17. The UMUC Daily News reported that the color distribution for plain M&M’s was
ID: 3131062 • Letter: 1
Question
17.
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
(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
21
12
7
18
Explanation / Answer
17. let p1 be the proportion of brown plain M&M,p2 be the proportion of yellow plain M&M,p3 be the proportion of orange plain M&M , p4 be the proportion of green plain M&M and p5 be the proportion of tan plain M&M
so the population is divided into k=5 mutually exclusive and exhaustive classes
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
we want to test the claim
a) hence the null hypothesis is H0: p1=0.4 p2=0.2 p3=0.2 p4=0.1 p5=0.1 vs alternative hypothesis H1: not H0
b) sample size=n=100 which is quite large
hence PEARSONIAN CHi-square test can be used.
for the sample we have
Color
BROWN
YELLOW
ORANGE
GREEN
TAN
NUMBER
42
21
12
7
18
let f1=number of brown plain M&M in the sample=42
f2=number of yellow plain M&M in the sample=21
f3=number of orange plain M&M in the sample=12
f4=number of green plain M&M in the sample=7
f5=number of tan plain M&M in the sample=18
hence the test statistic is given by
T=(f1-0.4n)2/(0.4n)+(f2-0.2n)2/(0.2n)+(f3-0.2n)2/(0.2n)+(f4-0.1n)2/(0.1n)+(f5-0.1n)2/(0.1n) which under H0 follows a chi square distribution with degrees of freedom=k-1=4
hence th value of the test statistic is t=(42-0.4*100)2/(0.4*100)+(21-0.2*100)2/(0.2*100)+(12-0.2*100)2/(0.2*100)+(7-0.1*100)2/(0.1*100)+(18-0.1*100)2/(0.1*100)=10.65
c) H0 is rejected iff t>chi-sqk-1;alpha where ch-sqk-1;alpha is the upper alpha point of a chi-sq k-1 distribution.
hence the rejection region is right sided.
so the p value is p=P[T>t] where T~chisquare with df=4
now t=10.65
so p=P[T>10.65]=1-P[T<10.65]=1-0.969208=0.030792 [using MINITAB]
d) level of significance is taken as alpha=0.05
so p<alpha
hence H0 is rejected
hence there is not sufficient evidence to support the claim that the published color distribution is correct [answer]
Color
BROWN
YELLOW
ORANGE
GREEN
TAN
NUMBER
42
21
12
7
18
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