Question

Chapter 8, Section 8.5, Problem 75a x] Incorrect. An independent bank concerned about its customer base decided to conduct a survey of bank customers. Out of 503 customers who returned the survey form, 208 rated the overall bank services as excellent. Find a 90% confidence interval for the proportion of customers who would rate the overall bank services as excellent. Round endpoints of the interval to four decimal places. Interval = ( 1.3775 .3775 1.4495 .4495

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 208 / 503 = 0.4135

1 - $\hat p$ = 1 - 0.4135 = 0.5865

Z$\alpha$/2 = Z0.05 = 1.645

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

= 1.645 ($\sqrt$((0.4135 * 0.5865) / 503)

= 0.0361

A 90% confidence interval for population proportion p is ,

$\hat p$ ± E

= 0.4135  ± 0.0361

= ( 0.3774, 0.4496 )