Case Study - Comparing Two Means

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Is it true that tweaking a car engine improves gas mileage? A total of eight cars were tested to establish their mileage in miles per gallon. The cars were then tuned up and driven again to determine mileage for the second time. The outcomes are shown in the table below. Enter this

data into two separate columns in Minitab. Sample 1 will be the data for the automobiles

“After Tune-up” (entered in Column C1) and Sample 2 will be the data for the automobiles

“Before Tune-up” (entered in Column C2). Then test the claim that the mean mileage is higher

after the tune-up. Use a significance level of 0.05.

Automobile

1

2

3

4

5

6

7

8

After Tune-up

35.44

35.17

31.07

31.57

26.48

23.11

25.18

32.39

Before Tune-up

33.76

34.30

29.55

30.90

24.92

21.78

24.30

31.25

a. Are these Independent samples or Paired samples? (Highlight the correct choice below.)

Independent Paired

b. Highlight the appropriate test to use in Minitab: 2-Sample t Paired t

c. Let µd denote the population mean difference After – Before. Write the null and alternative hypotheses: H0: µd = 0 HA: µd > 0

d. What is the rejection rule?

If the p-value is less than or equal to 0.05, reject H0

If the p-value is greater than 0.05, do not reject H0.

e. Using Minitab, compute the test statistic? T = 9.14

f. Using Minitab, compute the P-value? P-value = 0.000

g. Based on your P-value and rejection rule, what is the conclusion about the null hypothesis?

(Highlight the correct conclusion below.)

Reject H0 Fail to reject H0

h. What do you conclude about the claim that the gas mileage is higher after the tune-up?

Since we reject the null hypothesis, we conclude that gas mileage is higher after the tune-up. Therefore, tuning a car engine improves gas mileage.

2.

The National Assessment of Educational Progress (NAEP) tested a sample of students who had used a computer in their math classes, and another sample of students who had not used a computer. The sample mean score for students using the computer was 309, with a sample standard deviation of 29. For students not using a computer, the sample mean was 303, with a sample standard deviation of 32. Assume there were 60 students in the computer sample, and 40 students in the sample that hadn’t used a computer. Can you conclude that the population mean scores differ? Use a significance level of 0.05. Let the sample who used the computer be Sample 1, and the sample that didn’t use the computer be Sample 2.a. Are these Independent samples or Paired samples? (Highlight the correct choice below.)

Independent Paired

b. Highlight the appropriate test to use in Minitab: 2-Sample t Paired t

c. Write the null and alternative hypotheses: H0: µ1 – µ2 = 0

HA: µ1 – µ2 ≠ 0

d. What is the rejection rule?

If the p-value is less than or equal to 0.05, reject H0

If the p-value is greater than 0.05, do not reject H0.

e. Using Minitab, compute the test statistic? T = 0.95

f. Using Minitab, compute the P-value? P-value = 0.343

g. Based on your P-value and rejection rule, what is the conclusion about the null hypothesis?

(Highlight the correct conclusion below.)

Reject H0 Fail to reject H0

h. What do you conclude about the claim that the mean scores differ?

Since we do not reject the null hypothesis, we conclude that the mean scores do not differ. Therefore, the mean score the students who used computers is equal to the mean score for the students who did not use computers.