Question

A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal​ range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute.
Complete parts​ (a) through​ (c) below.
a. Identify the null and alternative hypotheses.
b. Compute the test statistic. χ2 = ___ (round to three decimals)
c. Find the​ P-value. ​P-value=____​(Round to four decimal places as​ needed.)
d. State the conclusion. (reject null/ eccept null)

Answer 1

Answer:

(a) Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute  

     Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute  

(b) The value of chi-square test statistics is 35.704.

(c) P-value = 0.4360.

(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

If the range rule of thumb is applied to that normal​ range, the result is a standard deviation of 10 beats per minute.

Let $\sigma$ = population standard deviation for the pulse rates of men.

(a) So, Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute      {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute      {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is One-sample chi-square test for standard deviation;

                             T.S.  =  $\frac{(n-1)\times s^{2} }{\sigma^{2} }$  ~  $\chi^{2}__n_-_1$  

where, s = sample standard deviation = 10.1 beats per minute

            n = sample of men = 36

So, the test statistics =  $\frac{(36-1)\times 10.1^{2} }{10^{2} }$  ~ $\chi^{2}__3_5$

                                   =  35.70  4

(b) The value of chi-square test statistics is 35.704.

(c) Also, the P-value of the test statistics is given by;

                    P-value = P($\chi^{2}__3_5$ > 35.704) = 0.4360

(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.