M&M\'s Color Distribution: Suppose the makers of M&M candies give the following
ID: 3225542 • Letter: M
Question
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
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 12. You suspect that the maker's claim is not true.
Observed Counts by Color (n = 200)
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 = 1/6
H0: pbrown = 0.30, pyellow = 0.20, pred = 0.20, porange = 0.10, pgreen = 0.10, and pblue = 0.10.
H0: The probabilities are not all equal to 1/6.
H0: At least one of the probabilities doesn't fit the company's statement.
(b) What is the value of the test statistic? Round to 3 decimal places unless your software automatically rounds to 2 decimal places.
2 =
(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 H0
fail 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.
Brown Yellow Red Orange Green Blue Percent 30% 20% 20% 10% 10% 10%Explanation / Answer
a)H0: pbrown = 0.30, pyellow = 0.20, pred = 0.20, porange = 0.10, pgreen = 0.10, and pblue = 0.10.
b)
X2 =5.1417
p value=0.3988
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.
observed Expected Chi square Probability O E=total*p =(O-E)^2/E Brown 0.30 64.000 60.00 0.27 Yellow 0.20 35.000 40.00 0.63 red 0.20 44.000 40.00 0.40 Orange 0.10 22.000 20.00 0.20 green 0.10 23.000 20.00 0.45 Blue 0.10 12.000 20.00 3.20 200 200 5.1417Related Questions
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