Question

What is the point estimate of the proportion of all recent clients who were “dissatisfied?” Develop the 92% confidence interval for the proportion of all recent clients who were “dissatisfied.” Interpret what the confidence interval tells you about the proportion of all recent clients who were “dissatisfied.” What is the corresponding margin of error? How can the margin of error be decreased?

Sample size=200

Poor room quality: P = 42/200 = 0.21

Poor food quality: P = 47/200 = 0.235

Poor service quality: P= 54/200 = 0.27

Answer 1

sOL:

Dissatisfied customers are poor room quality +poor food quality +poor service quality

=42+47+54

=143

Proportion of customers dissatisfied=p^=143/200=0.715

z critial for 92% confidence level= 1.750686

margin of error=zcrit*sqrt(p^*(1-p^)/n)

=  1.750686*sqrt(0.715*(1-0.715)/200)

= 0.05588158

92% confidence interval for the proportion of all recent clients who were “dissatisfied.

=sample proportion-margin of error,sample proportion+margin of error

=0.715- 0.05588158,0.715+ 0.05588158

= 0.6591184,  0.7708816

92% confidence interval for p is

0.6591184 and 0.7708816