Question

A manufacturer of storm windows randomly sampled 14 new (less than five years of age) homes and found that 8 of them had storm windows. Another random sample of 320 older (at least five years of age) homes was taken; 150 of them had storm windows. (a) With α 0.07, test whether more new homes have storm windows that older ones (b) Find a 99% confidence interval for the difference between the proportion of new homes with storm windows and the proportion of old er homes with storm windows (c) Repeat part (b) if the manufacturer randomly sampled more new homes, finding that altogether 142 of 250 have storm windows.

Answer 1

Part a

H₀: p₁ = p₂ versus H_a: p₁ > p₂

We are given α = 0.07

X1=8, N1=14, X2=150, N2=320

P1=X1/N1=8/14= 0.571428571

P2=X2/N2 =150/320 = 0.46875

Test statistic for z test for two population proportions is given as below:

Z = (P₁ – P₂) / sqrt[(P₁*(1 – P₁)/n₁) + (P₂*(1 – P₂)/n₂)]

Z = (0.571428571 - 0.46875) / sqrt[(0.571428571*(1 - 0.571428571)/14) + (0.46875*(1 - 0.46875)/320)]

Z = 0.102678571/ sqrt[(0.571428571*(1 - 0.571428571)/14) + (0.46875*(1 - 0.46875)/320)]

Z = 0.102678571/0.1352

Z = 0.759456886

P-value = 0.2238

P-value > α = 0.07

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that more new homes have storm windows that older ones.

Part b

Confidence interval = (P1_hat – P2_hat) ± Z*sqrt[(P1_hat*(1 – P1_hat)/N1) + (P2_hat*(1 – P2_hat)/N2)]

Z = 2.5758 (by using z-table)

Confidence interval = 0.102678571 ± 2.5758*sqrt[(0.571428571*(1 - 0.571428571)/14) + (0.46875*(1 - 0.46875)/320)]

Confidence interval = 0.102678571 ± 2.5758*0.1352

Confidence interval = 0.102678571 ± 0.3482

Lower limit = 0.102678571 - 0.3482 = -0.2455

Upper limit = 0.102678571 + 0.3482 = 0.4509

Part c

X1=142, N1=250, X2=150, N2=320

P1=X1/N1=142/250= 0.568

P2=X2/N2 =150/320 = 0.46875

P1 – P2 = 0.568 - 0.46875 = 0.09925

Confidence interval = 0.09925± 0.3482

Lower limit = 0.09925- 0.3482 = -0.0088

Upper limit = 0.09925+ 0.3482 = 0.2073