Questions: In carrying out a study on views of capital punishment, a student asked a question the two ways shown below. Complete parts a through c. 1. With persuasion: "My brother has been accused of murder and he is innocent. If he is found guilty, he might suffer capital punishment. Now do you support or oppose capital punishment?" 2. Without persuasion "Do you support or oppose capital punishment? a. What percentage of those questioned with persuasion support capital punishment? 36% b. Find the percentage of those questioned without persuasion who support capital punishment. %% Support of capital punishment based on gender. Men With Persuasion No Persuasion For capital punishment 6 12 Against capital punishment 9 3 Women With Persuasion No Persuasion For capital punishment 3 6 Against capital punishment 7 4

Solution Steps

To find the percentage of those questioned with persuasion who support capital punishment, we calculate:

\[ \text{Total with persuasion} = 6 + 9 + 3 + 7 = 25 \] \[ \text{Support with persuasion} = 6 + 3 = 9 \] \[ \text{Percentage with persuasion} = \frac{9}{25} \times 100 = 36.00\% \]

Next, we calculate the percentage of those questioned without persuasion who support capital punishment:

\[ \text{Total without persuasion} = 12 + 3 + 6 + 4 = 25 \] \[ \text{Support without persuasion} = 12 + 6 = 18 \] \[ \text{Percentage without persuasion} = \frac{18}{25} \times 100 = 72.00\% \]

We perform a Chi-Square Test of Independence to determine if persuasion affects support for capital punishment. The observed frequencies are:

\[ \text{Observed} = \begin{bmatrix} 6 & 9 \\ 12 & 3 \\ 3 & 7 \\ 6 & 4 \end{bmatrix} \]

The expected frequencies are calculated as follows:

• For cell (1, 1): \[ E = \frac{15 \times 27}{50} = 8.1 \]
• For cell (1, 2): \[ E = \frac{15 \times 23}{50} = 6.9 \]
• For cell (2, 1): \[ E = \frac{15 \times 27}{50} = 8.1 \]
• For cell (2, 2): \[ E = \frac{15 \times 23}{50} = 6.9 \]
• For cell (3, 1): \[ E = \frac{10 \times 27}{50} = 5.4 \]
• For cell (3, 2): \[ E = \frac{10 \times 23}{50} = 4.6 \]
• For cell (4, 1): \[ E = \frac{10 \times 27}{50} = 5.4 \]
• For cell (4, 2): \[ E = \frac{10 \times 23}{50} = 4.6 \]

The expected frequencies are:

\[ \text{Expected} = \begin{bmatrix} 8.1 & 6.9 \\ 8.1 & 6.9 \\ 5.4 & 4.6 \\ 5.4 & 4.6 \end{bmatrix} \]

The Chi-Square Test Statistic (\(\chi^2\)) is calculated as:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} = 7.7295 \]

The critical value at \(\alpha = 0.05\) for 3 degrees of freedom is:

\[ \chi^2_{\alpha, df} = 7.8147 \]

The p-value is:

\[ P = P(\chi^2 > 7.7295) = 0.0519 \]

Final Answer