Question

A decision maker is interested in estimating a population proportion. A sample of size n150 yields 115 successes. Based on these sample​ data, construct a​ 90% confidence interval estimate for the true population proportion. A. ​(0.714, 0.826) B. ​(0.717, 0.823) C. ​(0.750, 0.790) D. ​(0.737, 0.803)

Answer 1

A. ​(0.714, 0.826)

Step-by-step explanation:

Given that:

The sample size n = 150

The number of success(sample proportion) x = 115

The population proportion $\hat p$ =  $\dfrac{x}{n}$

The population proportion $\hat p$ = $\dfrac{115}{150}$

The population proportion $\hat p$ = 0.767

At 90% confidence interval level;

$Z_{\alpha /2} = Z_{0.05} = 1.645$

Thus, the confidence interval estimate for the true population proportion can be computed by using the formula:

$=\hat p \ \pm \ z_{\alpha /2} \sqrt{\dfrac{\hat p(1- \hat p)}{n} }$

$= 0.767 \ \pm 1.645 * \sqrt{\dfrac{0.767(1- 0.767)}{150} }$

$= 0.767 \ \pm 1.645 * \sqrt{\dfrac{0.767(0.233)}{150} }$

$= 0.767 \ \pm 1.645 * \sqrt{\dfrac{0.178711}{150} }$

$= 0.767 \ \pm 1.645 * \sqrt{.0011914}$

$= 0.767 \ \pm 1.645 * 0.03452$

$= 0.767 \ \pm 0.0568$

= (0.767 - 0.0568,  0.767 + 0.0568)

= (0.7102 , 0.8238 )

Due to approximation;  the confidence interval estimate for the true population proportion  is $\simeq$  ​(0.714, 0.826)