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 | 72 | 31 | 50 | 15 | 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 0.05 significance level.
(a) What is the null hypothesis for this test in terms of the probabilities of the outcomes?
H_{0}: The probabilities are not all equal to 1/6.
H_{0}: p_{brown} = p_{yellow} = p_{red} = p_{orange} = p_{green} = p_{blue} = 1/6
H_{0}: At least one of the probabilities doesn't fit the company's statement.
H_{0}: p_{brown} = 0.30, p_{yellow} = 0.20, p_{red} = 0.20, p_{orange} = 0.10, p_{green} = 0.10, and p_{blue} = 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.
χ^{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 H_{0}
fail to reject H_{0}
(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.
a)H0: pbrown = 0.30, pyellow = 0.20, pred = 0.20, porange = 0.10, pgreen = 0.10, and pblue = 0.10.
b)
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 | 5.1417 |
X^{2} =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.
Get Answers For Free
Most questions answered within 1 hours.