Question

A simple random sample of 43 men from a normally distributed population results in a standard deviation of 9.9 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. Complete parts (a) through (d) below.

Identify the null and alternative hypotheses for this test.

Identify the test statistic for this hypothesis test.

Identify the P-value for this hypothesis test.

Identify the conclusion for this hypothesis test.

Answer #1

Given that

sample standard deviation (s) = 9.9

population standard deviation (sigma) = 10

sample size n = 43

(A) We have to test whether the standard deviation is equal to 10 or not. So, it is a left tailed hypothesis testing

(B) test statistic = [(n-1)*s^2]/sigma^2

= [(43-1)*9.9^2]/10^2

= 4116.42/100

= 41.164

(C) degree of freedom =n-1

= 43-1

= 42

using chi square distribution table with df(42) and test statistic(41.164)

we get

p value= 0.9849

(D) p value is greater than 0.05 significance level, so we failed to reject the null hypothesis

Therefore, we can conclude that there is insufficient evidence to warrant the rejection of claim that the standard deviation is equal to 10

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