Question

A simple random sample of 28 wait times while pumping gas at 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. Could anyone show me how to do this?

Answer 1

The t value for the test statistic is (22.314-22)/(20/√28)=0.0831 approx. The number of degrees of freedom is 28-1=27. The critical value for a significance level of 0.05 for this is 1.703 for a 1-tail test (because we are testing an inequality standard deviation<22 seconds, the null hypothesis claim). Null hypothesis is the stated claim; the alternative hypothesis is that the standard deviation is greater than or equal to 22 seconds. 0.0831 is less than the critical value, so we reject the null hypothesis and by default conclude that the standard deviation is greater than or equal to 22 seconds with a confidence of 95%.