In: Statistics and Probability
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 | 21 |
Orange | 15 | Green | 16 | Blue | 14 |
Using the ?2 goodness of fit test to determine if the
proportion of M&Ms is what is claimed, what is the test
statistic?
The distribution of colors in a bag of M&M as claimed by Mars Inc is
Color | Probability |
Brown | 0.30 |
Orange | 0.10 |
Red | 0.20 |
Green | 0.10 |
Yellow | 0.20 |
Blue | 0.10 |
The observed frequency of colors in a randomly selected bag of M&M is
Color | Probability | Observed Frequency (fo) |
Brown | 0.30 | 23 |
Orange | 0.10 | 15 |
Red | 0.20 | 21 |
Green | 0.10 | 16 |
Yellow | 0.20 | 21 |
Blue | 0.10 | 14 |
We want to test the hypothesis that the distribution of colors claimed by Mars is a good description of the color found in a bag of M&M randomly selected from the grocery store shelf.
That is we want to test the following hypotheses
The total M&Ms in the bag is
If the claims of Mars inc were true (that is null hypothesis were true), then the number of Brown M&M in the bag should be 110*0.30 = 33
This is called expected frequency
The chi-square statistics is calculated as
the following tables shows the calculations
The values are
The value of test statistics is 7.67
The degrees of freedom is (number of groups -1) = (6-1) = 5
The chi-square critical value for alpha = 0.05 is obtained from the chi-square table for df=5 is 11.070
The test statistics of 7.67 is less than the critical value 11.070. That means we accept the null hypothesis.
We conclude that there is suffcient evidence to support the claim that the proportion of M&Ms is what is claimed by Mars Inc