Questions: Use the following information to answer Questions 45-47. A study examines the personal goals of children in grades 4, 5, and 6. A random sample of students was selected for each of the grades 4,5 , and 6 from schools in Georgia. The students received a questionnaire regarding achievement of personal goals. They were asked what they would most like to do at school: make good grades, be good at sports, or be popular. Results are presented in the following table by the sex of the child. Boys Girls Make good grades 192 590 Be popular 64 90 Be good in sports 188 80 Which hypotheses are being tested by the chi-square test? The null hypothesis is that personal goals and sex are independent, and the alternative is that they are dependent. The null hypothesis is that the mean personal goal is the same for boys and girls, and the alternative is that the means differ. The distribution of personal goals is different for boys and girls. The distribution of sex is different for the three different personal goals.

Solution Steps

The observed frequencies from the study regarding personal goals of children by sex are as follows:

\[ \begin{array}{c|c|c} & \text{Boys} & \text{Girls} \\ \hline \text{Make good grades} & 192 & 590 \\ \text{Be popular} & 64 & 90 \\ \text{Be good in sports} & 188 & 80 \\ \end{array} \]

To calculate the expected frequencies for each cell in the contingency table, we use 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 grand total.

The expected frequencies are calculated as follows:

• For cell (1, 1): \[ E = \frac{782 \times 444}{1204} = 288.3787 \]

• For cell (1, 2): \[ E = \frac{782 \times 760}{1204} = 493.6213 \]

• For cell (2, 1): \[ E = \frac{154 \times 444}{1204} = 56.7907 \]

• For cell (2, 2): \[ E = \frac{154 \times 760}{1204} = 97.2093 \]

• For cell (3, 1): \[ E = \frac{268 \times 444}{1204} = 98.8306 \]

• For cell (3, 2): \[ E = \frac{268 \times 760}{1204} = 169.1694 \]

Thus, the expected frequencies are:

\[ \begin{array}{c|c|c} & \text{Boys} & \text{Girls} \\ \hline \text{Make good grades} & 288.3787 & 493.6213 \\ \text{Be popular} & 56.7907 & 97.2093 \\ \text{Be good in sports} & 98.8306 & 169.1694 \\ \end{array} \]

The Chi-Square test statistic (\(\chi^2\)) is calculated 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): \[ O = 192, \quad E = 288.3787, \quad \frac{(192 - 288.3787)^2}{288.3787} = 32.2106 \]

• For cell (1, 2): \[ O = 590, \quad E = 493.6213, \quad \frac{(590 - 493.6213)^2}{493.6213} = 18.8178 \]

• For cell (2, 1): \[ O = 64, \quad E = 56.7907, \quad \frac{(64 - 56.7907)^2}{56.7907} = 0.9152 \]

• For cell (2, 2): \[ O = 90, \quad E = 97.2093, \quad \frac{(90 - 97.2093)^2}{97.2093} = 0.5347 \]

• For cell (3, 1): \[ O = 188, \quad E = 98.8306, \quad \frac{(188 - 98.8306)^2}{98.8306} = 80.4527 \]

• For cell (3, 2): \[ O = 80, \quad E = 169.1694, \quad \frac{(80 - 169.1694)^2}{169.1694} = 47.0013 \]

Summing these values gives:

\[ \chi^2 = 32.2106 + 18.8178 + 0.9152 + 0.5347 + 80.4527 + 47.0013 = 179.9323 \]

The critical value at \(\alpha = 0.05\) for a Chi-Square distribution with 2 degrees of freedom is:

\[ \chi^2_{\alpha, df} = \chi^2_{(0.05, 2)} = 5.9915 \]

The p-value associated with the calculated Chi-Square statistic is:

\[ P = P(\chi^2 > 179.9323) = 0.0 \]

Since the p-value \(0.0\) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis. This indicates that personal goals and sex are dependent.

Final Answer