Question

To study whether the students at a university are satisfied with the food service, a sample of 185 students revealed that 63% of them said they were satisfied with the food service. Develop the LOWER bound of the 99% confidence interval for the proportion of students who are satisfied with the food service. (please express your answer AS A DECIMAL using 4 decimal places)

Answer 1

Level of Significance,   α =    0.01   
Number of Items of Interest,   x =   116.55          
Sample Size,   n =    185          
                  
Sample Proportion ,    p̂ = x/n =    0.630          
z -value =   Zα =    2.326   [excel formula =NORMSINV(α)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0355          
margin of error , E = Z*SE =    2.326   *   0.0355   =   0.0826
                  
99%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.630   -   0.0826   =   0.5474

(-∞ , 0.5474)