The UMUC Daily News reported that the color distribution for plain M&M’s was: 40
ID: 3130928 • Letter: T
Question
The UMUC Daily News reported that the color distribution for plain M&M’s was: 40% brown, 20% yellow, 20% orange, 10% green, and 10% tan. Each piece of candy in a random sample of 100 plain M&M’s was classified according to color, and the results are listed below. Use a 0.05 significance level to test the claim that the published color distribution is correct. Show all work and justify your answer.
Color
Brown
Yellow
Orange
Green
Tan
Number
42
21
12
7
18
1.Identify the null hypothesis and the alternative hypothesis.
2. Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.
3. Determine the P-value. Show all work; writing the correct P-value, without supporting work, will receive no credit.
4. Is there sufficient evidence to support the claim that the published color distribution is correct? Justify your answer.
Color
Brown
Yellow
Orange
Green
Tan
Number
42
21
12
7
18
Explanation / Answer
1.Null hypothesis: The proportion of brown, yellow, orange, green and tan M&Ms are is 40%, 20%,20%,10%,10% respectively.
The alternative hypothesis is: At least one of the proportions in the null hypothesis is false.
2.The degrees of freedom=k-1=5-1=4
Now expected value is Ei=n*pi
E1=100*0.40=40
E2=100*0.20=20
E3=100*0.20=20
E4=100*0.10=10
E5=100*0.10=10
The test statistic chi square is:
Sum over(Oi-Ei)^2/Ei
=(42-40)^2/40+…+(18-10)^2/10
=10.65
3.The P value is determined using a P value calculator: 0.03
4.The P value is less than 0.05. Thus reject the null hypothesis. Thus there is not sufficient evidence to support the claim that the published color distribution is correct.
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