section 11.3 assignment\nquestion 1\na poll w...
section 11.3 assignment\nquestion 1\na poll was conducted on capital punishment for murder. of 580 protestants surveyed, 180 were opposed. of 600 catholics, 238 were opposed. is there evidence that a different proportion of people in these groups oppose capitol punishment? use the 1% level of significance.\nuse \protestants\ as group 1 and \catholics\ as group 2.\nstate the p - value\n0.000004435\n0.0009694\n0.001939\n0.7378\n0.9990
Answer
# Explanation:
## Step1: Calculate sample proportions
For Protestants (group 1), $n_1 = 580$, $x_1=180$, so $\hat{p}_1=\frac{x_1}{n_1}=\frac{180}{580}\approx 0.3103$. For Catholics (group 2), $n_2 = 600$, $x_2 = 238$, so $\hat{p}_2=\frac{x_2}{n_2}=\frac{238}{600}\approx0.3967$.
## Step2: Calculate pooled proportion
$\hat{p}=\frac{x_1 + x_2}{n_1 + n_2}=\frac{180+238}{580 + 600}=\frac{418}{1180}\approx0.3542$.
## Step3: Calculate the test - statistic
The test - statistic $z=\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}$.
Substitute the values: $\hat{p}_1-\hat{p}_2=0.3103 - 0.3967=- 0.0864$, $\hat{p}(1 - \hat{p})=0.3542\times(1 - 0.3542)=0.3542\times0.6458\approx0.2289$, $\frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{580}+\frac{1}{600}=\frac{600 + 580}{580\times600}=\frac{1180}{348000}\approx0.0034$.
$z=\frac{-0.0864}{\sqrt{0.2289\times0.0034}}=\frac{-0.0864}{\sqrt{0.00077826}}\approx\frac{-0.0864}{0.0279}\approx - 3.097$.
## Step4: Calculate the p - value
Since this is a two - tailed test, the p - value is $2P(Z\lt - 3.097)$. Looking up in the standard normal table, $P(Z\lt - 3.097)\approx0.0009694$, so the p - value is $2\times0.0009694 = 0.001939$.
# Answer:
0.001939