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Chapter 9: Problem 8

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 \(p_{2} \approx 0.21\)

Short Answer

Expert verified
80% Confidence: 1025 (unknown), 1550 (known); 90% Confidence: 541 (unknown), 265 (known); 95% Confidence: 96 (unknown), 98 (known).

Step by step solution

01

Determine the Formula

To estimate the minimum sample size necessary to capture the difference in proportions \( p_1 - p_2 \) with given confidence and precision, the formula to use is:\[n = \left( \frac{Z^2 \cdot ( p_1(1-p_1) + p_2(1-p_2) )}{d^2} \right)\]where: * \( n \) is the sample size for each group (since \( n_1 = n_2 \)), * \( Z \) is the Z-value corresponding to the desired confidence level,* \( d \) is the margin of error (half the width of the confidence interval),* \( p_1 \) and \( p_2 \) are the estimated proportions of the two groups.
02

Calculate Specific Z-values

For specific confidence levels:* \( 80\% \) confidence corresponds to \( Z = 1.28 \),* \( 90\% \) confidence corresponds to \( Z = 1.645 \),* \( 95\% \) confidence corresponds to \( Z = 1.96 \).
03

Calculate Sample size for 80% Confidence, d=0.02, No Prior Knowledge

Since there is no prior knowledge, we assume the worst-case scenario where both \( p_1 \) and \( p_2 \) are \( 0.5 \).\[n = \left( \frac{1.28^2 \cdot (0.5 \times (1-0.5) + 0.5 \times (1-0.5))}{0.02^2} \right) = \left( \frac{1.6384 \cdot 0.5}{0.0004} \right) = 1025\]Thus, \( n = 1025 \).
04

Calculate Sample Size for 80% Confidence, d=0.02, Known Proportions

Given \( p_1 \approx 0.78 \) and \( p_2 \approx 0.65 \), we use the known values:\[n = \left( \frac{1.28^2 \cdot (0.78 \times 0.22 + 0.65 \times 0.35)}{0.02^2} \right) = \left( \frac{1.6384 \cdot 0.3796}{0.0004} \right) \approx 1550\]Thus, \( n = 1550 \).
05

Calculate Sample Size for 90% Confidence, d=0.05, No Prior Knowledge

Assuming \( p_1 = p_2 = 0.5 \) for worst-case:\[n = \left( \frac{1.645^2 \cdot (2 \times 0.5 \times (1 - 0.5))}{0.05^2} \right) = \left( \frac{2.706025 \cdot 0.5}{0.0025} \right) \approx 541\]Thus, \( n = 541 \).
06

Sample Size for 90% Confidence, d=0.05, Known Proportions

With \( p_1 \approx 0.12 \) and \( p_2 \approx 0.24 \):\[n = \left( \frac{1.645^2 \cdot (0.12 \times 0.88 + 0.24 \times 0.76)}{0.05^2} \right) = \left( \frac{2.706025 \cdot 0.2448}{0.0025} \right) \approx 265\]Thus, \( n = 265 \).
07

Calculate Sample Size for 95% Confidence, d=0.1, No Prior Knowledge

Assuming worst case with \( p_1 = p_2 = 0.5 \):\[n = \left( \frac{1.96^2 \cdot (2 \times 0.5 \times (1 - 0.5))}{0.1^2} \right) = \left( \frac{3.8416 \cdot 0.5}{0.01} \right) = 96\]Thus, \( n = 96 \).
08

Sample Size for 95% Confidence, d=0.1, Known Proportions

Given \( p_1 \approx 0.14 \) and \( p_2 \approx 0.21 \):\[n = \left( \frac{1.96^2 \cdot (0.14 \times 0.86 + 0.21 \times 0.79)}{0.1^2} \right) = \left( \frac{3.8416 \cdot 0.2551}{0.01} \right) \approx 98\]Thus, \( n = 98 \).

