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.

Spoken Words Couples were recruited for a study of how many words people speak in a day. A random sample of 56 males resulted in a mean of 16,576 words and a standard deviation of 7871 words. Use a 0.01 significance level to test the claim that males have a standard deviation that is greater than the standard deviation of 7460 words for females (based on Data Set 24 “Word Counts”).

The standard deviation of the words of 56 males is 7871.

The level of significance is 0.01.

The standard deviation for females is 7460.

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 males have a standard deviation of words greater than the standard deviation of 7460 words of females.

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

$$\begin{gathered} {\text{H}}_0:\sigma =7460 \\ {\text{H}}_1:\sigma >7460 \end{gathered}$$

Here, $\sigma$is the standard deviation of the words of males.

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

$$\begin{gathered} {\chi }^2=\frac{\left(\text{n}-1\right)\times s^2}{{\sigma }^2} \\ =\frac{\left(56-1\right)\times {7871}^2}{{7460}^2} \\ =61.227 \end{gathered}$$

Thus, the value of the test statistic is 61.227

The test is right-tailed.

The critical value is computed as follows.

$$\begin{gathered} P\left({\chi }^2>{\chi }_{\alpha }^2\right)=\alpha \\ P\left({\chi }^2>{\chi }_{0.01}^2\right)=0.01 \end{gathered}$$

Referring to the chi-square table, for the critical value, the -ailed area 0.01 corresponding to a degree of freedom of 55, the value is 82.292.

The decision rule for the test is stated below.

If the test statistic is greater than the critical value, reject the null hypothesis at the given level of significance.

As the observed value ${\chi }^2=61.227 < {\chi }_{\alpha \left(n-1\right)}^2=82.291$, the null hypothesis is failed to be rejected.

Thus, there is not enough evidence to support the claim that males have a standard deviation of words greater than the standard deviation of 7460 words of females.