Answer to Question #125504 in Statistics and Probability for Jse

Question

The 2004 presidential election exit polls from the critical state of Ohio provided the following results. There were 2020 respondents in the exit polls and 768 were college graduates. Of the college graduates, 412 voted for George Bush.

(a) Calculate a 99% confidence interval for the proportion of college graduates in Ohio that voted for George Bush. Round the answers to 3 decimal places.

(b) Calculate a 95% lower confidence bound for the proportion of college graduates in Ohio that voted for George Bush. Round the answer to 3 decimal places.

Answer 1

Make sure, that normal distribution is possible here. Check if $n\widetilde{p} > 5$ and $n(1-\widetilde{p}) > 5$

$\widetilde{p} = \frac{412}{768}, n = 768$

So $n\widetilde{p} = 412 > 5, n(1-\widetilde{p}) = 356 > 5$

(a)

Solution

Confidence inteval bounds are $\widetilde{p} \pm z\sqrt{\frac{\widetilde{p}(1-\widetilde{p})}{n}}$ . For the 99% confidence interval $z = 2.576$

$l = \frac{412}{768}-2.576\sqrt{\frac{412\cdot356}{768^3}} \approx 0.536 - 2.576\cdot0.018 \approx 0.489$

$r = \frac{412}{768}+2.576\sqrt{\frac{412\cdot356}{768^3}} \approx 0.536+2.576\cdot0.018 \approx 0.582$

Answer: 0.489 $\leq$ p $\leq$ 0.582

(b)

Solution

Lower confidence bound is $\widetilde{p} - z\sqrt{\frac{\widetilde{p}(1-\widetilde{p})}{n}}$ . For the 95% lower confidence bound $z = 1.645$

$l = \frac{412}{768}-1.645\sqrt{\frac{412\cdot356}{768^3}} \approx 0.536 - 1.645\cdot0.018 \approx 0.506$

Answer: p $\geq$ 0.506