Question

A recent report claims that non-college graduates get married at an earlier age than college graduates. To support the claim, random samples of size 100 were selected from each group, and the mean age at the time of marriage was recorded. The mean and standard deviation of the non-college graduates were 22.5 years and 1.4 years respectively, while the mean and standard deviation of college graduates were 23 years and 1.8 years respectively. Test the hypothesis (the claim being made) of the report.

Answer 1

Solution:

Here, we have to use two sample t test for the difference between two population means assuming equal population variances. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: There is no significant difference in the average married age for the non-college graduates and college graduates.

Alternative hypothesis: H_a: The non-college graduates get married at an earlier age than college graduates.

H₀: µ₁ = µ₂ versus H_a: µ₁ < µ₂

This is a lower tailed or left tailed test.

We assume level of significance = α = 0.05

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

WE are given

X1bar = 22.5

X2bar = 23

S1 = 1.4

S2 = 1.8

n1 = 100

n2 = 100

df = n1 + n2 – 2 = 100 + 100 – 2 = 198

α = 0.05

Critical value = -1.6526

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(100 – 1)*1.4^2 + (100 – 1)*1.8^2]/(100 + 100 – 2)

S_p² = 2.6

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (22.5 – 23) / sqrt[2.6*((1/100)+(1/100))]

t = -0.5 / 0.2280

t = -2.1926

P-value = 0.0147

(by using t-table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to claim that the non-college graduates get married at an earlier age than college graduates.