Daily News reported that the color distribution for plain M&M’s was: 40% brown,
ID: 3232680 • Letter: D
Question
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.
(a) Identify the null hypothesis and the alternative hypothesis.
(b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.
(c) Determine the P-value. Show all work; writing the correct P-value, without supporting work, will receive no credit.
(d) 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
18
15
7
18
Color
Brown
Yellow
Orange
Green
Tan
Number
42
18
15
7
18
Explanation / Answer
Answer:
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.
(a) Identify the null hypothesis and the alternative hypothesis.
Ho: the published color distribution is correct
H1: the published color distribution is not correct
(b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.
Goodness of Fit Test
observed
expected
O - E
(O - E)² / E
42
40.000
2.000
0.100
18
20.000
-2.000
0.200
15
20.000
-5.000
1.250
7
10.000
-3.000
0.900
18
10.000
8.000
6.400
Total
100
100.000
0.000
8.850
8.85
chi-square
4
df
.0650
p-value
(c) Determine the P-value. Show all work; writing the correct P-value, without supporting work, will receive no credit.
P=0.065
Excel function to get p value : =CHISQ.DIST.RT(8.85,4)
(d) Is there sufficient evidence to support the claim that the published color distribution is correct? Justify your answer.
Calculated chi square =8.85,.
P=0.0650 > 0.05 level. Ho is not rejected.
There is no sufficient evidence to reject the claim that the published color distribution is correct.
Goodness of Fit Test
observed
expected
O - E
(O - E)² / E
42
40.000
2.000
0.100
18
20.000
-2.000
0.200
15
20.000
-5.000
1.250
7
10.000
-3.000
0.900
18
10.000
8.000
6.400
Total
100
100.000
0.000
8.850
8.85
chi-square
4
df
.0650
p-value
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