Question

A simple random sample of 41men from a normally distributed population results in a standard deviation of 12.5 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 0.05 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute.

Answer 1

Given that a simple random sample of n = 41 men from a normally distributed population results in a standard deviation of s = 12.5 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.

With a 0.05 significance level, we test the claim that pulse rates of men have a standard deviation equal to = 10 beats per minute.

Based on the claim the hypotheses are:

Based on the hypothesis it will be a two-tailed test.

Rejection region:

based on the type of hypothesis and the given significance level the critical values for the rejection region are calculated using the excel formula for chi-square distribution which takes the significance level and the degree of freedom as the parameters, degree of freedom is calculated as df = n-1= 41-1= 40.

Since the test is two-tailed hence the area of both the tail will be 0.05/2 = 0.025.

The formula used is =CHISQ.INV.RT(1-0.025, 40) and =CHISQ.INV.RT(0.025, 40) thus the critical score computed as:

24.433 and 59.342 so reject the null hypothesis if:

Test statistic:

Decision:

Since we can see that the chi-square value is greater than the critical value which is 59.342 hence we can reject the null hypothesis.

Conclusion:

As we are able to reject the null hypothesis hence we conclude that at 0.05 level of significance there is sufficient evidence to warrant the rejection of the claim.