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

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

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} = (1.34)/(2) = 0.67\\\\x = (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 .