Question

1. An experiment is conducted to compare the starting salaries of male versus female college graduates who find jobs. Pairs are formed by choosing a male and a female with the same major and similar GPAs. Suppose a random sample of 10 pairs is formed in this manner and the starting annual salary of each person is recorded. The results are shown in the table below. Perform a hypothesis test to determine if starting salaries of male versus female college graduates are different from each other. Use a 5% significance level.

Starting Salary (in Thousand $)
Pair	1	2	3	4	5	6	7	8	9	10
Male	29.3	41.5	40.4	38.5	43.5	37.8	69.5	41.2	38.4	59.2
Female	28.8	41.6	39.8	38.5	42.6	38.0	69.2	40.1	38.2	58.5
Diff (Male – Female)

1. State the null (H₀) and alternate (H_A) hypotheses.

Ho: µd = 0

Ha: µ d ≠ 0

2. Check the assumptions

3. Give the test statistic.

4. Give the P-value and the conclusion reached about the null hypothesis based on the P-value.

5. Summarize the final conclusion in the context of the claim.

6. Also construct a 95% confidence interval for the mean difference in salaries of males versus females. Does this confidence interval include the value of 0? Should it? Explain

Answer 1

Male	Female	Difference
29.3	28.8	0.5
41.5	41.6	-0.1
40.4	39.8	0.6
38.5	38.5	0
43.5	42.6	0.9
37.8	38	-0.2
69.5	69.2	0.3
41.2	40.1	1.1
38.4	38.2	0.2
59.2	58.5	0.7

∑d = 4

∑d² = 3.3

n = 10

Mean , x̅d = Ʃd/n = 4/10 = 0.4000

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] = √[(3.3-(4)²/10)/(10-1)] = 0.4346

a)Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd ≠ 0

b) assumptions:

• The dependent variable must be continuous (interval/ratio).
• The observations are independent of one another.
• The dependent variable should be approximately normally distributed.
• The dependent variable should not contain any outliers.

c) Test statistic:

t = (x̅d)/(sd/√n) = (0.4)/(0.4346/√10) = 2.9104

d) df = n-1 = 9

p-value :

Two tailed p-value = T.DIST.2T(ABS(2.9104), 9) = 0.0173

Decision:

p-value < α, Reject the null hypothesis

e) Conclusion:

There is enough evidence to conclude that the starting salaries of male versus female college graduates are different from each other at 5% significance level.

f)

95% Confidence interval :

At α = 0.05 and df = n-1 = 9, two tailed critical value, t-crit = T.INV.2T(0.05, 9) = 2.262

Lower Bound = x̅d - t-crit*sd/√n = 0.4 - 2.262 * 0.4346/√10 = 0.0891

Upper Bound = x̅d + t-crit*sd/√n = 0.4 + 2.262 * 0.4346/√10 = 0.7109

0.0891 < µd < 0.7109

The confidence interval does not contain 0. so we reject the null hypothesis.