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 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 25, 28, 50, 39, 27, 31. Use a 0.01 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. (...) The test statistic is (Round to three decimal places as needed.) The critical value is (Round to three decimal places as needed.) State the conclusion. Ho. There to be equally likely, sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes to behave differently from a fair die. so the loaded die - X Chi-square distribution table Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 1 0.001 0.004 0.016 2.706 2 0.010 0.020 0.051 0.103 0.211 4.605 3 0.072 0.115 0.216 0.352 0.584 6.251 4 0.207 0.297 0.484 0.711 1.064 7.779 5 0.412 0.554 0.831 1.145 1.610 9.236 6 1.635 2.204 10.645 7 2.167 2.833 12.017 8 0.676 0.872 1.237 0.989 1.239 1.690 1.344 1.646 2.180 1.735 2.088 2.700 3.325 2.156 2.558 3.247 3.940 2.733 3.490 13.362 9 4.168 14.684 10 4.865 15.987
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 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 25, 28, 50, 39, 27, 31. Use a 0.01 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. (...) The test statistic is (Round to three decimal places as needed.) The critical value is (Round to three decimal places as needed.) State the conclusion. Ho. There to be equally likely, sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes to behave differently from a fair die. so the loaded die - X Chi-square distribution table Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 1 0.001 0.004 0.016 2.706 2 0.010 0.020 0.051 0.103 0.211 4.605 3 0.072 0.115 0.216 0.352 0.584 6.251 4 0.207 0.297 0.484 0.711 1.064 7.779 5 0.412 0.554 0.831 1.145 1.610 9.236 6 1.635 2.204 10.645 7 2.167 2.833 12.017 8 0.676 0.872 1.237 0.989 1.239 1.690 1.344 1.646 2.180 1.735 2.088 2.700 3.325 2.156 2.558 3.247 3.940 2.733 3.490 13.362 9 4.168 14.684 10 4.865 15.987
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Conducting a Chi-Square Hypothesis Test for a Loaded Die
**Objective**: Conduct the hypothesis test and provide the test statistic and the critical value, and then state the conclusion.
**Scenario**: A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 25, 28, 50, 39, 27, 31. Use a 0.01 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?
**Step-by-Step Solution Process**:
#### Step 1: State the Observed Frequencies
```
Outcome: 1, 2, 3, 4, 5, 6
Frequency: 25, 28, 50, 39, 27, 31
Total: 200 rolls
```
#### Step 2: Calculate the Expected Frequencies
For a fair die, the expected frequency for each outcome is 200 rolls / 6 sides = 33.33 rolls per outcome.
#### Step 3: State the Hypotheses
- Null Hypothesis (H₀): The die is fair (Outcomes are equally likely).
- Alternative Hypothesis (H₁): The die is not fair (Outcomes are not equally likely).
#### Step 4: Calculate the Chi-Square Test Statistic
The formula for the chi-square test statistic is:
\[ \chi² = \sum \frac{(O_i - E_i)²}{E_i} \]
where \( O_i \) represents observed frequencies and \( E_i \) represents expected frequencies.
#### Step 5: Obtain the Critical Value
Using a 0.01 significance level and degrees of freedom (df) = 5 (number of outcomes - 1), use the chi-square distribution table to find the critical value.
#### Completing the Template:
1. **Insert the Test Statistic and Critical Value**:
- The test statistic is: [value].
- The critical value is: [value from chi-square table].
2. **State the Conclusion**:
- Compare the test statistic to the critical value to determine whether to reject the null hypothesis.
Here is the Chi-S](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e2dce9f-bcdf-4f32-a081-371e2dab1fcb%2F4b7ed4a1-a327-490e-a7de-2d43be691228%2F0trfqqa_processed.png&w=3840&q=75)
Transcribed Image Text:### Conducting a Chi-Square Hypothesis Test for a Loaded Die
**Objective**: Conduct the hypothesis test and provide the test statistic and the critical value, and then state the conclusion.
**Scenario**: A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 25, 28, 50, 39, 27, 31. Use a 0.01 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?
**Step-by-Step Solution Process**:
#### Step 1: State the Observed Frequencies
```
Outcome: 1, 2, 3, 4, 5, 6
Frequency: 25, 28, 50, 39, 27, 31
Total: 200 rolls
```
#### Step 2: Calculate the Expected Frequencies
For a fair die, the expected frequency for each outcome is 200 rolls / 6 sides = 33.33 rolls per outcome.
#### Step 3: State the Hypotheses
- Null Hypothesis (H₀): The die is fair (Outcomes are equally likely).
- Alternative Hypothesis (H₁): The die is not fair (Outcomes are not equally likely).
#### Step 4: Calculate the Chi-Square Test Statistic
The formula for the chi-square test statistic is:
\[ \chi² = \sum \frac{(O_i - E_i)²}{E_i} \]
where \( O_i \) represents observed frequencies and \( E_i \) represents expected frequencies.
#### Step 5: Obtain the Critical Value
Using a 0.01 significance level and degrees of freedom (df) = 5 (number of outcomes - 1), use the chi-square distribution table to find the critical value.
#### Completing the Template:
1. **Insert the Test Statistic and Critical Value**:
- The test statistic is: [value].
- The critical value is: [value from chi-square table].
2. **State the Conclusion**:
- Compare the test statistic to the critical value to determine whether to reject the null hypothesis.
Here is the Chi-S
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