Does favorite food depend on your gender? A survey asks a group of college students about their favorite foods. It also groups the students by their gender. The data are recorded in the contingency table below, and a chi-square Test of Independence at the [tex]$5 \%$[/tex] significance level is performed.

\begin{tabular}{|c|c|c|c|c|}
\hline
& Sushi & Pizza & Hamburgers & Row Total \\
\hline
Men & 5 & 24 & 21 & 50 \\
\hline
Women & 17 & 21 & 8 & 46 \\
\hline
Column Total & 22 & 45 & 29 & 96 \\
\hline
\end{tabular}

(a) The null and alternative hypotheses are:
- [tex]$H_0$[/tex]: The two variables are independent, so gender does not affect favorite food.
- [tex]$H_a$[/tex]: The two variables are dependent, so gender does affect favorite food.

(b) Compute the test statistic, rounded to one decimal place. (Use expected frequencies that are also rounded to one decimal place.)



Answer :

To determine whether favorite food depends on gender, we will perform a Chi-Square Test of Independence using the provided contingency table. We'll go through the following systematic steps:

1. Formulate the Hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): Gender and favorite food are independent.
- Alternative hypothesis ([tex]\(H_a\)[/tex]): Gender and favorite food are dependent.

2. Construct the Contingency Table:

[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & \text{Sushi} & \text{Pizza} & \text{Hamburgers} & \text{Row Total} \\ \hline \text{Men} & 5 & 24 & 21 & 50 \\ \hline \text{Women} & 17 & 21 & 8 & 46 \\ \hline \text{Column Total} & 22 & 45 & 29 & 96 \\ \hline \end{array} \][/tex]

3. Compute the Expected Frequencies:
- The expected frequency for each cell is calculated using the formula:
[tex]\[ E_{ij} = \frac{(\text{Row Total for row } i) \times (\text{Column Total for column } j)}{\text{Grand Total}} \][/tex]

Calculating the expected frequencies for each cell:
- Expected frequency for Men who prefer Sushi ([tex]\(E_{11}\)[/tex]):
[tex]\[ E_{11} = \frac{50 \times 22}{96} \approx 11.5 \][/tex]
- Expected frequency for Men who prefer Pizza ([tex]\(E_{12}\)[/tex]):
[tex]\[ E_{12} = \frac{50 \times 45}{96} \approx 23.4 \][/tex]
- Expected frequency for Men who prefer Hamburgers ([tex]\(E_{13}\)[/tex]):
[tex]\[ E_{13} = \frac{50 \times 29}{96} \approx 15.1 \][/tex]
- Expected frequency for Women who prefer Sushi ([tex]\(E_{21}\)[/tex]):
[tex]\[ E_{21} = \frac{46 \times 22}{96} \approx 10.5 \][/tex]
- Expected frequency for Women who prefer Pizza ([tex]\(E_{22}\)[/tex]):
[tex]\[ E_{22} = \frac{46 \times 45}{96} \approx 21.6 \][/tex]
- Expected frequency for Women who prefer Hamburgers ([tex]\(E_{23}\)[/tex]):
[tex]\[ E_{23} = \frac{46 \times 29}{96} \approx 13.9 \][/tex]

4. Calculate the Chi-Square Test Statistic:
- The formula for the test statistic is:
[tex]\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
Where [tex]\(O_{ij}\)[/tex] is the observed frequency and [tex]\(E_{ij}\)[/tex] is the expected frequency.

Summing these values:
- Contribution from Men who prefer Sushi:
[tex]\[ \frac{(5 - 11.5)^2}{11.5} \approx 3.67 \][/tex]
- Contribution from Men who prefer Pizza:
[tex]\[ \frac{(24 - 23.4)^2}{23.4} \approx 0.02 \][/tex]
- Contribution from Men who prefer Hamburgers:
[tex]\[ \frac{(21 - 15.1)^2}{15.1} \approx 2.31 \][/tex]
- Contribution from Women who prefer Sushi:
[tex]\[ \frac{(17 - 10.5)^2}{10.5} \approx 4.01 \][/tex]
- Contribution from Women who prefer Pizza:
[tex]\[ \frac{(21 - 21.6)^2}{21.6} \approx 0.02 \][/tex]
- Contribution from Women who prefer Hamburgers:
[tex]\[ \frac{(8 - 13.9)^2}{13.9} \approx 2.35 \][/tex]

Adding these contributions gives us:
[tex]\[ \chi^2 \approx 12.4 \][/tex]

5. State the Conclusion:
- The Chi-Square test statistic [tex]\(\chi^2\)[/tex] is approximately 12.4.

Therefore, the computed test statistic, rounded to one decimal place, is [tex]\(\chi^2 = 12.4\)[/tex].