Question

In a survey of 3 comma 841 adults concerning complaints about​ restaurants, 1 comma 488 complained about dirty or​ ill-equipped bathrooms and 1 comma 166 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. Round to the nearest 4 digits. b). Construct a 95% confidence interval estimate of the population proportion of adults who complained about distracting diners at other tables.

Answer 1

a)
sample proportion, = 0.3874
sample size, n = 3841
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.3874 * (1 - 0.3874)/3841) = 0.0079

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0079
ME = 0.0155

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.3874 - 1.96 * 0.0079 , 0.3874 + 1.96 * 0.0079)
CI = (0.3719 , 0.4029)

b)

sample proportion, = 0.3036
sample size, n = 3841
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.3036 * (1 - 0.3036)/3841) = 0.0074

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0074
ME = 0.0145

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.3036 - 1.96 * 0.0074 , 0.3036 + 1.96 * 0.0074)
CI = (0.2891 , 0.3181)