Answer to Question #139283 in Statistics and Probability for Creole Moodley

Question

Construct a 95% confidence interval for the true proportion of all employees of Handy Glass Pty who are members of the company’s medical aid scheme, given that 75 out of a sample of 135 employees are members of the company’s medical aid scheme. 

Answer 1

$\text{The confidence interval for the }\\ \text{proportion can be calculated as follows:}\\ \hat p ±Z_{\frac{α}{2}}(\sqrt {\frac{\hat p (1-\hat p )}{n}}),\\ \hat p =\frac{75}{135}\approx0.56,\\ Z_{\frac{α}{2}}=Z_{\frac{0.05}{2}}=Z_{0.025}=1.96,\\ \text{the lower limit }= 0.56-1.96(\sqrt {\frac{0.56 (1-0.56 )}{135}})\\ ≈0.476,\\ \text{the upper limit }= 0.56+1.96(\sqrt {\frac{0.56(1-0.56)}{135}})\\ ≈0.644,\\ \text{so the confidence interval is}\\ 0. 476\leq \ p \leq 0.644$