A researcher purchased four 1-pound bags of plain M&Ms at different stores to co

Question

A researcher purchased four 1-pound bags of plain M&Ms at different stores to compare the color distribution to the one stated at the manufacturer's website. The observed results for the combined bags and the proportions alleged by the manufacturer are given in the following table.

State and test the appropriate hypotheses to determine whether it is reasonable to assume that the observed colors are a random sample from a population with the manufacturer's alleged proportions.

State the null and alternative hypotheses.
H0: p1 =    (brown), p2 =    (red), p3 =    (yellow), p4 =    (blue), p5 =    (orange), p6 =    (green) (manufacturer's claim)

Ha: not all pi are as specified in H0.

Find the expected counts for all results. (Give your answers correct to one decimal place.)

Compute the value of the chi-square test statistic, 2. (Give your answer correct to one decimal place.)
2 =  

What are the degrees of freedom, df, for this test statistic?
df =   

Find the p-value for the chi-square statistic. (Give your answer correct to three decimal places.)
p-value =  

State the conclusion.
---Select--- Do not reject Reject ( )  the null hypothesis and conclude that proportions  ( )are not are

as stated at M&M web site for the population from which these bags were drawn.

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Brown Red Yellow Blue Orange Green Total Observed 602
(0.290) 393
(0.189) 379
(0.183) 226
(0.109) 240
(0.116) 235
(0.113) 2075 Alleged
Proportions .30 .20 .20 .10 .10 .10

Explanation / Answer

Ans:

H0: p1 = 0.3 (brown), p2 =0.2 (red), p3 =0.2 (yellow), p4 = 0.1 (blue), p5 = 0.1 (orange), p6 =0.1 (green) (manufacturer's claim)

Ha: not all pi are as specified in H0.

Test statistic:

chi square score=15.3

Number of categories=6

Degree of freedom=6-1=5

p-value=CHIDIST(15.349,5)=0.009

Assume significance level=0.05

As,p-value<0.05,we reject null hypothesis

and conclude that proportions are not same as stated at M&M web site for the population from which these bags were drawn.

color Observed count(O) pi Expected count(E) (O-E)^2/E Brown 602 0.3 622.5 0.675 Red 393 0.2 415 1.166 Yellow 379 0.2 415 3.123 Blue 226 0.1 207.5 1.649 Orange 240 0.1 207.5 5.090 Green 235 0.1 207.5 3.645 Total 2075 1 2075 15.349

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