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 5is 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

Answer:

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.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 5 minutes

Sample mean, $\bar{x}$ = 9.130 minutes

Sample size, n = 10

Alpha, α = 0.01

Sample standard deviation, s = 1.4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 5\text{ minutes}\\H_A: \mu \neq 5\text{ minutes}$

We have to construct 99% confidence interval.

99% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.01} = \pm 3.249$  

$9.130 \pm 3.249(\dfrac{1.4}{\sqrt{10}} ) = 9.130 \pm 1.4383 = (7.6917,10.5683)$

99% Confidence interval:  (7.6917,10.5683)

Since the population mean does not lie in he calculated confidence interval, thus, we fail to accept the null hypothesis and accept the alternate hypothesis.

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.