Answer to Question #127578 in Statistics and Probability for jessica

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 90% 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

a) we have to construct 90% confidence interval for population proportion.We have been provided with following information .

X= 412

N= 768

The sample proportion is computed as follows, based on the sample size N = 768 and the number of favorable cases X = 412

$\widehat{p}= \frac{X}{N}$ = $\frac{412}{768}$ = 0.536

Using z table , the critical value for α=0.1 is $z_c = z_{1-\alpha/2} = 1.645$ . The corresponding confidence interval is computed as shown below

$[\widehat{p}- {Z_{c}*}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}},\widehat{p}+{Z_{c}*}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}]$

$[0.536- {1.645*}\sqrt{\frac{0.536(1-0.536)}{768}},0.536+ {1.645*}\sqrt{\frac{0.536(1-0.536)}{768}}]$

[0.507,0.566]

b)we have to construct 95% lower confidence bound for population proportion.We have been provided with following information .

X= 412

N= 768 and $\widehat{p}$ = 0.536

Using Z table the critical value for α=0.05 is $z_c = z_{1-\alpha/2} = 1.96$

Lower confidence bound is given by

$\widehat{p}- {Z_{c}*}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}} < p$

$0.536- {1.96*}\sqrt{\frac{0.536(1-0.536)}{768}} < p$

0.501 <p

Hence the lower bound of the confidence interval is 0.501