The UMUC Daily News reported that the color distribution for plain M&M’s was: 40
ID: 3233281 • 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 18 15 7 18 (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
Explanation / Answer
Ho = The published color distribution for plain M&M’s is correct
H1 = The published color distribution for plain M&M’s is wrong
Colors
Observed
Expected
Chi - Square
Brown
42
0.40*100=40
0.1
Yellow
18
0.20*100=20
0.2
Orange
15
0.20*100=20
1.25
Green
7
0.1*100=10
0.9
Tan
18
0.1*100=10
6.4
100
100
8.85
The chi-square value = (O-E)2/E = 8.85
Degrees of freedom = n(number of colors) – 1= 5-1= 4
20.5,4 = 9.488
P-value = 0.06496
Since 2cal = 8.85< 20.5,4 =9.488 and P-value > =0.05, we accept the null hypothesis and hence conclude that the published color distribution for plain M&M’s is correct
Colors
Observed
Expected
Chi - Square
Brown
42
0.40*100=40
0.1
Yellow
18
0.20*100=20
0.2
Orange
15
0.20*100=20
1.25
Green
7
0.1*100=10
0.9
Tan
18
0.1*100=10
6.4
100
100
8.85
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.