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.

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 (\(H_0\))**: The color distribution of the candies is as claimed by the company.
- **Alternative Hypothesis (\(H_1\))**: 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: \( O_{Red} = 14 \)
- Orange: \( O_{Orange} = 25 \)
- Yellow: \( O_{Yellow} = 10 \)
- Brown: \( O_{Brown} = 6 \)
- Blue: \( O_{Blue} = 27 \)
- Green: \( O_{Green} = 18 \)

#### 3. Expected Frequencies:
The expected frequencies for each color, based on the company's claim, are:
- Red: \( E_{Red} = 0.15 \times 100 = 15 \)
- Orange: \( E_{Orange} = 0.21 \times 100 = 21 \)
- Yellow: \( E_{Yellow} = 0.16 \times 100 = 16 \)
- Brown: \( E_{Brown} = 0.10 \times 100 = 10 \)
- Blue: \( E_{Blue} = 0.23 \times 100 = 23 \)
- Green: \( E_{Green} = 0.15 \times 100 = 15 \)

#### 4. Calculate the chi-square test statistic:
The chi-square test statistic is calculated using the formula:
\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]
Where \( O_i \) are the observed frequencies and \( E_i \) are the expected frequencies.

Calculations:
\[
\begin{align*}
\chi^2 &= \frac{(14 - 15)^2}{15} + \frac{(25 - 21)^2}{21} + \frac{(10 - 16)^2}{16} + \frac{(6 - 10)^2}{10} + \frac{(27 - 23)^2}{23} + \frac{(18 - 15)^2}{15} \\
&= \frac{1}{15} + \frac{16}{21} + \frac{36}{16} + \frac{16}{10} + \frac{16}{23} + \frac{9}{15} \\
&= 0.067 + 0.762 + 2.250 + 1.600 + 0.696 + 0.600 \\
&= 5.975
\end{align*}
\]

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.

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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.

Learn how to conduct a hypothesis test on candy color distribution using the chi-square test. Calculate the test statistic and critical value, and determine the conclusion. Detailed solution with observed and expected frequencies provided.

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