Question

In a survey of 3,531 adults concerning complaints about restaurants, 1,469 complained about dirty or ill-equipped bathrooms and 1,168 complained about loud or distracting diners at other tables. Complete parts (a) through (c) below. a. Construct a 95% confidence interval estimate of the population proportion of adults who complained about dirty or ill-equipped bathrooms sas (Round to four decimal places as needed.)

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 1469 / 3531 = 0.416

1 - $\hat p$ = 1 - 0.416 = 0.584

Z$\alpha$/2 = 1.96

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

= 1.96 * ($\sqrt$((0.416 * 0.584) / 3531)

Margin of error = E = 0.0163

A 95% confidence interval for population proportion p is ,

$\hat p$ - E $\leq$$\pi$$\leq$$\hat p$ + E

0.416 - 0.0163  $\leq$$\pi$$\leq$ 0.416 + 0.0163

0.3997 $\leq$$\pi$$\leq$ 0.4323