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 . 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?

Required:
a. State the null and alternative hypotheses to answer the question of interest.
b. Check conditions for inference. List the conditions and state whether they are met.
c. Calculate the test statistic. Show work.
d. What is the p-value for the test? Is it one or two sided?
e. Calculate a 90% confidence interval.

Step-by-step explanation:

Hello!

The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.

For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.

The established variable has a normal distribution with a standard deviation of σ= 2 min.

From the taken sample an average time difference of X[bar]= 0.77 was obtained.

If the app is correct, the true mean should be around cero, symbolically: μ=0

a. The hypotheses are:

H₀:μ=0

H₁:μ≠0

b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.

c.

d. This hypothesis test is two-tailed and so is the p-value.

p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058

e. 90% CI

X[bar] ±

0.77 ± 1.645 *

[0.098;1.442]

I hope this helps!

24/56=hx^7

step-by-step explanation:

the answer is 24

step-by-step explanation: