Question

1. A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 4% margin of error at a 95% confidence level, what size of sample is needed? Give your answer in whole people.

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 415 drivers and find that 286 claim to always buckle up. Construct a 95% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5]

Answer 1

1)

without prior estimate, let
sample proportion ,   p̂ =    0.5                          
sampling error ,    E =   0.04                          
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   /   0.04   ) ² *   0.50   * ( 1 -   0.50   ) =    600.2
                                  
                                  
so,Sample Size required=       601                          

2)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   286          
Sample Size,   n =    415          
                  
Sample Proportion ,    p̂ = x/n =    0.689          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0227          
margin of error , E = Z*SE =    1.960   *   0.0227   =   0.0445
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.689   -   0.0445   =   0.6446
Interval Upper Limit = p̂ + E =   0.689   +   0.0445   =   0.7337
                  
95%   confidence interval is (   0.6446   < p <    0.7337   )