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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}


Learn more about this topic:

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Using the t Distribution to Find Confidence Intervals

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 6
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