Question

The mean number of children among a sample of 15 low-income households is 2.8. The mean number of children among a sample of 19 high-income households is 2.4. The standard deviations for low- and high-income households are found to be 1.6 and 1.7, respectively. Test the hypothesis of no difference against the alternative that high-income households have fewer children. Use a = 0.05 and a pooled estimate of the variance.

Answer 1

Solution:

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

Null hypothesis: H₀: The high-income households have same number of children as compared to low-income households.

Alternative hypothesis: H_a: The high-income households have fewer children than low-income households.

H₀: µ₁ = µ₂ versus H_a: µ₁ < µ₂

This is a lower tailed test.

We assume

µ₁ = population mean for high-income households

µ₂ = population mean for low-income households

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)

From given data, we have

X1bar = 2.4

X2bar = 2.8

S1 = 1.7

S2 = 1.6

n1 = 19

n2 = 15

df = n1 + n2 – 2 = 19 + 15 – 2 = 32

α = 0.05

Critical value = -1.6939

(by using t-table)

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

S_p² = [(19 – 1)*1.7^2 + (15 – 1)*1.6^2]/(19 + 15 – 2)

S_p² = 2.7456

(X1bar – X2bar) = 2.4 – 2.8 = -0.4

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

t = -0.4 / sqrt[2.7456*((1/19)+(1/15))]

t = -0.4/0.5723

t = -0.6989

Test statistic = t = -0.6989

P-value = 0.2448

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that the high-income households have fewer children than low-income households.