Question

A researcher is interested in estimating the difference in two population proportions. A sample of 400 from each population results in sample proportions of .61 and .64. A 90% confidence interval for the difference in the population proportions is _______.

Answer 1

Answer: (-0.086, 0.026)

Step-by-step explanation:

The confidence interval for the difference of two population proportion is given by :-

$p_1-p_2\pm\ z_{\alpha/2}\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}$

Given : Significance level : $\alpha: 1-0.90=0.10$

Critical value : $z_{\alpha/2}=1.645$

Sample size : $n_1=400,\ n_2=400$

$p_1=0.61,\ p_2=0.64$

The 90% confidence interval for the difference in the population proportions will be :-

$0.61-0.64\pm\ (1.645)\sqrt{\dfrac{0.61(1-0.61)}{400}+\dfrac{0.64(1-0.64)}{400}}\\\\\approx-0.03\pm0.056=(-0.03-0.056,-0.03+0.056)\\\\=(-0.086, 0.026)$