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In a poll taken among college students, 300 of 500 fraternity men favored a certain proposition whereas 64 of 100 nonfraternity men favored it. Estimate the difference in the proportions favoring the proposition and place a 2 -standard-deviation bound on the error of estimation.

Short Answer

Expert verified
Difference: -0.04; Confidence Interval: \(-0.1454\) to \(0.0654\).

Step by step solution

01

Understand the Problem

We need to find the estimated difference in proportions of fraternity and nonfraternity men favoring a proposition and determine the confidence interval. Fraternity men: 300 out of 500. Nonfraternity men: 64 out of 100.
02

Calculate the Proportions

Calculate the proportion of fraternity men in favor: \( p_1 = \frac{300}{500} = 0.6 \). Calculate the proportion of nonfraternity men in favor: \( p_2 = \frac{64}{100} = 0.64 \).
03

Estimate the Difference in Proportions

Find the difference between the two proportions: \( \hat{p}_1 - \hat{p}_2 = 0.6 - 0.64 = -0.04 \).
04

Find the Standard Deviation of the Difference

The standard deviation of the difference of two proportions is given by \( \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \). Substitute the values: \( \sqrt{\frac{0.6(1-0.6)}{500} + \frac{0.64(1-0.64)}{100}} = \sqrt{\frac{0.24}{500} + \frac{0.2304}{100}} = \sqrt{0.00048 + 0.002304} = \sqrt{0.002784} \approx 0.0527 \).
05

Calculate the Error Bound

Calculate the 2-standard-deviation bound (error of estimation) using \( 2 \times 0.0527 = 0.1054 \).
06

Determine the Confidence Interval

Subtract and add the error estimation from the estimated difference: \(-0.04 \pm 0.1054\). This yields the interval \(-0.1454\) to \(0.0654\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportions
Proportions are a way to represent a part of a whole. In statistics, they are often used to indicate a fraction of a sample with a particular characteristic, such as the proportion of people in favor of a proposal. To calculate a proportion, divide the number of favorable outcomes by the total number of observations. Here, the proportion of fraternity men favoring a proposition is found by dividing the number of supporters by the total fraternity men: \( p_1 = \frac{300}{500} = 0.6 \), meaning 60% support. Similarly, for nonfraternity men, the proportion is \( 0.64 \), or 64% support. Using proportions allows comparisons between groups of different sizes, helping us understand how views or behaviors vary across different segments of a population. In our example, while both groups have many supporters, nonfraternity men show slightly more support proportionally.
Confidence Interval
A confidence interval offers a range of values, providing a way to express the uncertainty or variability of an estimate. In statistics, it helps determine how reliable a statistical estimate might be. The intervals are usually calculated at a 95% confidence level, meaning if we repeated the sample many times, 95% of the calculated intervals would include the true population parameter.

In our exercise, we are dealing with differences in proportions, and we want to find out how confident we can be about our estimate of this difference. The estimated difference is \(-0.04\), meaning nonfraternity men support the proposition slightly more. We calculate the confidence interval by taking into account the standard deviation of this difference. With a 2-standard-deviation method, we find our confidence range by adding and subtracting the bound from our difference: \(-0.04 \pm 0.1054\). This gives an interval from \(-0.1454\) to \(0.0654\). This means that while we estimate a small difference favoring nonfraternity men, the interval suggests this difference could range from fraternity men favoring it slightly to nonfraternity men favoring it.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, whereas a high standard deviation indicates a wide spread of values. It's crucial in statistics for quantifying uncertainty and is often used in calculating confidence intervals.

In the context of our problem, we compute the standard deviation of the difference between two proportions. Using the formula for this, standard deviation helps understand not just the estimate itself but how much variation exists around this estimate: \( \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \).

After substituting values, we find it is approximately \(0.0527\). This step is critical, as it directly influences the confidence interval's breadth. Therefore, understanding the standard deviation helps in assessing how reliable or variable our difference estimate might be.

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Most popular questions from this chapter

In Exercise 8.81 , we gave the carapace lengths of ten mature Thenus orientalis lobsters caught in the seas in the vicinity of singapore. For your convenience, the data are reproduced here. Suppose that you wished to describe the variability of the carapace lengths of this population of lobsters. Find a \(90 \%\) confidence interval for the population variance \(\sigma^{2}\). $$\begin{array}{lcccccccccc}\text { Lobster Field Number } & \text {A061} & \text {A062} & \text {A066} & \text {A070} & \text {AO67} & \text {A069} & \text {A064} & \text {A068} & \text {A065} & \text {AO63} \\ \hline \text { Carapace Length (mm) } & 78 & 66 & 65 & 63 & 60 & 60 & 58 & 56 & 52 & 50\end{array}$$

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