Question

QUESTION 5

A bus company advertised a mean time of 150 minutes for a

trip between two cities. A consumer group had reason to

believe that the mean time was more than 150 minutes. A

sample of 40 trips showed a mean ˉx=153 minutes and a

standard deviation s=7.5 minutes. At the 0 .05 level of

significance, test the consumer group’s belief.

Answer 1

Given, Level of significance(a) = 0.05 sample Size(n) = 40 Standard Deviation(s) = 7.5 Sample Mean(c) = 153 Null Hypothesis Ho : p = 150 . Alternative Hypothesis H1 : > 150 T-H Test Statistic(t) 153 - 150 7.5/40 = 2.53 s/n Degree of freedom = n -1 = 40 – 1= 39 P – value P(t > 2.53) 0.0078 (use excel function :=TDIST(2.53, 39, 1)) as p-value is less than significance levella = 0.05) Reject Null Hypothesis. Conclusion : Hence, there is Enough evidence to support the claim that the mean time was more than 150 minutes.