The UMUC Daily News reported that the color distribution for plain M&M’s was: 35
ID: 3324947 • Letter: T
Question
The UMUC Daily News reported that the color distribution for plain M&M’s was: 35% brown, 30% yellow, 15% orange, 10% green, and 10% tan. Each piece of candy in a random sample of 200 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
80
45
30
30
15
(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.
The UMUC Daily News reported that the color distribution for plain M&M’s was: 35% brown, 30% yellow, 15% orange, 10% green, and 10% tan. Each piece of candy in a random sample of 200 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
80
45
30
30
15
Explanation / Answer
Here we use the goodness of fit test.
The Hypothesis:
H0: The published colour distribution is correct., i.e There is no difference between the observed and expected frequencies.
Ha: The published colour distribution is incorrect. i.e There is a difference in observed and expected frequencies.
(a) The Test statistic: The below table show the calculation of the test Statistic.
2test = 11.43
(b) The Critical Value: The 2 critical at = 0.05, df = n -1 = 5 - 1 = 4 is 9.4877
The p Value: The p value at 2test = 3.78 , df = 3 is
p value = 0.0221
The Decision Rule: If 2test is > 2 Critical, Then Reject H0
Also, If p value is < , then Reject H0.
The Decision: Since 2test(11.43) is > 2 Critical(9.4877), We Reject H0
Since If p value (0.0221) is < (0.05), We Reject H0.
The Conclusion: There is sufficient evidence at the 95% significance level to conclude that the published colour distribution is incorrect.
Observed Expected % Expected (O-E)2 (O-E)2/E Brown 80 35 70 100 1.428571 Yellow 45 30 60 225 3.75 Orange 30 15 30 0 0 Green 30 10 20 100 5 Tan 15 10 20 25 1.25 Total 200.00 100.00 200.00 450.00 11.43Related Questions
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