A simple random sample of 28 wait times while pumping gas at a gas station has a standard deviation of 20 seconds. The test statistic for the random sample is 22.314 when found to test a claim that the standard deviation for all wait times while pumping gas at a gas station is less than 22 seconds. Using a 0.05 significance level, find the critical value and state the initial conclusion.

In order to calculate the p value we need to have in count the degrees of freedom , on this case 27. And since is a left tailed test the p value would be given by:

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

"=CHISQ.DIST(22.314,27,TRUE)"

The critical value can be founded with the following code:

=CHISQ.INV(0.05,27)

And we got :

If we compare the p value and the significance level provided we see that so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And we can conclude that the true deviation is not lower than 22 at 5% of significance.

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"

represent the sample size

represent the confidence level  

represent the sample deviation obtained

represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation specification is lower than 22, so the system of hypothesis would be:

Null Hypothesis:

Alternative hypothesis:

Calculate the statistic  

For this test we can use the following statistic:

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

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 27. And since is a left tailed test the p value would be given by:

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

"=CHISQ.DIST(22.314,27,TRUE)"

The critical value can be founded with the following code:

=CHISQ.INV(0.05,27)

And we got :

Conclusion

If we compare the p value and the significance level provided we see that so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And we can conclude that the true deviation is not lower than 22 at 5% of significance.

answer is 67%

step-by-step explanation: