A manufacturer of colored candies states that 13% of the candies in a bag should

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 alpha = 0.05 level of significance. Determine the null and alternative hypotheses. Choose the correct answer below. H_0: The distribution of colors is not the same as stated by the manufacturer. H_1: The distribution of colors is the same as stated by the manufacturer. 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. None of these. 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 alpha = 0.05 level of significance. 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 H_0. There is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer. Do not reject H0. There is not sufficient evidence that the distribution of colors is not the same as stated by the manufacturer. Reject H_0. There is sufficient evidence that the distribution of colors is not the same as stated by the manufacturer. Reject H_0. There is not sufficient evidence that the distribution of colors is not the same as stated by the manufacturer

Explanation / Answer

Null and alternative hypotheses: Option B.

Expected frequency

52.26 [402*0.13], 56.28 (402*0.14), 52.26, 96.48, 80.40 and 64.32.

X^2=Summation (Observed-Expected)^2/Expected

=(60-52.26)^/52.26+...+(64-64.32)^2/64.32

=21.044

at df=5, the p value is 0.001

Conclusion Option c) [p value is less than alpha=0.05]

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