Questions: The statistical decision made in a chi-squared test for homogeneity of proportions at the 0.01 level of significance is: . A. Accept Ho B. Fail to Reject Ho C. Reject Ho A survey is done to determine if there is a statistically significant difference among the proportions of 9th, 10th, 11th, and 12th grade students who are satisfied with the food services at their school. Independent random samples of size 30 are taken from each grade level. The following table summarizes the results: 9th 10th 11th 12th --------------- Satisfied 20 24 17 10 Not Satisfied 10 6 13 20

Solution Steps

To perform the Chi-Square Test for Homogeneity of Proportions, we first calculate the expected frequencies for each cell in the contingency table. The expected frequency \( E \) for each cell is calculated using the formula:

\[ E = \frac{R_i \times C_j}{N} \]

where \( R_i \) is the total for row \( i \), \( C_j \) is the total for column \( j \), and \( N \) is the total number of observations.

The expected frequencies are as follows:

• For cell (1, 1): \( E = \frac{71 \times 30}{120} = 17.75 \)
• For cell (1, 2): \( E = \frac{71 \times 30}{120} = 17.75 \)
• For cell (1, 3): \( E = \frac{71 \times 30}{120} = 17.75 \)
• For cell (1, 4): \( E = \frac{71 \times 30}{120} = 17.75 \)
• For cell (2, 1): \( E = \frac{49 \times 30}{120} = 12.25 \)
• For cell (2, 2): \( E = \frac{49 \times 30}{120} = 12.25 \)
• For cell (2, 3): \( E = \frac{49 \times 30}{120} = 12.25 \)
• For cell (2, 4): \( E = \frac{49 \times 30}{120} = 12.25 \)

Thus, the expected frequencies are:

\[ \begin{bmatrix} 17.75 & 17.75 & 17.75 & 17.75 \\ 12.25 & 12.25 & 12.25 & 12.25 \end{bmatrix} \]

Next, we calculate the Chi-Square test statistic \( \chi^2 \) using the formula:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

where \( O \) is the observed frequency and \( E \) is the expected frequency. The calculations for each cell are as follows:

• For cell (1, 1): \( O = 20, E = 17.75 \) \[ \frac{(20 - 17.75)^2}{17.75} = 0.2852 \]
• For cell (1, 2): \( O = 24, E = 17.75 \) \[ \frac{(24 - 17.75)^2}{17.75} = 2.2007 \]
• For cell (1, 3): \( O = 17, E = 17.75 \) \[ \frac{(17 - 17.75)^2}{17.75} = 0.0317 \]
• For cell (1, 4): \( O = 10, E = 17.75 \) \[ \frac{(10 - 17.75)^2}{17.75} = 3.3838 \]
• For cell (2, 1): \( O = 10, E = 12.25 \) \[ \frac{(10 - 12.25)^2}{12.25} = 0.4133 \]
• For cell (2, 2): \( O = 6, E = 12.25 \) \[ \frac{(6 - 12.25)^2}{12.25} = 3.1888 \]
• For cell (2, 3): \( O = 13, E = 12.25 \) \[ \frac{(13 - 12.25)^2}{12.25} = 0.0459 \]
• For cell (2, 4): \( O = 20, E = 12.25 \) \[ \frac{(20 - 12.25)^2}{12.25} = 4.9031 \]

Summing these values gives:

\[ \chi^2 = 0.2852 + 2.2007 + 0.0317 + 3.3838 + 0.4133 + 3.1888 + 0.0459 + 4.9031 = 14.4524 \]

The critical value for the Chi-Square distribution at \( \alpha = 0.01 \) with 3 degrees of freedom is:

\[ \chi^2_{\alpha, df} = \chi^2_{(0.01, 3)} = 11.3449 \]

The p-value corresponding to the calculated Chi-Square statistic is:

\[ P = P(\chi^2 > 14.4524) = 0.0023 \]

We compare the p-value with the significance level \( \alpha \):

• Since \( P = 0.0023 < 0.01 \), we reject the null hypothesis \( H_0 \).

Final Answer