Question

3. 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 10. You suspect that the maker's claim is not true. Observed Counts by Color (n = 200) Brown Yellow Red Orange Green Blue Count 66 35 44 22 23 10 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? H0: pbrown = pyellow = pred = porange = pgreen = pblue = 1/6 H0: At least one of the probabilities doesn't fit the company's statement. 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. (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 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.

Answer 1

This is test for goodness of fit test. Here we have been given a pre test distribution and set of observed values. So we carry the test to check whether these values follow the expected distribution. So we use a chi-squre test. Here one condtion is that all the expected values should be at least 5.

Expected : Expected % * Total freq

Eg: Brown = 30% * 200 = 60

Colour	Observed	Expected %	Expected
Brown	66	30%	60	6	36	0.6
Yellow	35	20%	40	-5	25	0.625
Red	44	20%	40	4	16	0.4
Orange	22	10%	20	2	4	0.2
Green	23	10%	20	3	9	0.45
Blue	10	10%	20	-10	100	5
Total	200	1	200			7.275

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

H0: pbrown = 0.30, pyellow = 0.20, pred = 0.20, porange = 0.10, pgreen = 0.10, and pblue = 0.10.

The distribution is a good fit.

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

Test Stat =

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 =

=

Assuming level of significance = 5%

p -value > 5%

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

Fail to reject H0

It means that there sufficient evidence to conclude that the data fits the claim

(e) Choose the appropriate concluding statement.

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

We have proven that the distribution of candy colors fits the maker's claim. : Even though this statement gives same conclusion but it is a test with some error so we need to conclude mentioning about the evidence.

The rest of the answers mean the same that the distribution does not fit.