In: Statistics and Probability
A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n = 1046 and x = 584 who said "yes." Use a 99% confidence level.
Level of Significance,   α =   
0.01          
Number of Items of Interest,   x =  
584          
Sample Size,   n =    1046  
       
          
       
Sample Proportion ,    p̂ = x/n =   
0.558          
z -value =   Zα/2 =    2.576   [excel
formula =NORMSINV(α/2)]      
          
       
Standard Error ,    SE = √[p̂(1-p̂)/n] =   
0.0154          
margin of error , E = Z*SE =    2.576  
*   0.0154   =   0.0396
          
       
99%   Confidence Interval is  
           
Interval Lower Limit = p̂ - E =    0.558  
-   0.0396   =   0.5188
Interval Upper Limit = p̂ + E =   0.558  
+   0.0396   =   0.5979
          
       
99%   confidence interval is (  
0.5188   < p <    0.5979  
)
Please revert in case of any doubt.
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