Question

A manufacturer of colored candies states that 13​% of the candies in a bag should be​ brown, 14​% ​yellow, 13​% ​red, 24​% ​blue, 20​% ​orange, and 16​% green. A student randomly selected a bag of colored candies. He counted the number of candies of each color and obtained the results shown in the table. Test whether the bag of colored candies follows the distribution stated above at the α=0.05 level of significance.

Colored Candies in a bag

Color	Brown	Yellow	Red	Blue	Orange	Green
Frequency	61	63	53	63	82	65
Claimed Proportion	0.13	0.14	0.13	0.24	0.20	0.16

1. Determine the null and alternative hypotheses. Choose the correct answer below.

A .H0​: The distribution of colors is not the same as stated by the manufacturer.

H1​: The distribution of colors is the same as stated by the manufacturer.

B. H0​: The distribution of colors is the same as stated by the manufacturer.

H1​: The distribution of colors is not the same as stated by the manufacturer.

C . None of these.

2. Compute the expected counts for each color.

Color	Frequency	Expected Count
Brown	61
Yellow	63
Red	53
Blue	63
Orange	82
Green	65
		​(Round to two decimal places as​ needed.)

3. What is the test​ statistic?_______(Round to three decimal places as​ needed.)

4. ​P-value=_______What is the​ P-value of the​ test?

​(Round to three decimal places as​ needed.)

5. Based on the​ results, do the colors follow the same distribution as stated in the​ problem?

A. Reject H0. There is not sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.

B. Do not reject H0. There is not sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.

C. Reject H0. There is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.

D. Do not reject H0. There is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.

Answer 1

1)

B. H0​: The distribution of colors is the same as stated by the manufacturer.

H1​: The distribution of colors is not the same as stated by the manufacturer.

2)

  
expected frequncy,E = expected proportions*total frequency  
total frequency=   387

category	observed frequencey, O	expected proportion	expected frequency,E			(O-E)²/E
brown	61	0.130	50.31			2.271
yellow	63	0.140	54.18			1.436
red	53	0.130	50.31			0.144
blue	63	0.240	92.88			9.613
orange	82	0.200	77.40			0.273
green	65	0.160	61.92			0.153

3)

chi square test statistic,X² = Σ(O-E)²/E =   13.890

4)

level of significance, α=   0.05              
Degree of freedom=k-1=   6   -   1   =   5
                  
P value =   0.016 [ excel function: =chisq.dist.rt(test-stat,df) ]  

5)

  
Decision: P value < α, Reject Ho                  
C. Reject H0. There is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.