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Key Concepts

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

Confidence Intervals
In statistics, confidence intervals are used to estimate the range within which a population parameter lies, based on a sample statistic. They provide a way to quantify the uncertainty around this estimate. For example, when estimating a population proportion, a confidence interval offers a range that is expected to capture the true proportion a specified percentage of the time. This percentage is known as the confidence level, and it reflects the degree of certainty in the interval estimate. A higher confidence level indicates greater certainty about the parameter lying within the interval. However, this also means the interval becomes wider. For instance, an 80% confidence level typically results in a narrower interval compared to a 95% confidence level. The width of the interval depends on several factors:
  • The variability in the data.
  • The sample size.
  • The selected confidence level.
For practical applications in sample size estimation, understanding how confidence intervals work is crucial. Setting the interval at the start guides the sample size needed to ensure accurate estimations.
Difference in Proportions
When comparing two groups, researchers often want to assess whether there is a substantial difference in proportions between them. This can happen in fields like medicine, social sciences, and marketing, where determining if one group has a higher or lower likelihood of a specific outcome can inform decisions. For example, if one group is subjected to a treatment and another is not, determining the effectiveness of the treatment can rely on understanding the difference in proportions of desired outcomes between the two groups. To estimate this, it is essential to have reliable data from both groups. Often, researchers use statistical tests to determine if observed differences are significant, or if they might have occurred by chance. This commonly includes:
  • Applying formulas to calculate expected differences.
  • Considering variation in the two group proportions.
  • Using confidence intervals to assess estimation accuracy.
Accurate estimation aids in determining whether there is enough evidence to suggest a real difference, which is crucial in forming concrete conclusions based on research findings.
Margin of Error
The margin of error in statistics highlights the amount of random sampling error in a survey's results. It indicates how much the sample statistic is expected to vary from the true population parameter. It is typically expressed as a percentage and helps to determine the precision of the estimate. Calculating the margin of error involves certain steps, primarily focusing on:
  • The desired confidence level.
  • Variability of the data.
  • Sample size.
The larger the margin of error, the less precise the results. This is why achieving a smaller margin of error is often a goal, and this can be achieved by using a larger sample size. In sample size estimation, the margin of error defines how close the researcher wants the sample estimate to be from the actual population parameter. Known as the precision of the estimate, adjusting the margin of error provides a balance between reliability and resource constraints, especially when sample sizes must be realistically and practically determined.

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Most popular questions from this chapter

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.

A sociologist surveys 50 randomly selected citizens in each of two countries to compare the mean number of hours of volunteer work done by adults in each. Among the 50 inhabitants of Lilliput, the mean hours of volunteer work per year was 52 , with standard deviation 11.8 . Among the 50 inhabitants of Blefuscu, the mean number of hours of volunteer work per year was \(37,\) with standard deviation 7.2 . a. Construct the \(99 \%\) confidence interval for the difference in mean number of hours volunteered by all residents of Lilliput and the mean number of hours volunteered by all residents of Blefuscu. b. Test, at the \(1 \%\) level of significance, the claim that the mean number of hours volunteered by all residents of Lilliput is more than ten hours greater than the mean number of hours volunteered by all residents of Blefuscu. c. Compute the observed significance of the test in part (b).

An automotive tire manufacturer wishes to estimate the difference in mean wear of tires manufactured with an experimental material and ordinary production tire, with \(90 \%\) confidence and to within \(0.5 \mathrm{~mm}\). To eliminate extraneous factors arising from different driving conditions the tires will be tested in pairs on the same vehicles. It is known from prior studies that the standard deviations of the differences of wear of tires constructed with the two kinds of materials is \(1.75 \mathrm{~mm}\). Estimate the minimum number of pairs in the sample necessary to meet these criteria.

A university administrator wishes to know if there is a difference in average starting salary for graduates with master's degrees in engineering and those with master's degrees in business. Fifteen recent graduates with master's degree in engineering and 11 with master's degrees in business are surveyed and the results are summarized below. \begin{tabular}{|c|c|c|c|} \hline & \(n\) & \(=\) & \(s\) \\ \hline Engineering & 15 & 68,535 & 1627 \\ \hline Business & 11 & 63,230 & 2033 \\ \hline \end{tabular} a. Construct the \(90 \%\) confidence interval for the difference in the population means based on these data. b. Test, at the \(10 \%\) level of significance, whether the data provide sufficient evidence to conclude that the average starting salaries are different.

Large Data Sets \(6 \mathrm{~A}\) and \(6 \mathrm{~B}\) record results of a random survey of 200 voters in each of two regions, in which they were asked to express whether they prefer Candidate Afor a U.S. Senate seat or prefer some other candidate. Let the population of all voters in region 1 be denoted Population 1 and the population of all voters in region 2 be denoted Population 2. Let \(p_{1}\) be the proportion of voters in Population 1 who prefer Candidate \(A,\) and \(p_{2}\) the proportion in Population 2 who do. a. Find the relevant sample proportions \(p^{\wedge} 1\) and \(p^{\prime}\) ?. b. Construct a point estimate for \(p 1-p_{2}\). c. Construct a \(95 \%\) confidence interval for \(p 1-p 2\). d. Test, at the \(5 \%\) level of significance, the hypothesis that the same proportion of voters in the two regions favor Candidate \(A\), against the alternative that a larger proportion in Population 2 do.
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