Question

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Amanufacturer 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 a= 0.05 level of significance. Click the icon to view the table. Determine the null and alternative hypotheses. Choose the correct answer below. Observed Distribution of Colors - X O A. Ho: The distribution of colors is the same as stated by the manufacturer. H: The distribution of colors is not the same as stated by the manufacturer. OB. Ho: 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 O C. None of these Color Frequency Claimed Proportion Colored Candies in a bag Brown Yellow Red Blue 62 67 54 61 0.13 0.14 0.13 0.24 Orange Green 85 63 0.20 0.16 Print Done Compute the expected counts for each color. Color Frequency Expected Count Brown 62 Yellow 67 Red 54 Blue 61 Orange 85 Green 63 (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? O A. Do not reject Ho. There is not sufficient evidence that the distribution of colors is not the same as stated by the manufacturer. O B. Do not reject Ho. There is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer oc. Reject Ho. There is not sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.

Answer 1

Null and alternative hypotrhesis are

Ho: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.

Category	Pi	Oi=frequency	Ei=Expected	R2i=(Oi-Ei)2/Ei
Brown	0.13	62	50.96	2.391711146
Yellow	0.14	67	54.88	2.67664723
Red	0.13	54	50.96	0.181350078
Blue	0.24	61	94.08	11.63144558
Orange	0.2	85	78.4	0.555612245
Green	0.16	63	62.72	0.00125
total	1	392	392	17.43801628

#test stattistics

#

#$\chi^2_0=\sum(Oi-Ei)^2/Ei$=17.4381

$\chi^2_0=17.438$

P-value=P($\chi^2_0$>17.438)=0.00374

from excel: chidist(17.438,5)

#P-value=0.004

P-value<0.05 hence we reject Ho

Conclusion:

Option c is correct