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 o 0.05 level of significance EEB Click the icon to view the table Determine the null and alternative hypotheses. Choose the correct answer below. OA Ho: The distribution of colors is the same as stated by the manufacturer B. Ho: The distribution of colors is not the same as stated by the manufacturer ° C. None of these Compute the expected counts for each color H: The distribution of colors is not the same as stated by the manufacturer H: The distribution of colors is the same as stated by the manufacturer. ColorObserved Count Expected Count Brown Yellow 60 67 56 62 94 63 Red Blue Orange Green (Round to two decimal places as needed.) What is the test statistic? (Round to three decimal places as needed.) What is the P-value of the test? P-value(Round to three decimal places as needed.) Based on the results, do the colors follow the same distribution as stated in the problem? Do not reject Ho. There is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer A. 0 B. Reject Ho-There is not sufficient evidence that the distribution of colors is not the same as stated by the manufacturer ° C. Do not reject Ho. There is not sufficient evidence that the distribution of colors is not the same as stated by the manufacturer O D. Reject Ho. There is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.

Answer 1

SOLUTION:

From given data,

The null and alternative hypotheses.

$H_{0}$ : The distribution of colors is the same as stated by the manufacturer.

$H_{1}$ : The distribution of colors is not the same as stated by the manufacturer.

The expected counts for each color.

Color	Observed count (O)	claimed prop	Expected count (E)	(O-E)2 / E
Brown	60	0.13	402*0.13=52.26	(60-52.26)2 / 52.26 =1.146337
Yellow	67	0.14	402*0.14=56.28	(67-56.28)2 / 56.28 = 2.041904
Red	56	0.13	402*0.13=52.26	(56-52.26)2 / 52.26 = 0.267654
Blue	62	0.24	402*0.24=96.48	(62-96.48)2 / 96.48 = 12.322454
Orange	94	0.20	402*0.20=80.4	(94-80.4)2 / 80.4 = 2.300497
Green	63	0.16	402*0.16=64.32	(63-64.32)2 / 64.32 = 0.027089
Total	402		402	18.105935

Test statistics:

The chi-squared statistic is computed as follows:

$\chi ^{2}$ = $\Sigma (O-E)^{2} /E$ (From above table)

$\chi ^{2}$ =1.146337+2.041904+0.267654+12.322454+2.300497+0.027089

$\chi ^{2}$ = 18.105935

$\chi ^{2}$$\approx$ 18.106

P-value :

Degree of freedom

df =n-1=6-1 = 5

P-value at df=5 and  $\chi ^{2}$ = 18.1059

P-value = 0.0028

Significance level = $\alpha$ = 0.05

Since ,

P-value < $\alpha$

Hence we reject the null hypothesis , there is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer

Answer: option(D) is correct.

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