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Chapter 9: Problem 11

A journalist plans to interview an equal number of members of two political parties to compare the proportions in each party who favor a proposal to allow citizens with a proper license to carry a concealed handgun in public parks. Let \(p_{1}\) and \(p_{2}\) be the true proportions of members of the two parties who are in favor of the proposal. Suppose it is desired to find a \(95 \%\) confidence interval for estimating \(p_{1-p 2}\) to within 0.05. Estimate the minimum equal number of members of each party that must be sampled to meet these criteria.

Short Answer

Expert verified
385 members from each party should be sampled.

Step by step solution

01

Understand the Problem

We need to find the minimum number of members to sample from two political parties to be able to estimate the difference in proportions who favor a policy, with a specified confidence interval width and level.
02

Set up the Formula for Sample Size

To determine the sample size, we use the formula for the difference in proportions' confidence interval: \[ n = \left( \frac{Z_{(\alpha/2)}^2 (p_1 (1-p_1) + p_2 (1-p_2))}{E^2} \right), \] where \(E\) is the margin of error, and \(Z_{(\alpha/2)}\) is the Z-score for a 95% confidence level.
03

Determine the Z-score and Margin of Error

For a 95% confidence level, the Z-score \(Z_{0.025}\) is 1.96. The margin of error \(E\) is given as 0.05.
04

Assume Maximum Variability

Since we do not have estimates for \(p_1\) and \(p_2\), assume the maximum possible variability, which makes \(p_1 = p_2 = 0.5\). This gives the largest sample size.
05

Insert Into Formula and Calculate

Substitute \(Z=1.96\), \(E=0.05\), and \(p_1 = p_2 = 0.5\) into the formula: \[ n = \left( \frac{1.96^2 (0.5 \times 0.5 + 0.5 \times 0.5)}{0.05^2} \right) = 384.16. \]Since \(n\) must be a whole number, round up to the nearest whole number, which is 385.

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Key Concepts

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

Sample Size Calculation
When designing a study or survey, it's crucial to determine the number of participants needed for confident and accurate results. This is known as sample size calculation. In the context of our exercise, we are trying to estimate the difference in proportions between two groups.

To do this, we use a special formula:
\[n = \left( \frac{Z_{(\alpha/2)}^2 (p_1 (1-p_1) + p_2 (1-p_2))}{E^2} \right),\]where:
  • \(Z_{(\alpha/2)}\) is a Z-score reflecting our desired confidence level,
  • \(p_1\) and \(p_2\) are the proportions of successes in each group,
  • and \(E\) represents the margin of error we are willing to tolerate.
For a 95% confidence level, the \(Z\)-score is 1.96. If we lack specific proportions \(p_1\) and \(p_2\), we assume maximum variability by setting them both to 0.5. This scenario optimizes the sample size, ensuring our results remain robust even with unknown distributions.
Difference of Proportions
The difference of proportions is a statistical measure that helps us compare how two groups differ on a particular characteristic. In simple terms, it shows if there's a significant difference in the way two groups behave or respond.

In our scenario, the goal is to understand how different the two political parties are when it comes to supporting a proposal. We use the true proportions, \(p_1\) and \(p_2\), which represent the fraction in each group that supports the proposal.

Our task is to estimate the variation \(p_1 - p_2\), aiming to determine if this difference is real or just due to random sampling errors. This estimation becomes more reliable as we increase the sample size, reducing the impact of random chance and providing clearer insight into the true nature of the populations. The confidence interval gives us a range where we believe the true difference lies, aiding in informed decision-making.
Margin of Error
The margin of error is a key component within any statistical estimate, describing the range of uncertainty around a measured sample statistic. It's particularly important when constructing confidence intervals because it sets the width of the interval.

A smaller margin of error suggests more precision in your estimate, giving you a tight range in which the true proportion difference likely falls. In our exercise, the desired margin of error was set at 0.05.
  • This means that the interval will span a total of 0.10, centered around the estimated difference, ensuring our final report is within 5 percentage points on either side of the estimated value.
  • This precision level ensures that conclusions drawn from the data are reliable, influencing decisions with greater accuracy and confidence.
The margin of error is dependent on both the sample size and the level of confidence we wish to achieve, playing a pivotal role in determining the thickness of the confidence interval.

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Most popular questions from this chapter

Estimate the minimum equal sample sizes \(n 1=n 2\) necessary in order to estimate \(p 1-p_{2}\) as specified. a. \(80 \%\) confidence, to within 0.02 (two percentage points) a. when no prior knowledge of \(p_{1}\) or \(p_{2}\) is available b. when prior studies indicate that \(p 1 \approx 0.78\) and \(p 2 \approx 0.65\) b. \(90 \%\) confidence, to within 0.05 (two percentage points) a. when no prior knowledge of \(p_{1}\) or \(p_{2}\) is available b. when prior studies indicate that \(p 1 \approx 0.12\) and \(p 2 \approx 0.24\) c. \(95 \%\) confidence, to within 0.10 (ten percentage points) a. when no prior knowledge of \(p_{1}\) or \(p_{2}\) is available b. When prior studies indicate that \(p_{1} \approx 0.14\) and \(p_{2} \approx 0.21\)

Eight golfers were asked to submit their latest scores on their favorite golf courses. These golfers were each given a set of newly designed clubs. After playing with the new clubs for a few months, the golfers were again asked to submit their latest scores on the same golf courses. The results are summarized below. $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text { Golfer } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text { Own clubs } & 77 & 80 & 69 & 73 & 73 & 72 & 75 & 77 \\ \hline \text { New clubs } & 72 & 81 & 68 & 73 & 75 & 70 & 73 & 75 \\ \hline \end{array}$$ a. Compute \(\bar{d}\) and \(s_{d}\). b. Give a point estimate for \(\mu_{1}-\mu_{2}-\mu_{\mathrm{d}}\). c. Construct the \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}-\mu_{\mathrm{d}}\) from these data. d. Test, at the \(1 \%\) level of significance, the hypothesis that on average golf scores are lower with the new clubs.

Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were randomly selected to investigate whether there was any difference in the mean time in years that they were kept by the original owner before being sold. For cars the mean was 5.3 years with standard deviation 2.2 years. For pickup trucks the mean was 7.1 years with standard deviation 3.0 years. a. Construct the \(95 \%\) confidence interval for the difference in the means based on these data. b. Test the hypothesis that there is a difference in the means against the null hypothesis that there is no difference. Use the \(1 \%\) level of significance. c. Compute the observed significance of the test in part (b).

To assess to the relative happiness of men and women in their marriages, a marriage counselor plans to administer a test measuring happiness in marriage ton randomly selected married couples, record the their test scores, find the differences, and then draw inferences on the possible difference. Let \(\mu 1\) and \(\mu_{2}\) be the true average levels of happiness in marriage for men and women respectively as measured by this test. Suppose it is desired to find a \(90 \%\) confidence interval for estimating \(\mu d=\mu 1-\mu 2\) to within two test points. Suppose further that, from prior studies, it is known that the standard deviation of the differences in test scores is \(\sigma d \approx 10 .\) What is the minimum number of married couples that must be included in this study?

A university administrator wishes to estimate the difference in mean grade point averages among all men affiliated with fraternities and all unaffiliated men, with \(95 \%\) confidence and to within \(0.15 .\) It is known from prior studies that the standard deviations of grade point averages in the two groups have common value \(0.4 .\) Estimate the minimum equal sample sizes necessary to meet these criteria.
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