Questions: To investigate the claim that Men are less likely than Women to support a presidential candidate, two random samples are taken. The results are shown in the table. Pop # surveyed # who support --- --- --- 1. Men 137 63 2. Women 107 65 Test the claim at a 1% level of significance. Note: you will not be able to complete parts 2 and 3 of this problem until the previous parts are correct or you have run out of tries on all parts. On an Exam, you will need to Submit each part before continuing. In each case, you cannot go back and change answers to previous parts. 1. Set up the hypotheses and state the sample size requirement. If needed, select from the Vars and Inequality tabs for the appropriate symbols. Ho: □ Ha: □ The sample size requirement is: Select an answer 2. Identify the correct distribution to use and the value of alpha. Select from the Vars tab for the distribution. This will be a □ test with alpha= □

Solution Steps

We set up the following hypotheses for the Chi-Square Test of Independence:

• Null Hypothesis (\(H_0\)): The proportion of support is the same for men and women.
• Alternative Hypothesis (\(H_a\)): The proportion of support is different for men and women.

We calculate the expected frequencies for each cell in the contingency table based on the row and column totals:

• For cell (1, 1): \[ E = \frac{R_1 \times C_1}{N} = \frac{137 \times 128}{244} = 71.8689 \]

• For cell (1, 2): \[ E = \frac{R_1 \times C_2}{N} = \frac{137 \times 116}{244} = 65.1311 \]

• For cell (2, 1): \[ E = \frac{R_2 \times C_1}{N} = \frac{107 \times 128}{244} = 56.1311 \]

• For cell (2, 2): \[ E = \frac{R_2 \times C_2}{N} = \frac{107 \times 116}{244} = 50.8689 \]

The expected frequencies are: \[ \begin{bmatrix} 71.8689 & 65.1311 \\ 56.1311 & 50.8689 \end{bmatrix} \]

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.

Calculating for each cell:

• For cell 1: \[ O = 63, \quad E = 71.8689 \quad \Rightarrow \quad \frac{(63 - 71.8689)^2}{71.8689} = 1.0944 \]

• For cell 2: \[ O = 74, \quad E = 65.1311 \quad \Rightarrow \quad \frac{(74 - 65.1311)^2}{65.1311} = 1.2077 \]

• For cell 3: \[ O = 65, \quad E = 56.1311 \quad \Rightarrow \quad \frac{(65 - 56.1311)^2}{56.1311} = 1.4013 \]

• For cell 4: \[ O = 42, \quad E = 50.8689 \quad \Rightarrow \quad \frac{(42 - 50.8689)^2}{50.8689} = 1.5463 \]

Summing these values gives: \[ \chi^2 = 1.0944 + 1.2077 + 1.4013 + 1.5463 = 4.6744 \]

The critical value at \(\alpha = 0.01\) for a Chi-Square distribution with 1 degree of freedom is: \[ \chi^2_{\alpha, df} = \chi^2_{(0.01, 1)} = 6.6349 \]

The p-value associated with the calculated Chi-Square statistic is: \[ P = P(\chi^2 > 4.6744) = 0.0306 \]

We compare the p-value with the significance level:

• Since \(P = 0.0306 > 0.01\), we fail to reject the null hypothesis.

Final Answer