Question

A certain railway company claims that its trains run late 5 minutes on the average. The actual times (minutes) that 10 randomly selected trains ran late were provided giving a sample mean = 9.130 and sample standard deviation s = 1.4 . In testing the company’s claim, (2-sided test) at the significance level of 0.01 and assuming normality Find a 99% confidence interval and state which of the following is true about trains belonging to this railway company?

A. We conclude that the trains run late an average of 9.13 minutes.

B. The correct answer is not among the choices.

C. We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.

D. We do not have enough evidence to reject the claim that the trains run late an average of 5 minutes since the value of 5 is included in the 99% confidence interval.

E. We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of less than 5 minutes since 5 exceeds the entire 99% confidence interval.

Answer 1

H0: Null Hypothesis: = 5 (Claim)

HA: Alternative Hypothesis: 5

SE = s/

= 1.4/

= 0.4427

Test Statistic is given by:
t = (9.13 - 5)/0.4427

= 9.3287

= 0.01

ndf = n - 1 = 10 - 1 = 9

From Table, critical values of t = 3.2498

Since calculated value of t = 9.3287 is greater than critical value of t = 3.2498, the difference is significant. Reject null hypothesis.

The data do not support the claim that the trains run late an average of 5 minutes.

Confidence Interval:

9.13 (3.2498 X 0.4427)

= 9.13 1.4387)

= ( 7.6913 ,10.5687)

Confidence Interval:

7.6913 <   < 10.5687

Correct option:

C. We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.