13. The color distribution of plain M&M;\'s varies by the factory in which they

Question

13. The color distribution of plain M&M;'s varies by the factory in which they were made. The Hackettstown, New Jersey plant uses the following color distribution for plain M&M;'s: 12.5% red, 25% orange, 12.5% yellow, 12.5% green, 25% blue, and 12.5% brown. Each piece ofcandy in a random sample of 100 plain M&M;'s from the Hackettstown factory was classified according to color, and the results are listed below. Use a 0.05 significance level to test the claim that the Hackettstown factory color distribution is correct. Describe method used for calculating answer Color Number Orange 28 Yellow 20 Red Green ue Brown 20 12 (a) Identify the appropriate hypothesis test and explain the reasons why it is appropriate for analyzing this data. (b) Identify the null hypothesis and the alternative hypothesis. (c) Determine the test statistic. (Round your answer to two decimal places) (d) Determine the p-value. (Round your answer to two decimal places) (e) Compare p-value and significance level a. What decision should be made regarding the null hypothesis (e.g., reject or fail to reject) and why? (f) Is there sufficient evidence to support the claim that the Hackettstown factory color distribution is correct? Justify your answer

Explanation / Answer

a) we use chi square test of fit

as we want to if given data fits the proportion under null hypothesis

b)

1 = red , 2 = orange,..6 = brown

null : p1 = 0.125,p2 = 0.25 ,...p6 = 0.125

alternate : not Ho

c)

TS = 19.81602374= 19.81

d)

p-value = 0.001353062 = 0.00

e)

p-value< 0.05

we reject the null hypothesis

f) No, there is not sufficient evidence that distribution is correct as we reject tht null

but there is sufficient evidence the distribution is not correct

p Oi Ei (Oi-Ei)^2/Ei 0.125 88 84.25 0.166913947 0.25 154 168.5 1.247774481 0.125 83 84.25 0.018545994 0.125 115 84.25 11.22329377 0.25 138 168.5 5.520771513 0.125 96 84.25 1.638724036 1 674 674 TS 19.81602374 5% critical value 11.07049769 p-value 0.001353062

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