Question

100 people were asked if they had health insurance, and 13 said they did not. What is the 95% confidence interval estimate for the proportion of people who do not have health insurance?

13 ± 0.066

0.13 ± 0.022

0.13 ± 0.066

0.87 ± 0.066

MiraCosta College wishes to estimate the proportion of students who take online classes with a margin of error of 0.01 with 95% confidence. There is no pilot data. What is the required sample size?

9,715

10,021

9,604

7,241

Answer 1

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   13          
Sample Size,   n =    100          
                  
Sample Proportion ,    p̂ = x/n =    0.130          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0336          
margin of error , E = Z*SE =    1.960   *   0.0336   =   0.06
                  
95%   Confidence Interval is 0.13 ± 0.066

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without prior estimate, let    sample proportion ,   p̂ =    0.5000                          
   sampling error ,    E =   0.01                          
   Confidence Level ,   CL=   0.95                          
                                      
   alpha =   1-CL =   0.05                          
   Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]                      
                                      
   Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.960   /   0.01   ) ² *   0.50   * ( 1 -   0.50   ) =    9603.65
                                      
                                      
   so,Sample Size required=       9604