Question

Use the distribution below to test that Plain M&Ms follow the stated distribution. Discuss your choice of ?. Would a different ? have changed your conclusion?
M&M states the following distribution for Plain M&Ms:
Red = 13%, Orange = 20%, Yellow = 13%, Green = 20%, Blue = 20%, Brown = 14%

Total amount of Plain M&Ms: 678

Red=69 Orange=216 Yellow=75 Green=61 Blue=187 Brown=70

Answer 1

We would be using the chi square goodness of fit test here to test the claim whether the distribution follows the stated distribution.

The expected frequencies for each colour is computed here as:
= Proportion of that colour expected * Total Sample size

E_i = p_i*678

The computations are thus made here as:

Colour	Observed Freq	Expected Freq	(O_i - E_i)^2/E_i
Red	69	88.14	4.1563
Orange	216	135.6	47.6708
Yellow	75	88.14	1.9589
Green	61	135.6	41.0410
Blue	187	135.6	19.4835
Brown	70	94.92	6.5424
	678	678	120.8530

The chi square test statistic now is computed as shown in the last column of the above table as:

For n - 1 = 5 degrees of freedom, the p-value here is obtained from the chi square distribution tables as:

As the p-value here is approx. negligible, therefore the test is significant here and we can reject the null hypothesis here. Therefore we have sufficient evidence here that the given observed frequencies are not in line with the stated distribution.