Question

WS 21.12 Gun control. A December 2014 Pew Research Center poll asked a sample of 1507 adults whether they thought it was important to control gun ownership. A total of 693 of the poll respondents were in support of controls on gun ownership. Although the samples in national polls are not SRSs, they are similar enough that our method gives approximately correct confidence intervals. (a) Say in words what the population proportion p is for this poll. (b) Find a 95% confidence interval for p. (c) The Pew Research Center announced a margin of error of plus or minus 2.9 percentage points for this poll result. How well does your work in part (b) agree with this margin of error?

Question B and C

Answer 1

Given that, n = 1507 and x = 693

=> sample proportion $(\hat p)$ = 693/1507 = 0.4599

b) A 95% confidence level has significance level of 0.05 and critical value is,

Za/2 = 1.96

The 95% confidence interval for population proportion (p) is,

– 2012 * V *(1 - ) <p<ộ+Za/12* * (1 – é)

0.4599 * (1 -0.4599) 0.4599-1.96*V = - 1507 0.4599 * (1 - 0.4599) <p < 0.4599+1.96*1/ 1507

0.4599 - 0.0252 <p <0.4599 +0.0252

0.4347<p <0.4851

Therefore, the 95% confidence interval is (0.4347, 0.4851).

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c) In part (b), the margin of error was found to be 2.5 percentage points. Compared to 2.9 percentage points, this is fairly close, but this is a smaller margin of error than what the Pew Research Center announced.