Answer:
The minimum sample size required is 49.
Step-by-step explanation:
The (1 - α)% confidence interval for the difference between two proportions is:
[tex]CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\ \sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}}[/tex]
*The sample size is considered equal in this case.
The width of the interval is at most 0.40.
Then the margin of error of the interval will be:
MOE = Width ÷ 2 = 0.20
The formula of the margin of error is:
[tex]MOE= z_{\alpha/2}\ \sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}}[/tex]
Assume that the two sample proportion values are 0.50.
The critical value of z for 95% confidence level is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Compute the sample size required as follows:
[tex]MOE= z_{\alpha/2}\ \sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}}[/tex]
[tex]n=[\frac{z_{\alpha/2}\times \sqrt{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}}{MOE}]^{2}[/tex]
[tex]=[\frac{1.96\times \sqrt{0.50(1-0.50)+0.50(1-0.50)}}{0.20}]^{2}\\\\=48.02\\\\\approx 49[/tex]
Thus, the minimum sample size required is 49.
