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Chapter 8: Q6BSC (page 356)

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then the conclusion about the null hypothesis, as well as the final conclusion that address the original claim. Assume that a simple random sample is selected from a normally distributed population.

Pulse Rates of Women Repeat the preceding exercise using the pulse rates of women listed in Data Set 1 鈥淏ody Data鈥� in Appendix B. For the sample of pulse rates of women, n= 147 and s= 12.5. See the accompanying JMP display that results from using the original list of pulse rates instead of the summary statistics. (Hint:The bottom three rows of the display provide P-values for a two-tailed test, a left-tailed test, and a right-tailed test, respectively.) What do the results indicate about the effectiveness of using the range rule of thumb with the 鈥渘ormal range鈥� from 60 to 100 beats per minute for estimating s in this case?

Short Answer

Expert verified

The hypothesis is stated as follows.

\(\begin{array}{l}{H_0}:\sigma = 10\\{H_1}:\sigma \ne 10{\rm{ }}\end{array}\)

The test statistic\({\chi ^2} = 229.68\), and the\(p\)-value = 0.0001. Reject\({{\rm{H}}_{\rm{0}}}\).

There is insufficient evidence to warrant the claim that the pulse rates of women have a standard deviation of 10 bpm.

It indicates that the range rule of thumb is not effective in estimating the standard deviation \(\sigma \).

Step by step solution

01

Given information

The summary of the results for the pulse rates of women is given.

The sample standard deviation for 147 sampled women is 12.5 bpm.

The normal range of pulse rates is 60 to 100 beats per minute, used to estimate the standard deviation.

02

State the hypothesis

Using the normal range, the estimate for the standard deviation of pulse rates is obtained as follows.

\(\begin{array}{c}{\sigma _0} = \frac{{Range}}{4}\\ = \frac{{100 - 60}}{4}\\ = 10\end{array}\)

To test the significance of theclaim that the pulse rates of women have a standard deviation equal to 10 beats per minute, the null and alternative hypothesis is formulated as follows.

\(\begin{array}{l}{H_0}:\sigma = 10\\{H_1}:\sigma \ne 10\end{array}\)

Here, \(\sigma \) is the actual standard deviation of all women.

03

State the test statistic

The test statistic is obtained from the output as 229.7176\(\left( {{\chi ^2}} \right)\)with the degree of freedom (DF) as 146. The chi-square test is a right-tailed test, and the p-value for the test is 0.0001.

04

State the decision rule

For the given level of significance, if the p-value is lesser than 0.05, the null hypothesis will be rejected. Otherwise, fail to reject the null hypothesis.

In this case, the p-value is lesser than 0.05. So, reject the null hypothesis.

05

Conclusion of the test

As the null hypothesis is rejected, it can be concluded that there is not enough evidence to support the claim that the pulse rates of women have a standard deviation of 10 bpm.

It indicates that the effectiveness of using the range rule of thumb with a normal range of 60 to 100 beats per minute is not a good standard deviation for estimating \(\sigma \), as it does not support the sampled results.

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