A manufacturer of colored chocolate candies specifies the proportion for each co

Question

A manufacturer of colored chocolate candies specifies the proportion for each color on its website. A sample of randomly selected 109 candies was taken, with the following result:

(a) Which hypotheses should be used to test if the sample is consistent with the company's specifications:
H0: --- p1 = 0.25, p2 = 0.12, p3 = 0.17, p4 = 0.19, p5 = 0.14, p6 = 0.13

p1 = p2 = p3 = p4 = p5 = p6 = 0 p1 = p2 = p3 = p4 = p5 = p6 = 1/6

p1 = 15, p2 = 21, p3 = 9, p4 = 13, p5 = 23, p6 = 28 ,

implying the sample is  --- (inconsistent/ consistent) ---- with the company's specifications.
Ha: --- (at least one of the p's is different from the specified values /all of the p's are different from the specified values)--- , implying the sample is  --- (inconsistent /consistent)-- with the company's specifications.

(b) Calculate the expected count for each color: (use 2 decimals.)
Blue:  
Brown:  
Green:  
Orange:  
Red:  
Yellow:  

(c) Are the conditions met for doing a chi-squared test? (check box)

None of the conditions are met.

It it reasonable to assume the candies are independent of each other.

The sample was randomly selected.

The expected count is at least 5 for every cell.

(d) How many degrees of freedom are there?  

(e) Calculate the contribution to the test statistic for Blue: (use 2 decimals.)  

(f) For this sample, the P-value is 0.001. What is the conclusion of the hypothesis test, for = 0.05?
--- (Reject H0/ Do not reject H0) .There is  ---( sufficient /insufficient )evidence the sample is  --- (consistent/ inconsistent) with the company's specifications.

Color Blue Brown Green Orange Red Yellow Specified % 25% 12% 17% 19% 14% 13% Observed count 15 21 9 13 23 28

Explanation / Answer

a) Ho:p1 = 0.25, p2 = 0.12, p3 = 0.17, p4 = 0.19, p5 = 0.14, p6 = 0.13

implying the sample is  consistent with the company's specifications.

Ha: at least one of the p's is different from the specified values  implying the sample is  inconsistent  with the company's specifications.

b)

applying chi square test of goodness of fit:

expected count are given column under that heading

c)

The expected count is at least 5 for every cell.

d) degrees of freedom =categories-1 =6-1=5

e) from above able contribution to the test statistic for Blue =5.51

f) Reject H0 There is sufficient evidence the sample is consistent with the company's specifications

observed Expected Chi square category Probability O E=total*p =(O-E)^2/E blue 0.250 15.000 27.25 5.51 brown 0.120 21.000 13.08 4.80 green 0.170 9.000 18.53 4.90 Orange 0.190 13.000 20.71 2.87 red 0.140 23.000 15.26 3.93 Yellow 0.130 28.000 14.17 13.50 1 109 109 35.4980

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