Question

In a survey of 2543 adults in a recent​ year, 1291 say they have made a New​ Year's resolution. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

Answer 1

Solution :

(A)Given that,

n = 2543

x = 1291

Point estimate = sample proportion = $\hat p$ = x / n = 1291 / 2543=0.508

1 - $\hat p$ = 1 - 0.508=0.492

At 90% confidence level

$\alpha$ = 1 - 90%  

$\alpha$ = 1 - 0.90 =0.10

$\alpha$/2 = 0.05

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

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.645 ($\sqrt$((0.508*0.492) /2543 )

= 0.0163

A 90% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.508 - 0.0163 < p < 0.508 +0.0163

0.4917< p < 0.5243

The 90% confidence interval for the population proportion p is : 0.4917 ,0.5243

(B)Solution :

Given that,

n = 2543

x = 1291

Point estimate = sample proportion = $\hat p$ = x / n = 1291 / 2543=0.508

1 - $\hat p$ = 1 - 0.508=0.492

At 95% confidence level

$\alpha$ = 1 - 95%  

$\alpha$ = 1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.96}

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.96 ($\sqrt$((0.508*0.492) /2543 )

= 0.0194

A 95% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.508 - 0.0194 < p < 0.508 +0.0194

0.4886< p < 0.5274

The 95% confidence interval for the population proportion p is : 0.4886 ,0.5274

part b is larger than part a