Question

A survey asked 900 adult Americans how much it would bother them to stay in a room on the 13^th floor of a hotel. Interestingly, 15% said yes. Explain the meaning of the result!

a. What is the Standard Error of the estimate of the population proportion?

b. What is the Margin of Error at a C.L. of 95%?

c. What is the Margin of Error at a C.L. of 92%?

d. At a 98% C.L., if the researcher wants to limit the error within one percent range, what size of the sample is needed?

Answer 1

Sample Size,   n =    900
      
Sample Proportion ,    p̂ =    0.15

Number of Items of Interest,   x = 0.15*900 = 135

it means in a survey 135 people out of 900 adult Americans say that it would bother them to stay in a room on the 13^thfloor of a hotel

a)

std error of estimate = √[p̂(1-p̂)/n] = √(0.15*(1-0.15)/900) = 0.0119

b)

α =    0.05
z -value =   "Zα/2 = "   1.9600   [excel formula =NORMSINV(α/2)]

std error of estimate,SE = 0.0119

margin of error ,    E = Z*SE =    0.0233

c)

α = 1-0.92 = 0.08

z -value =   "Zα/2 = "   1.7507   [excel formula =NORMSINV(α/2)]

std error of estimate ,SE= 0.0119

  
margin of error ,    E = Z*SE =    0.0208  

d)

sample proportion ,   p̂ =    0.15                          
sampling error ,    E =   0.01                          
Confidence Level ,   CL=   98%                          
                                  
alpha =   1-CL =   2.00%                          
Z value =    Zα/2 =    2.326   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   2.326   /   0.01   ) ² *   0.15   * ( 1 -   0.15   ) =    6900.17
                                  
                                  
so,Sample Size required=       6901