# 1. In a poll of 600 voters in a campaign to eliminate non-returnable beverage containers, 210 of...

## Question:

1. In a poll of 600 voters in a campaign to eliminate non-returnable beverage containers, 210 of the voters were opposed.

Develop a 92% confidence interval estimate for the proportion of all the voters who opposed the container control bill.

2. A random sample of 87 airline pilots had an average yearly income of $99,400 with a standard deviation of$12,000. If we want to determine a 95% confidence interval for the average yearly income, what is the value of t?

Develop a 95% confidence interval for the average yearly income of all pilots.

## Confidence Intervals

The confidence interval consists of an upper bound and lower bound found by taking the product of the standard error and the critical value of the t-distribution and adding/subtracting it from the mean. The formula for the confidence interval is:

{eq}\text{Lower Bound: } \bar{x} - t_{crit}*s_x/\sqrt{n} \\ \text{Upper Bound: } \bar{x} + t_{crit}*s_x/\sqrt{n} {/eq}

Where {eq}\bar{x} {/eq} is the mean of the sample, {eq}s_x {/eq} is the sample standard deviation, {eq}n {/eq} is the sample size, and {eq}t_{crit} {/eq} is the critical value for the t-distribution for the given level of confidence.

Sample proportions are normally distributed, so the confidence interval for a proportion is constructed using the normal distribution. The formula for the confidence interval is:

{eq}\hat{p} \pm z_\alpha*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} {/eq}

Where {eq}\hat{p} {/eq} is the observed proportion, {eq}n {/eq} is the sample size, and {eq}z_\alpha {/eq} is the critical value for the confidence interval.

## Answer and Explanation:

1. In a poll of 600 voters in a campaign to eliminate non-returnable beverage containers, 210 of the voters were opposed. Develop a 92% confidence interval estimate for the proportion of all the voters who opposed the container control bill.

The sample proportion here can be computed as 210/600 = 0.35. The critical value for z at 92% confidence is 1.7507. Using this information we can compute the confidence interval with the equation:

{eq}\text{Lower Bound: } 0.35 - 1.7507*\sqrt{\frac{0.35(1-0.35)}{600}} = 0.3159 \\ \text{Upper Bound: } 0.35 + 1.7507*\sqrt{\frac{0.35(1-0.35)}{600}} = 0.3841 {/eq}

2. A random sample of 87 airline pilots had an average yearly income of $99,400 with a standard deviation of$12,000. If we want to determine a 95% confidence interval for the average yearly income, what is the value of t? Develop a 95% confidence interval for the average yearly income of all pilots.'

The critical value of the t-distribution for 95% confidence with 86 degrees of freedom is 1.9879. Using that value with the formula allows us to construct the confidence interval:

{eq}\text{Lower Bound: } 99400 - 1.9879*12000/\sqrt{87} = 96842.453 \\ \text{Upper Bound: } 99400 + 1.9879*12000/\sqrt{87} = 101957.547 {/eq}