Question

Each day a commuter takes a bus to work,
the transportation system has a phone app that tells her what
time the bus will arrive. She feels the app is often wrong, either
making her late or missing the bus all together. For 6 weeks,
the rider randomly picks four times a week to measure if the
bus was off from the estimated arrival time at her stop to work
(n =24). Each measurement is Actual Arrival Time – App
Estimated Time. If the bus was early, she recorded the number
as a negative time difference. If the bus was on time, she
recorded it as a positive time difference. From her sample, the average time difference of arrival versus the app is 0.77 minutes. For ease she assumes the population is normal and the standard deviation is ? = 2 ???????. If the bus app is consistently accurate than the overall average should be close to zero.

State: Is there evidence that the average time between actual arrival time and the bus app time is more than 0 minutes?

Plan:

1. (1 point) State the null and alternative hypotheses to answer the question of interest.

2. (2 points) Check conditions for inference. List the conditions and state whether they are met.

Solve:

c. (2 points) Calculate the test statistic. Show work.
d. (1 point) What is the p-value for the test? Is it one or two sided?

e. (3 points) Calculate a 90% confidence interval for

Conclude:

μ. Show work. [1] 0.7720833

f. Write a four-part conclusion describing the results.

•  (1 point) Provide a statement in terms of the alternative hypothesis.

•  (1 point) State whether the result is statistically significant. That is, whether, or not, to reject the null.

•  (1 point) State whether the result is practically significant or not.

•  (1 point) Give an interpretation of the point and interval estimate.

•  (1 point) Include context.

Answer 1

Answer Date: 03/03/2020 To test the hypothesis is that the average time between actual arrival time and the bus app time is more than 0 minutes at 5% significance level. The null and alternative hypothesis is, Họ: A = 0 H:\>0 The standard normal sampling distribution will used. Since it assumes x has a normal distribution with known population standard deviation is 2 conditions for inference of one-sample z-test is satisfied.

The z-test statistics is, Zr-u (g/n) 0.77-0 (2//24) = 1.886 The Z-test statistics is 1.886. d) This is one-tailed due to right-tailed alternative hypothesis is that the average time between actual arrival time and the bus app time is more than 0 minutes. The p-value for this test is, From the standard normal distribution table, the associated probability is 0.9704. Since find the area to the right, so subtract from 1 to get,

P(2>1.886)=1-(z<1.886) = 1 -0.9704 = 0.0296 The p-value for this test is 0.0296 e) A 90% confidence interval for the average time between actual arrival time and the bus app time is, First, compute z-critical value then find confidence interval. The z-critical value is, From the standard normal table, the z-critical value at 90% confidence is 1.645. 90% C.I. = utz, xo/In = 0.77 +1.645 x 2/24 = 0.77 +0.67 = 0.10 to 1.44

A 90% confidence interval for the average time between actual arrival time and the bus app time is 0.10 to 1.44. Decision Part 1 The conclusion is that the significant p-value in this context is less than 0.05 which is 0.0296, so the null hypothesis is rejected at 5% level of significance. Part 2 The result is practically statistically due to the null hypothesis is rejected at 5% significance level. Part 3

A 90% confidence interval for the average time between actual arrival time and the bus app time is between 0.10 and 1.44 does not include 0, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the average time between actual arrival time and the bus app time is more than 0 minutes indicate that the result is statistically significant. Part 4 Conclusion in terms of context There is sufficient evidence to indicate that the average time between actual arrival time and the bus app time is more than 0 minutes indicate that the result is statistically significant.