Question

Claim: The standard deviation of pulse rates of adult males is less than 11 bpm. For a random sample of 132 adult​ males, the pulse rates have a standard deviation of 10.7 bpm.. Find the value ofthe test statistic.

Answer 1

Answer:

$\chi^2 =\frac{132-1}{121} 114.49 =123.952$

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=132$ represent the sample size

$\alpha$ represent the confidence level  

$s^2 =10.7^2 =114.49$ represent the sample variance obtained

$\sigma^2_0 =11^2=121$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is violated, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 121$

Alternative hypothesis: $\sigma^2$

Calculate the statistic  

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{132-1}{121} 114.49 =123.952$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 131. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(123.952,131,TRUE)"