Testing Claims About Variation. In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population.

Statistics Test Scores Tests in the author’s statistics classes have scores with a standard deviation equal to 14.1. One of his last classes had 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this class has less variation than other past classes. Does a lower standard deviation suggest that this last class is doing better?

The standard deviation of 27 test scores is 9.3. The classes in the past have scores with a standard deviation of 14.1.

The level of significance is 0.01.

The claim states that the class with 27 scores has less variation than the classes in the past.

For applying the hypothesis test, first, set up a null and an alternative hypothesis.

The null hypothesis is the statement about the value of a population parameter, which is equal to the claimed value. It is denoted by\({H_0}\).

The alternate hypothesis is a statement that the parameter has a value opposite to the null hypothesis. It is denoted by\({H_1}\).

The claim states that the last class has a variation\(\left( \sigma \right)\)lesser than the other past classes.

From the claim, the null and alternative hypotheses are as follows.

\(\begin{array}{l}{{\rm{H}}_0}:\sigma = 14.1\,\,\,\\{{\rm{H}}_1}:\sigma < 14.1\end{array}\)

Here, \(\sigma \) is the population standard deviation measure of the last class.

To conduct a hypothesis test of a claim about a population standard deviation\(\sigma \) or population variance\({\sigma ^2}\),the test statistic is as follows.

\(\begin{array}{c}{\chi ^2} = \frac{{\left( {{\rm{n}} - 1} \right) \times {s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {27 - 1} \right) \times {{9.3}^2}}}{{{{14.1}^2}}}\\ = 11.3110\end{array}\).

Thus, the value of the test statistic is 11.3110.

The degree of freedom is as follows.

\(\begin{array}{c}df = n - 1\\ = 27 - 1\\ = 26\end{array}\)

As the hypothesis is left-tailed, the critical value is obtained as follows.

\(\begin{array}{c}P\left( {{\chi ^2} < \chi _\alpha ^2} \right) = \alpha \\1 - P\left( {{\chi ^2} > \chi _\alpha ^2} \right) = \alpha \\P\left( {{\chi ^2} > \chi _\alpha ^2} \right) = 1 - \alpha \\P\left( {{\chi ^2} > \chi _{0.01}^2} \right) = 0.99\end{array}\)

Referring to the chi-square table, the critical value of 12.198 is obtained at the intersection of the row with a degree of freedom of 26 and the column with a value of 0.99.

Thus, the rejection region is\(\left( {{\chi ^2}:{\chi ^2} < 12.198} \right)\).

The decision rule for the test is stated as follows.

If the test statistic is lesser than the critical value, reject the null hypothesis at the given level of significance; otherwise, fail to reject the null hypothesis.

It is observed that the test statistic 11.311 lies in the rejection region.

Thus, the null hypothesis is rejected.

Therefore, there is enough evidence to conclude that thestandard deviation of the last class is lesser than that of the past classes.

A lower standard deviation of the last class suggests that the test scores do not vary greatly as compared to those of the past classes.