Math Problem Statement

Conduct the hypothesis test and provide the test statistic and the critical​ value, and state the conclusion.

A company claims that its packages of 100 candies are distributed with the following color​ percentages:

1515​%

​red,

2121​%

​orange,

1616​%

​yellow,

1010​%

​brown,

2323​%

​blue, and

1515​%

green. Use the given sample data to test the claim that the color distribution is as claimed. Use a

0.100.10

significance level.

Color Number_in_Package

Red 14

Orange 25

Yellow 10

Brown 6

Blue 27

Green 18

The test statistic is

enter your response here.

​(Round to two decimal places as​ needed.)

Part 2

The critical value is

enter your response here.

​(Round to three decimal places as​ needed.)

Part 3

State the conclusion.

▼   Do not reject

Reject

Upper H 0H0.

There

▼   is

is not

sufficient evidence to warrant rejection of the claim that the color distribution is as claimed.

Solution

To test the company's claim about the color distribution of candies, we will perform a chi-square goodness-of-fit test.

Step-by-Step Solution

1. State the hypotheses:

  • Null Hypothesis (H0H_0): The color distribution of the candies is as claimed by the company.
  • Alternative Hypothesis (H1H_1): The color distribution of the candies is not as claimed by the company.

2. Observed Frequencies:

The observed frequencies for each color are given as:

  • Red: ORed=14O_{Red} = 14
  • Orange: OOrange=25O_{Orange} = 25
  • Yellow: OYellow=10O_{Yellow} = 10
  • Brown: OBrown=6O_{Brown} = 6
  • Blue: OBlue=27O_{Blue} = 27
  • Green: OGreen=18O_{Green} = 18

3. Expected Frequencies:

The expected frequencies for each color, based on the company's claim, are:

  • Red: ERed=0.15×100=15E_{Red} = 0.15 \times 100 = 15
  • Orange: EOrange=0.21×100=21E_{Orange} = 0.21 \times 100 = 21
  • Yellow: EYellow=0.16×100=16E_{Yellow} = 0.16 \times 100 = 16
  • Brown: EBrown=0.10×100=10E_{Brown} = 0.10 \times 100 = 10
  • Blue: EBlue=0.23×100=23E_{Blue} = 0.23 \times 100 = 23
  • Green: EGreen=0.15×100=15E_{Green} = 0.15 \times 100 = 15

4. Calculate the chi-square test statistic:

The chi-square test statistic is calculated using the formula: χ2=(OiEi)2Ei\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} Where OiO_i are the observed frequencies and EiE_i are the expected frequencies.

Calculations:

\chi^2 &= \frac{(14 - 15)^2}{15} + \frac{(25 - 21)^2}{21} + \frac{(10 - 16)^2}{16} + \frac{(6 - 10)^2}{10} + \frac{(27 - 23)^2}{23} + \frac{(18 - 15)^2}{15} \\ &= \frac{1}{15} + \frac{16}{21} + \frac{36}{16} + \frac{16}{10} + \frac{16}{23} + \frac{9}{15} \\ &= 0.067 + 0.762 + 2.250 + 1.600 + 0.696 + 0.600 \\ &= 5.975 \end{align*}$$ The test statistic is $$\chi^2 = 5.975$$. #### 5. Determine the critical value: For a chi-square test with $$k - 1$$ degrees of freedom (where $$k$$ is the number of categories), and a significance level of 0.10, we look up the critical value in the chi-square distribution table. Degrees of freedom ($$df$$) = $$6 - 1 = 5$$ Using the chi-square distribution table, the critical value for $$\alpha = 0.10$$ and $$df = 5$$ is approximately $$\chi^2_{crit} = 9.236$$. #### 6. Compare the test statistic to the critical value and make a conclusion: - If $$\chi^2 > \chi^2_{crit}$$, we reject the null hypothesis. - If $$\chi^2 \leq \chi^2_{crit}$$, we do not reject the null hypothesis. Since $$\chi^2 = 5.975$$ is less than $$\chi^2_{crit} = 9.236$$, we do not reject the null hypothesis. #### Conclusion: There is not sufficient evidence to warrant rejection of the claim that the color distribution is as claimed. ### Summary 1. **Test Statistic**: $$\chi^2 = 5.975$$ (rounded to two decimal places) 2. **Critical Value**: $$\chi^2_{crit} = 9.236$$ (rounded to three decimal places) 3. **Conclusion**: Do not reject $$H_0$$. There is not sufficient evidence to warrant rejection of the claim that the color distribution is as claimed. --- Do you want details or have any questions? Here are some questions you can ask: 1. How do I calculate the chi-square test statistic? 2. What is the significance level in hypothesis testing? 3. How do I interpret the chi-square distribution table? 4. What are the steps to perform a chi-square goodness-of-fit test? 5. Can you explain the degrees of freedom in this context? **Tip**: When conducting a chi-square goodness-of-fit test, ensure that the expected frequencies are not too small. Ideally, they should be at least 5 in each category to apply the chi-square test reliably.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Chi-square Test
Goodness-of-Fit Test

Formulas

Chi-square Test Statistic

Theorems

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Suitable Grade Level

Advanced Undergraduate