e As in the example done in class imagine we sampled 594 MMs

(e) As in the example done in class, imagine we sampled 594 M&M’s and found that 116 were blue. Compute a 95% CI for p, the true proportion of blue M&M’s. Assume all of the CI assumptions are correct.

(f) Compute a 99% CI for p, the true proportion of blue M&M’s using the same data as in (e). Assume all of the CI assumptions are correct.

(j) For the interval you created in (e) write the non-classical i.e subjective interpretation in English

Solution

e)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.195286195          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.016265345          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.03187949          
lower bound = p^ - z(alpha/2) * sp =   0.163406705          
upper bound = p^ + z(alpha/2) * sp =    0.227165685          
              
Thus, the confidence interval is              
              
(   0.163406705   ,   0.227165685   ) [ANSWER]

********************

f)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.195286195          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.016265345          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.041896752          
lower bound = p^ - z(alpha/2) * sp =   0.153389443          
upper bound = p^ + z(alpha/2) * sp =    0.237182947          
              
Thus, the confidence interval is              
              
(   0.153389443   ,   0.237182947   ) [ANSWER]

********************

g)

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.005  
       
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
E =    0.03187949 (margin of error in part e)  
p =    0.195286195  
      
Thus,      
      
n =    1025.945817  
      
Rounding up,      
      
n =    1026  

Thus, he needs 1026 - 594 = 432 more. [ANSWER]

******************

j)

We are 95% confident that the true proportion of blue M&Ms is between 0.163406705 and 0.227165685. [ANSWER]