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

H0: 360 (Since 6 min = 6 * 60 = 360 seconds)

Ha: > 360

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 370.69 - 360 ) / ( 24.36 / √(26) )
t = 2.2376

Test Criteria :-
Reject null hypothesis if t > t(α, n-1)
Critical value t(α, n-1) = t(0.05 , 26-1) = 1.708
t > t(α, n-1) = 2.2376 > 1.708
Result :- Reject null hypothesis

We conclude that we have sufficient evidence to support the claim that the data contradict prior belief.