist blower to apply an herbicide treatment. She wants to make sure that treatment has a consistent application rate, in other words, low variability not exceeding 0.25 gal./acre (0.06 gal. 2 ). She collects sample data (n = 11) on this type of mist blower and gets a sample variance of 0.064 gal. 2 Using a 5% level of significance, test the claim that the variance is significantly greater than 0.06 gal. 2
H 0 : σ 2 = 0.06
H 1 : σ 2 >0.06
The critical value is 18.307. Any test statistic greater than this value will cause you to reject the null hypothesis.
The test statistic is
We fail to reject the null hypothesis. The forester does NOT have enough evidence to support the claim that the variance is greater than 0.06 gal. 2 You can also estimate the p-value using the same method as for the student t-table. Go across the row for degrees of freedom until you find the two values that your test statistic falls between. In this case going across the row 10, the two table values are 4.865 and 15.987. Now go up those two columns to the top row to estimate the p-value (0.1-0.9). The p-value is greater than 0.1 and less than 0.9. Both are greater than the level of significance (0.05) causing us to fail to reject the null hypothesis.
(referring to Ex. 16)
Test and CI for One Variance
Method | ||
Null hypothesis | Sigma-squared | = 0.06 |
Alternative hypothesis | Sigma-squared | > 0.06 |
The chi-square method is only for the normal distribution.
Test | |||
Method | Statistic | DF | P-Value |
Chi-Square | 10.67 | 10 | 0.384 |
Excel does not offer 1-sample χ 2 testing.
To test a claim about μ when σ is known.
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Biological vs. Statistical Null Hypotheses. It is important to distinguish between biological null and alternative hypotheses and statistical null and alternative hypotheses. "Sexual selection by females has caused male chickens to evolve bigger feet than females" is a biological alternative hypothesis; it says something about biological processes, in this case sexual selection.
CIE. Spanish Language & Literature. Past Papers. Other Subjects. Revision notes on 5.1.1 Hypothesis Testing for the AQA A Level Maths: Statistics syllabus, written by the Maths experts at Save My Exams.
Once the null hypothesis is stated, a statistical test is then chosen. This either supports or fails to support the null hypothesis. Chi-squared test. All chi-squared tests are concerned with counts of things (frequencies) that you can put into categories. For example, you might be investigating flower colour and have counted the
To compare the means of two unrelated sets of data, of which the data has been measured. Give the formula for calculating the Student-T Test. (meanA-meanB) / √ ( (s²A/nA) + (s²B/nB)) How is the Student T-test result calculated? For each data set, square the standard deviation and divide by number of values. Then add these values for each ...
Summary. One of the main goals of statistical hypothesis testing is to estimate the P value, which is the probability of obtaining the observed results, or something more extreme, if the null hypothesis were true. If the observed results are unlikely under the null hypothesis, your reject the null hypothesis. Alternatives to this "frequentist ...
If the chi-squared value is greater than the critical value - reject the null hypothesis. The difference between observed and expected data is NOT due to chance. This means we would get this chi-squared value in less than 5% of cases, which is very unlikely. The chi-squared test is used in genetics to compare the goodness of fit of observed ...
1.111.97.99.9Spearman's RankSpearman's rank correlation is a statistical test that is carried out in order to assess the degree of association between diffe. ent measurements from the same sample. That is, if you are looking for a positive or nega. orrelation between two variables. Positive correlation.
A hypothesis test is the means by which we generate a test statistic that directs us to either reject or not reject the null hypothesis. The test statistic is a "summary" of the collected data, and should have a sampling distribution specified by the null hypothesis. A Level AQA Edexcel OCR.
Responding to Changes in Environment. Topic 7: Genetics, Populations, Evolution and Ecosystems. Topic 8: Control of Gene Expression. Practical Skills. All Topics, 1-8. Practical Skills. Revision for AQA Biology AS and A Level Papers, including summary notes, worksheets and past exam questions for each topic.
This resource summarises the four statistical tests required for A level biology (Standard Deviation, T-test, Spearman Rank, Chi-squared). There is a practice question and mark scheme for each test. The questions are Edexcel SNAB but hopefully they're relevant to other boards. Excellent resource! Report this resource to let us know if it ...
Example 1: Biology. Hypothesis tests are often used in biology to determine whether some new treatment, fertilizer, pesticide, chemical, etc. causes increased growth, stamina, immunity, etc. in plants or animals. For example, suppose a biologist believes that a certain fertilizer will cause plants to grow more during a one-month period than ...
Learn how to and when to use the student T test statistic for A-level Biology, how to write a null hypothesis and conclusion. In this video I go through wor...
3. Compare your t value to the critical value. To work out the critical value from a table of critical values, you first need to work out how many degrees of freedom you have. The degrees of freedom is calculated by doing (n1 + n2) - 2. In our experiment, there is eight individuals in each group so we do (8 + 8) - 2 = 14.
Hypothesis Tests. A hypothesis test consists of five steps: 1. State the hypotheses. State the null and alternative hypotheses. These two hypotheses need to be mutually exclusive, so if one is true then the other must be false. 2. Determine a significance level to use for the hypothesis. Decide on a significance level.
5. Phrase your hypothesis in three ways. To identify the variables, you can write a simple prediction in if…then form. The first part of the sentence states the independent variable and the second part states the dependent variable. If a first-year student starts attending more lectures, then their exam scores will improve.
One of the statistical tests used in A Level Biology, the Chi-squared test is used to compare observed results with expected results.Expected results from te...
A hypothesis proposes a relationship between the independent and dependent variable. A hypothesis is a prediction of the outcome of a test. It forms the basis for designing an experiment in the scientific method.A good hypothesis is testable, meaning it makes a prediction you can check with observation or experimentation.
Step 5: Phrase your hypothesis in three ways. To identify the variables, you can write a simple prediction in if … then form. The first part of the sentence states the independent variable and the second part states the dependent variable. If a first-year student starts attending more lectures, then their exam scores will improve.
Step 2) State the level of significance. We will choose a level of significance of 5% (α = 0.05). Step 3) Compute the test statistic. For this problem, the test statistic is: The p-value is two times the area of the absolute value of the test statistic (because the alternative claim is "not equal").
Practical Skills in Biology. Module 2: Foundations in Biology. Module 4: Biodiversity, Evolution and Disease. Module 6: Genetics, Evolution and Ecosystems. Advertisement. Revision for OCR Biology (A) AS and A Level Papers, including summary notes, worksheets and past exam questions for each topic.