Question

anufacturer 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 random y selected a bag of colored candies. He counted the numbe 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. 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 not the same as stated by the manufacturer B. Hj : The distribution of colors is the same as stated by the manufacturer H,: The distribution of colors is the same as stated by the manufacturer Observed Distribution of Colors H1: The distribution of colors is not the same as stated by the manufacturer C. None of these. Colored Candies in a ba BrownYellow RedBlue Or Compute the expected counts for each color olor ange Green 92 65 0.13 0.14 0.130.24 0.20 0.16 60 63 57 60 Frequen Claimed Color Brown Yellow Red Blue Orange Green Count ortion 60 63 57 60 92 65 Print Done Round to two decimal places as needed.)

Answer 1

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The distribution of colors is same as stated by the manufacturer.

Alternative hypothesis: The distribution of colors is not same as stated by the manufacturer.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 6 - 1
D.F = 5
(E_i) = n * p_i

$\chi ^{2}=\sum \left [\frac{(O_{r,c}-E_{r,c})^{2}}{E_{r,c}} \right ]$
X² = 18.015

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 5 degrees of freedom is more extreme than 19.58.

We use the Chi-Square Distribution Calculator to find P(X² > 18.015) = 0.00293.

Interpret results. Since the P-value (0.00293) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we do not have sufficient evidence in the favour of the claim that he distribution of colors is same as stated by the manufacturer.