Question

A man who moves to a new city sees that there are two routes he could take to work. A neighbor who has lived there a long time tells him Route A will average 5 minutes faster than Route B. The man decides to experiment. Each day, he flips a coin to determine which way to go, driving each route 20 days. He finds that Route A takes an average of 40 minutes, with standard deviation 3 minutes, and Route B takes an average of 43 minutes, with standard deviation 2 minutes. Histograms of travel times for the routes are roughly symmetric and show no outliers. a) Find a 95% confidence interval for the difference in average commuting time for the two routes. (From technology, df = 33.1) b) Should the man believe the old-timer 's claim that he can save an average of 5 minutes a day by always driving Route A? Explain.

Answer 1

Answer:

Given

A will take 5 more average minutes faster than B

A drives each route 40 days

Standard deviation=3

Confidence interval=95%

From the given data man should claim that she should save average of 5 minutes a day by driving.

Answer 2

Answer:

(a) Confidence Interval for Route A is (38.61, 41.39)min

Confidence Interval for Route B is (42.07, 43.93)min

(b) No

Step-by-step explanation:

(a) Route A

Mean=40min, sd=3min, df=33.1, t=2.0675, n=20

CI = (mean + or - t×sd/√n)

CI = (40 + 2.0675×3/√20) = 40+1.39 = 41.39min

CI = (40 - 2.0675×3/√20) = 40 - 1.39 = 38.61min

CI for Route A is (38.61,41.39)min

Route B

Mean=43min, sd=2min, t=2.0675, n=20

CI = (43 + 2.0675×2/√20) = 43+ 0.93 = 43.93min

CI = (43 - 2.0675×2/√20) = 43 - 0.93 = 42.07min

CI for Route B is (42.07,43.93)min

(b) He should not believe the old timer's claim. The man saves an average of 3 minutes a day by driving Route A