Question

In a survey of 3586 adults concerning complaints about​ restaurants, 1430 complained about dirty or​ ill-equipped bathrooms and 1211 complained about loud or distracting diners at other tables. Complete parts ​(a) through​ (c) below. a. Construct a 90​% confidence interval estimate of the population proportion of adults who complained about dirty or​ ill-equipped bathrooms.

Answer 1

Solution :

Given that,

n = 3586

x = 1430

Point estimate = sample proportion = = x / n = 1430 / 3586 = 0.399

1 - = 1 - 0.399 = 0.601

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.399 * 0.601) / 3586)

= 0.013

A 90% confidence interval for population proportion p is ,

± E

= 0.399  ± 0.013

= ( 0.386, 0.412 )