Question

A hotel conducts a satisfaction survey on 400 randomly chosen clients. It is found that 40 of these 400 clients are not satisfied by the service quality, and the rest are satisfied. a. Estimate the percentage of the clients who are satisfied at %99 confidence level. b. Estimate the percentage of the clients who are not satisfied at %90 confidence level.

Answer 1

A) $\widehat p$ = 360/400 = 0.9

At 99% confidence interval the critical value is z_0.005 = 2.58

The 99% confidence interval for population proportion is

$\widehat p$ +/- z_0.005 * sqrt($\widehat p$(1 - $\widehat p$)/n)

= 0.9 +/- 2.58 * sqrt(0.9 * 0.1/400)

= 0.9 +/- 0.0387

= 0.8613, 0.9387

= 86.13%, 93.87%

B) $\widehat p$ = 40/400 = 0.1

At 90% confidence interval the critical value is z_0.05 = 1.645

The 90% confidence interval for population proportion is

$\widehat p$ +/- z_0.05 * sqrt($\widehat p$(1 - $\widehat p$)/n)

= 0.1 +/- 1.645 * sqrt(0.1 * 0.9/400)

= 0.1 +/- 0.025

= 0.075, 0.125

= 7.5%, 12.5%