Question

A manufacturer of colored candies states that 13​% of the candies in a bag should be​ brown, 14​% ​yellow, 13​% ​red,24​% blue20​%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

alpha equalsα=0.05 level of significance. Using the level of significance

alpha equalsα=0.05, test whether the color distribution is the same.Click here to view the table.

Click here to view the table of critical values of the chi-square distribution

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

A.H0​:The distribution of colors is at least as uniform as stated by the manufacturer.

H1​:The distribution of colors is less uniform than stated by the manufacturer.

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.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.

D.H0​:The distribution of colors is at most as uniform as stated by the manufacturer.

H1​:The distribution of colors is more uniform than stated by the manufacturer.

Compute the expected counts for each color.

Color	Frequency	Expected Count
Brown	6161	nothing
Yellow	6666	nothing
Red	5555	nothing
Blue	6363	nothing
Orange	9393	nothing
Green	6666	nothing
		​(Round to two decimal places as​ needed.)

What is the test​ statistic?

chi Subscript 0 Superscript 2χ20	equals=	nothing
		​(Round to three decimal places as​ needed.)

What is the range of​ P-values of the​ test?

The range of​ P-values of the test is

▼

between 0.01 and 0.025.

between 0.025 and 0.05.

less than 0.01.

between 0.05 and 0.10.

greater than 0.10.

Based on the​ results, do the colors follow the same distribution as stated in the​ problem?

A.Do not reject Upper H 0Do not reject H0.

There insufficient sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.

B. Reject Upper H 0Reject H0. There is sufficient sufficientevidence that the distribution of colors is not the same as stated by the manufacturer.

C. Do not reject Upper H 0Do not reject H0.

There is not sufficient not sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.

D. Reject Upper H 0Reject H0.

There is not sufficientnot sufficient

evidence that the distribution of colors is not the same as stated by the manufacturer.

Answer 1

Solution:-

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

Null hypothesis: There is sufficient sufficient evidence that the distribution of colors is the same as stated by the manufacturer.

Alternative hypothesis: There is sufficient sufficient evidence that 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 = 16.827

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 16.827.

We use the Chi-Square Distribution Calculator to find P(X² > 16.827) = 0.0048.

The range of​ P-values of the test is less than 0.01.

Interpret results. Since the P-value (0.0048) is less than the significance level (0.05), we have to reject the null hypothesis.

B) Reject H₀. There is sufficient sufficient evidence that the distribution of colors is not the same as stated by the manufacturer.