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

	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	64	33	44	27	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 0.01 significance level.

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

H₀: At least one of the probabilities doesn't fit the company's statement.

H₀: The probabilities are not all equal to 1/6.   

H₀:  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.

H₀:  p_brown = p_yellow = p_red = p_orange = p_green = p_blue = 1/6.

(b) The table below is used to calculate the test statistic. Complete the missing cells.
Round your answers to the same number of decimal places as other entries for that column.

	Candy	Observed	Assumed	Expected
i	Color	Frequency (Oi)	Probability (pi)	Frequency (Ei)	(Oi − Ei)2 Ei
1	Brown	(?)	0.30	60	0.267
2	Yellow	33	(?)	40	1.225
3	Red	44	0.20	(?)	0.400
4	Orange	27	0.10	20	(?)
5	Green	23	0.10	20	0.450
6	Blue	9	0.10	20	6.050
Σ		n = 200			χ2 = (?)

(c) What is the value for the degrees of freedom?

(d) What is the critical value of

χ²? Use the answer found in the χ²-table or round to 3 decimal places.
t_α =  
(e) What is the conclusion regarding the null hypothesis?

reject H₀

fail to reject H₀    
(f) Choose the appropriate concluding statement.

We have proven that the distribution of candy colors fits the maker's claim.T

he 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

Ans:

a)

H₀:  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)

Candy	Observed	Assumed	Expected
i	Color	Frequency (Oi)	Probability (pi)	Frequency (Ei)	(Oi − Ei)2
					Ei
1	Brown	64	0.3	60	0.267
2	Yellow	33	0.2	40	1.225
3	Red	44	0.2	40	0.4
4	Orange	27	0.1	20	2.45
5	Green	23	0.1	20	0.45
6	Blue	9	0.1	20	6.05
Σ		n = 200	1	200	χ2 = 10.842

c)

df=6-1=5

d)

Critical chi square=CHIINV(0.01,5)=15.086

e)

fail to reject H₀

f)

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