Question

manufacturer of colored candies states that

1313​%

of the candies in a bag should be​ brown,

1414​%

​yellow,

1313​%

​red,

2424​%

​blue,

2020​%

​orange, and

1616​%

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 equalsα=0.050.05

level of significance.

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

A.

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.This is the correct answer.

B.

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.

C.

None of these.

Your answer is not correct.

Compute the expected counts for each color.

Color	Frequency	Expected Count
Brown	5959	49.66 49.66
Yellow	6464	53.48 53.48
Red	5555	49.66 49.66
Blue	6262	91.68 91.68
Orange	7878	76.40 76.40
Green	6464	61.12 61.12
		​(Round to two decimal places as​ needed.)

What is the test​ statistic?

chi Subscript 0 Superscript 2χ20

equals=

Answer 1

Solution:-

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

A)

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.

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


X² = 14.18

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 14.18.

We use the Chi-Square Distribution Calculator to find P(X² > 14.18) = 0.015.

Interpret results. Since the P-value (0.015) is less than the significance level (0.05), we have to reject the null hypothesis.