Question

The following information was accumulated from samples of new births taken from two provinces d Sample Statistic British Columbia Manitoba Sample size (n) 150 200 Number of low-birth-weight babies 18 22 Develop a 95% confidence interval estimate for the difference between the proportions of low-birth-weight babies in the two provinces Select one: a. (0.57 0.77) b. (-0.57.-0.771 C. (-0.057.0.077) d. (0.7.0.977)

Answer 1

Now , the estimates of the sample proportions are ,

$\hat{p_1}=\frac{X_1}{n_1}=\frac{18}{150}=0.12$

$\hat{p_2}=\frac{X_2}{n_2}=\frac{22}{200}=0.11$

The pooled estimate is ,

$\hat{P}=\frac{X_1+X_2}{n_1+n_2}=\frac{18+22}{150+200}=0.1143$

Now , the critical value is ,

$Z_{\alpha/2}=Z_{0.05/2}=1.96$ ; From Z-table

Therefore , the 95% confidence interval is ,

$(\hat{p_1}-\hat{p_2})\pm Z_{\alpha/2}\sqrt{\hat{P}(1-\hat{P})(\frac{1}{n_1}+\frac{1}{n_2})}$

$(0.12-0.11)\pm 1.96*\sqrt{0.1143(1-0.1143)(\frac{1}{150}+\frac{1}{200})}$

$0.01\pm 0.0674$