Question

A sample of 26 offshore oil workers took part in a simulated escape exercise, and their escape time (unit: second) were observed. The sample mean and sample standard deviation are 370.69 and 24.36, respectively. Suppose the investigators had believed a priori that true average escape time would be at most 6 minutes. Does the data contradict this prior belief? Assuming normality, test the appropriate hypotheses using the rejection region method at a significance level of 0.05.

Answer 1

Yes, it contradict this prior belief as there is enough evidence to support the claim that the true average escape time is significantly higher than 6 minutes.

Test statistic t=2.238>tc=1.708.

The null hypothesis is rejected.

Explanation:

This is a hypothesis test for the population mean.

The claim is that the true average escape time is significantly higher than 6 minutes (360 seconds).

Then, the null and alternative hypothesis are:

`H_0: \mu=360\\\\H_a:\mu> 360`

The significance level is 0.05.

The sample has a size n=26.

The sample mean is M=370.69.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=24.36.

The estimated standard error of the mean is computed using the formula:

`s_M=(s)/(√(n))=(24.36)/(√(26))=4.777`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(370.69-360)/(4.777)=(10.69)/(4.777)=2.238`

The degrees of freedom for this sample size are:

`df=n-1=26-1=25`

The critical value for a right-tailed test with a significance level of 0.05 and 25 degrees of freedom is tc=1.708. If the test statistic is bigger than 1.708, it falls in the rejection region and the null hypothesis is rejected.

As the test statistic t=2.238 is bigger than the critical value t=1.708, the effect is significant. The null hypothesis is rejected.

There is enough evidence to support the claim that the true average escape time is significantly higher than 6 minutes (360 seconds).