Question

A sample of 800 arthritis patients participate in a test of a new pain reliever. 624

of the participants rate their pain as ‘substantially reduced’. Give a 95% lower bound

confidence interval for the proportion of arthritis patients whose pain would be

substantially reduced when taking the new pain reliever. Interpret your result.

Answer 1

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   624          
Sample Size,   n =    800          
                  
Sample Proportion ,    p̂ = x/n =    0.780          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0146          
margin of error , E = Z*SE =    1.960   *   0.0146   =   0.0287
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.780   -   0.0287   =   0.7513
Interval Upper Limit = p̂ + E =   0.780   +   0.0287   =   0.8087
                  
95%   confidence interval is (   0.751   < p <    0.809   )

we are 95% confident that true proportion of arthritis patients whose pain would be substantially reduced when taking the new pain reliever will lie within confidence interval