What is the degree of freedom of the x² statistic? What is the P-value of the Independence test? (Round to 3 decimals) Given the significance level of 0.1, what can she conclude from the test? Zip code and diet are independent of one another. O Zip code and diet are dependent on one another.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.1: Measures Of Center
Problem 8PPS
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### Independence Test with Chi-Square Statistic

A researcher conducted an Independence test using data from two categorical variables: Zip Code and Diet. Her data is organized into a 5 by 4 contingency table. The calculated chi-square (\(\chi^2\)) test statistic is 11.25.

##### Degrees of Freedom Calculation
The degrees of freedom for a chi-square test on a contingency table is calculated using the formula:

\[ \text{Degrees of Freedom} = (r - 1) \times (c - 1) \]

where \(r\) is the number of rows and \(c\) is the number of columns. For this table:

\[ r = 5, \quad c = 4 \]
\[ \text{Degrees of Freedom} = (5 - 1) \times (4 - 1) = 4 \times 3 = 12 \]

The degree of freedom for the chi-square statistic is: {\input field}.

##### P-Value Calculation
The P-value for the Independence test is obtained from the chi-square distribution table corresponding to the degrees of freedom and the test statistic (\(\chi^2 = 11.25\)).

The P-value of the Independence test (rounded to 3 decimals) is: {\input field}.

##### Conclusion
With a significance level of \(0.1\), the conclusion of the test can be drawn based on the P-value:

- If the P-value \( < 0.1 \), we reject the null hypothesis and conclude that Zip code and diet are dependent on one another.
- If the P-value \( \geq 0.1 \), we do not reject the null hypothesis and conclude that Zip code and diet are independent of one another.

**Given the significance level of 0.1, what can she conclude from the test?**

- ☐ Zip code and diet are independent of one another.
- ☐ Zip code and diet are dependent on one another.

This exercise allows students to interpret the results of chi-square tests and make informed conclusions about the relationship between two categorical variables.
Transcribed Image Text:### Independence Test with Chi-Square Statistic A researcher conducted an Independence test using data from two categorical variables: Zip Code and Diet. Her data is organized into a 5 by 4 contingency table. The calculated chi-square (\(\chi^2\)) test statistic is 11.25. ##### Degrees of Freedom Calculation The degrees of freedom for a chi-square test on a contingency table is calculated using the formula: \[ \text{Degrees of Freedom} = (r - 1) \times (c - 1) \] where \(r\) is the number of rows and \(c\) is the number of columns. For this table: \[ r = 5, \quad c = 4 \] \[ \text{Degrees of Freedom} = (5 - 1) \times (4 - 1) = 4 \times 3 = 12 \] The degree of freedom for the chi-square statistic is: {\input field}. ##### P-Value Calculation The P-value for the Independence test is obtained from the chi-square distribution table corresponding to the degrees of freedom and the test statistic (\(\chi^2 = 11.25\)). The P-value of the Independence test (rounded to 3 decimals) is: {\input field}. ##### Conclusion With a significance level of \(0.1\), the conclusion of the test can be drawn based on the P-value: - If the P-value \( < 0.1 \), we reject the null hypothesis and conclude that Zip code and diet are dependent on one another. - If the P-value \( \geq 0.1 \), we do not reject the null hypothesis and conclude that Zip code and diet are independent of one another. **Given the significance level of 0.1, what can she conclude from the test?** - ☐ Zip code and diet are independent of one another. - ☐ Zip code and diet are dependent on one another. This exercise allows students to interpret the results of chi-square tests and make informed conclusions about the relationship between two categorical variables.
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