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

A person drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 

200200 

times. Here are the observed frequencies for the outcomes of​ 1, 2,​ 3, 4,​ 5, and​ 6, respectively: 

2727​, 

3232​, 

4444​, 

3737​, 

2626​, 

3434. 

Use a 

0.0250.025 

significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair​ die?

Click here to view the chi-square distribution table.

LOADING...

Question content area bottom

Part 1

The test statistic is 

enter your response here. 

​(Round to three 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 not

is
 
sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes 

▼
 
appear

do not appear
 
to be equally​ likely, so the loaded die 

▼
 
does not appear

appears
 
to behave differently from a fair die.

To test the hypothesis that the outcomes of the loaded die are not equally likely, we will perform a chi-square goodness-of-fit test.

### Step-by-Step Solution

#### 1. State the hypotheses:
- **Null Hypothesis (\(H_0\))**: The outcomes of the die are equally likely.
- **Alternative Hypothesis (\(H_1\))**: The outcomes of the die are not equally likely.

#### 2. Observed Frequencies:
The observed frequencies for the outcomes 1, 2, 3, 4, 5, and 6 are given as:
- \( O_1 = 27 \)
- \( O_2 = 32 \)
- \( O_3 = 44 \)
- \( O_4 = 37 \)
- \( O_5 = 26 \)
- \( O_6 = 34 \)

#### 3. Expected Frequencies:
If the die is fair, each outcome should appear with equal probability. For 200 rolls, the expected frequency for each outcome is:
\[ E_i = \frac{200}{6} \approx 33.33 \]

#### 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:
\[ \chi^2 = \frac{(27 - 33.33)^2}{33.33} + \frac{(32 - 33.33)^2}{33.33} + \frac{(44 - 33.33)^2}{33.33} + \frac{(37 - 33.33)^2}{33.33} + \frac{(26 - 33.33)^2}{33.33} + \frac{(34 - 33.33)^2}{33.33} \]

\[
\begin{align*}
\chi^2 &= \frac{(-6.33)^2}{33.33} + \frac{(-1.33)^2}{33.33} + \frac{(10.67)^2}{33.33} + \frac{(3.67)^2}{33.33} + \frac{(-7.33)^2}{33.33} + \frac{(0.67)^2}{33.33} \\
&= \frac{40.0689}{33.33} + \frac{1.7689}{33.33} + \frac{113.6689}{33.33} + \frac{13.4689}{33.33} + \frac{53.7289}{33.33} + \frac{0.4489}{33.33} \\
&= 1.202 + 0.053 + 3.411 + 0.404 + 1.612 + 0.013 \\
&= 6.695
\end{align*}
\]

The test statistic is \( \chi^2 = 6.695 \).

#### 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 0.025, 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 \( \alpha = 0.025 \) and \( df = 5 \) is approximately \( \chi^2_{crit} = 12.833 \).

#### 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 \( \chi^2 \) ≤ \( \chi^2_{crit} \), we do not reject the null hypothesis.

Since \( \chi^2 = 6.695 \) is less than \( \chi^2_{crit} = 12.833 \), we do not reject the null hypothesis.

#### Conclusion:
There is not sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes appear to be equally likely, so the loaded die does not appear to behave differently from a fair die.

### Summary
1. **Test Statistic**: \( \chi^2 = 6.695 \) (rounded to three decimal places)
2. **Critical Value**: \( \chi^2_{crit} = 12.833 \) (rounded to three decimal places)
3. **Conclusion**: Do not reject \( H_0 \). There is not sufficient evidence to support the claim that the outcomes are not equally likely. The loaded die does not appear to behave differently from a fair die.

---

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**: Always ensure that your observed and expected frequencies are based on the same total number of trials when conducting chi-square tests.

Conduct a hypothesis test using the chi-square test to analyze the outcomes of a loaded die rolled 200 times. Calculate the test statistic, critical value, and state the conclusion. Detailed step-by-step solution with observed frequencies and expected frequencies provided.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation