Question

In the 2009 General Social Survey, respondents were asked if they favored or opposed death penalty for people convicted of murder. The 95% confidence interval for the population proportion who were in favor (say, p) was (0.65, 0.69). For the above data, the 99% confidence interval for the true population proportion of respondents who were opposed to the death penalty would be narrower than the one your derived above

Answer 1

Answer:

The calculated 99% confidence interval is wider than the 95% confidence interval.      

Step-by-step explanation:

We are given the following in the question:

95% confidence interval for the population proportion

(0.65, 0.69)

Let $\hat{p}$ be the sample proportion

Confidence interval:

$p \pm z_{stat}(\text{Standard error})$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Let x be the standard error, then, we can write

$\hat{p} - 1.96x = 0.65\\\hat{p}+1.96x = 0.69$

Solving the two equations, we get,

$2\hat{p} = 0.65 + 0.69\\\\\hat{p} = \dfrac{1.34}{2} = 0.67\\\\x = \dfrac{0.69 - 0.67}{1.96} \approx 0.01$

99% Confidence interval:

$p \pm z_{stat}(\text{Standard error})$

$z_{critical}\text{ at}~\alpha_{0.01} = 2.58$

Putting values, we get,

$0.67 \pm 2.58(0.01)\\=0.67 \pm 0.0258\\=(0.6442,0.6958)$

Thus, the calculated 99% confidence interval is wider than the 95% confidence interval .