Question

A comparison is made between two bus lines to determine if arrival times of their regular buses from Denver to Durango are off schedule by the same amount of time. For 51 randomly selected runs, bus line A was observed to be off schedule an average time of 53 minutes, with standard deviation 17 minutes. For 61 randomly selected runs, bus line B was observed to be off schedule an average of 62 minutes, with standard deviation 11 minutes. Do the data indicate a significant difference in average off-schedule times? Use a 5% level of significance.

What are we testing in this problem?

difference of means

single proportion    

single mean

difference of proportions

paired difference

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: μ₁ = μ₂; H₁: μ₁ > μ₂

H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂    

H₀: μ₁ = μ₂; H₁: μ₁ < μ₂

H₀: μ₁ > μ₂; H₁: μ₁ = μ₂

(b) What sampling distribution will you use? What assumptions are you making?

The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.

The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.    

The standard normal. We assume that both population distributions are approximately normal with known standard deviations.

The Student's t. We assume that both population distributions are approximately normal with known standard deviations.

What is the value of the sample test statistic? (Test the difference μ₁ − μ₂. Round your answer to three decimal places.)


(c) Find (or estimate) the P-value.

P-value > 0.500

0.250 < P-value < 0.500   

0.100 < P-value < 0.250

0.050 < P-value < 0.100

0.010 < P-value < 0.050

P-value < 0.010

Sketch the sampling distribution and show the area corresponding to the P-value.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.    

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

(e) Interpret your conclusion in the context of the application.

There is sufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.

There is insufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.    

Answer 1

What are we testing in this problem?

difference of means

State the null and alternate hypotheses.

H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂    

(b) What sampling distribution will you use? What assumptions are you making?

The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.

What is the value of the sample test statistic? (Test the difference μ₁ − μ₂. Round your answer to three decimal places.)

Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

t = -3.376

(c) Find (or estimate) the P-value.

The p-value is p = 0.001

P-value < 0.010

Sketch the sampling distribution and show the area corresponding to the P-value.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.05 significance level.

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

(e) Interpret your conclusion in the context of the application.

There is sufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.