Question

In a survey of 3,748 ​travelers, 1,537 said that location was very important for choosing a hotel and 1,193 said that reputation was very important in choosing an airline. Complete parts ​(a) through​(c) below.

Construct a 95​% confidence interval estimate for the population proportion of travelers who said that reputation was very important in choosing an airline.

nothing less than or equals ≤piπ less than or equals ≤ nothing

​(Round to four decimal places as​ needed.)

Answer 1

Solution:

Here, we have to construct the confidence interval for the population proportion. The confidence interval formula is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Number of favourable observations = x = 1193

Sample size = n = 3748

Sample proportion = P = x/n = 1193/3748 = 0.318303095

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.318303095 ± 1.96* sqrt(0.318303095*(1 – 0.318303095)/3748)

Confidence Interval = 0.318303095 ± 1.96* 0.0076

Confidence Interval = 0.318303095 ± 0.0149

Lower limit = 0.318303095 - 0.0149 = 0.3034

Upper limit = 0.318303095 + 0.0149 = 0.3332

0.3034 ≤ ∏ ≤ 0.3332