Step 1 :The problem is asking to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute. The sample size is 144 and the sample standard deviation is 11.0 beats per minute. The significance level is 0.01. We can use the chi-square test for a single variance to test this claim. The null hypothesis is that the population standard deviation is equal to 10 beats per minute. The alternative hypothesis is that the population standard deviation is not equal to 10 beats per minute.
Step 2 :Given values are: sample size (n) = 144, sample standard deviation (s) = 11.0, population standard deviation (sigma) = 10, significance level (alpha) = 0.01.
Step 3 :Calculate the chi-square statistic using the formula: \(\chi^2 = (n-1)\frac{s^2}{\sigma^2}\). Substituting the given values, we get \(\chi^2 = 173.03\).
Step 4 :Calculate the critical chi-square values for the left and right tails of the distribution using the chi-square distribution table. The critical values are \(\chi^2_{left} = 103.19555083328622\) and \(\chi^2_{right} = 190.3063753701604\).
Step 5 :The test statistic is greater than the critical value on the left and less than the critical value on the right. Therefore, we do not reject the null hypothesis. This means that there is not enough evidence to support the claim that the population standard deviation is not equal to 10 beats per minute.
Step 6 :Final Answer: There is not sufficient evidence to reject the claim that pulse rates of men have a standard deviation equal to 10 beats per minute. The results indicate that using the range rule of thumb with the 'normal range' from 60 to 100 beats per minute for estimating \(\sigma\) is effective in this case. Therefore, the final conclusion is \(\boxed{\text{Do not reject the null hypothesis}}\).