Question

M&M's Color Distribution: Suppose the makers of M&M candies give the following average percentages for the...

M&M's Color Distribution: Suppose the makers of M&M candies give the following average percentages for the mix of colors in their bags of plain chocolate M&M's.

Stated Distribution of Colors

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

Now, you randomly select 200 M&M's and get the counts given in the table below. You expected about 20 blues but only got 9. You suspect that the maker's claim is not true.

Observed Counts by Color (n = 200)

Brown Yellow Red Orange Green Blue
Count   64     33     44     27     23     9  

The Test: Test whether or not the color of M&M's candies fits the distribution stated by the makers (Mars Company). Conduct this test at the 0.01 significance level.

(a) What is the null hypothesis for this test in terms of the probabilities of the outcomes?

H0:  pbrown = pyellow = pred = porange = pgreen = pblue = 1/6H0: At least one of the probabilities doesn't fit the company's statement.     H0: The probabilities are not all equal to 1/6.H0:  pbrown = 0.30, pyellow = 0.20, pred = 0.20, porange = 0.10, pgreen = 0.10, and pblue = 0.10.


(b) What is the value of the test statistic? Round to 3 decimal places unless your software automatically rounds to 2 decimal places.

χ2

=

(c) Use software to get the P-value of the test statistic. Round to 4 decimal places unless your software automatically rounds to 3 decimal places.
P-value =

(d) What is the conclusion regarding the null hypothesis?

reject H0fail to reject H0    


(e) Choose the appropriate concluding statement.

We have proven that the distribution of candy colors fits the maker's claim.The data suggests that the distribution of candy colors does not fit the maker's claim.    There is not enough data to suggest that the distribution of candy colors is different from what the makers claim.

Additional Materials

Homework Answers

Answer #1
Category pi Observed Freq. (Oi) Expected Freq. (Ei) (Oi - Ei)^2/Ei
Brown 0.3 64 60 0.267
Yellow 0.2 33 40 1.225
Red 0.2 44 40 0.400
Orange 0.1 27 20 2.450
Green 0.1 23 20 0.450
blue 0.1 9 20 6.05
200 10.842

b)
chi-square = 10.842

c)
p-value = 0.0546

d)
fail to reject H0

e)
There is not enough data to suggest that the distribution of candy colors is different from what the makers claim.


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