c. The test-statistic for this data is x . (Please show your answer to three decimal %3D places.) d. The p-value for this sample is (Please show your answer to four decimal places.) e. The p-value is (Select an answer a.
c. The test-statistic for this data is x . (Please show your answer to three decimal %3D places.) d. The p-value for this sample is (Please show your answer to four decimal places.) e. The p-value is (Select an answer a.
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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I need help with parts c-e
Transcribed Image Text:## Research Study on Living Situation and Pet Ownership
A researcher is interested in investigating whether the living situation and pet ownership are dependent. The table below shows the results of a survey:
### Frequencies of Living Situation and Pet Ownership
| Living Situation | Dog | Cat | Various Pets | None |
|------------------|-----|-----|--------------|------|
| Single | 83 | 118 | 38 | 50 |
| Family | 93 | 86 | 55 | 58 |
| Couple | 94 | 65 | 45 | 67 |
### Analysis Questions
1. **What can be concluded at the \( \alpha = 0.05 \) significance level?**
2. **a. What is the correct statistical test to use?**
- ☐ Homogeneity
- ☐ Goodness-of-Fit
- ☐ Independence
- ☐ Paired \( t \)-test
3. **b. What are the null and alternative hypotheses?**
**\( H_0 \):**
- ☐ Pet ownership and living situation are independent.
- ☐ Pet ownership and living situation are dependent.
- ☐ The distribution of pets is the same for each living situation.
- ☐ The distribution of pet ownership varies across living situations.
From the table and these questions, students can determine the appropriate statistical methods for testing relationships between categorical variables and develop hypotheses accordingly.
![### Chi-Square Test for Independence
This section provides an example of how to conduct a chi-square test for independence. Follow the steps outlined and fill in your calculations as required.
1. **Hypotheses Formulation:**
- \(\textbf{H}_0\): Null Hypothesis
- \(\underline{\hspace{1cm}}\) The distribution of pets is not the same for each living situation.
- \(\underline{\hspace{1cm}}\) Pet ownership and living situation are dependent.
- \(\textbf{H}_0\): Alternate Hypothesis
- \(\underline{\hspace{1cm}}\) The distribution of pets is the same for each living situation.
- \(\underline{\hspace{1cm}}\) Pet ownership and living situation are independent.
2. **Calculate the Chi-Square Test Statistic:**
- The test-statistic for this data is \(\chi^2 = \underline{\hspace{3cm}} \)
(Please show your answer to three decimal places.)
3. **Determine the p-value:**
- The p-value for this sample is \( \underline{\hspace{3cm}}\)
(Please show your answer to four decimal places.)
4. **Compare the p-value to the significance level (\(\alpha\)):**
- The p-value is \( \underline{\hspace{3cm}}\) \( \alpha \).
\( \bullet \) [Select an answer from the dropdown menu provided]
5. **Decision Rule:**
- Based on this, we should:
- \(\underline{\hspace{1cm}}\) accept the null.
- \(\underline{\hspace{1cm}}\) reject the null.
- \(\underline{\hspace{1cm}}\) fail to reject the null.
6. **Conclusive Statement:**
- Therefore, the final conclusion is:
- \(\underline{\hspace{1cm}}\) There is insufficient evidence to conclude that pet ownership and living situation are dependent.
- \(\underline{\hspace{1cm}}\) There is sufficient evidence to conclude that the distribution of pets is not the same for each living situation.
- \(\underline{\hspace{1cm}}\) There is sufficient evidence to conclude that pet ownership and living situation are independent.
- \(\underline{\hspace{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd16ddfa-c201-4e3f-90ee-e0e9ce8bf5de%2Fd5051f44-b5a9-425a-902a-f93fdb7003a3%2F6oxl3v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chi-Square Test for Independence
This section provides an example of how to conduct a chi-square test for independence. Follow the steps outlined and fill in your calculations as required.
1. **Hypotheses Formulation:**
- \(\textbf{H}_0\): Null Hypothesis
- \(\underline{\hspace{1cm}}\) The distribution of pets is not the same for each living situation.
- \(\underline{\hspace{1cm}}\) Pet ownership and living situation are dependent.
- \(\textbf{H}_0\): Alternate Hypothesis
- \(\underline{\hspace{1cm}}\) The distribution of pets is the same for each living situation.
- \(\underline{\hspace{1cm}}\) Pet ownership and living situation are independent.
2. **Calculate the Chi-Square Test Statistic:**
- The test-statistic for this data is \(\chi^2 = \underline{\hspace{3cm}} \)
(Please show your answer to three decimal places.)
3. **Determine the p-value:**
- The p-value for this sample is \( \underline{\hspace{3cm}}\)
(Please show your answer to four decimal places.)
4. **Compare the p-value to the significance level (\(\alpha\)):**
- The p-value is \( \underline{\hspace{3cm}}\) \( \alpha \).
\( \bullet \) [Select an answer from the dropdown menu provided]
5. **Decision Rule:**
- Based on this, we should:
- \(\underline{\hspace{1cm}}\) accept the null.
- \(\underline{\hspace{1cm}}\) reject the null.
- \(\underline{\hspace{1cm}}\) fail to reject the null.
6. **Conclusive Statement:**
- Therefore, the final conclusion is:
- \(\underline{\hspace{1cm}}\) There is insufficient evidence to conclude that pet ownership and living situation are dependent.
- \(\underline{\hspace{1cm}}\) There is sufficient evidence to conclude that the distribution of pets is not the same for each living situation.
- \(\underline{\hspace{1cm}}\) There is sufficient evidence to conclude that pet ownership and living situation are independent.
- \(\underline{\hspace{
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