Question

Mars Inc. claims that they produce M&Ms with the following distributions:

Brown	30%	Red	20%	Yellow	20%
Orange	10%	Green	10%	Blue	10%

A bag of M&Ms was randomly selected from the grocery store shelf, and the color counts were:

Brown	23	Red	21	Yellow	20
Orange	16	Green	15	Blue	14

Find the 95% confidence interval for the proportion of blue M&Ms in that bag.

a) [0.066, 0.191]

b) [0.116, 0.121]

c) [-0.014, 0.191]

d) [0.066, -0.009]

e) [-0.034, 0.141]

f) None of the above

Answer 1

total M&Ms = 23+16+21+15+20+14 = 109

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   14          
Sample Size,   n =    109          
                  
Sample Proportion ,    p̂ = x/n =    0.128          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0320          
margin of error , E = Z*SE =    1.960   *   0.0320   =   0.0628
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.128   -   0.0628   =   0.0656
Interval Upper Limit = p̂ + E =   0.128   +   0.0628   =   0.1913
                  
so, confidence interval is (   0.066   < p <    0.191   )