Question

Assume that a sample is used to estimate a population proportion p. Find the 99.5% confidence interval for a sample of size 289 with 254 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

< p <

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

Level of Significance,   α =    0.005
Number of Items of Interest,   x =   254          
Sample Size,   n =    289          
                  
Sample Proportion ,    p̂ = x/n =    0.879          
z -value =   Zα/2 =    2.807   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0192          
margin of error , E = Z*SE =    2.807   *   0.0192   =   0.0539
                  
100%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.879   -   0.0539   =   0.8250
Interval Upper Limit = p̂ + E =   0.879   +   0.0539   =   0.9328
                  
so, confidence interval is (   0.825 < p <    0.933 )