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Chapter 9: Problem 4

Estimate the number \(n\) of pairs that must be sampled in order to estimate \(\mu d=\mu 1-\mu 2\) as specified when the standard deviation \(s_{d}\) of the population of differences is as shown. a. \(90 \%\) confidence, to within 20 units, \(\sigma d=75.5\) b. \(95 \%\) confidence, to within 11 units, \(\sigma d=31.4\) c. \(99 \%\) confidence, to within 1.8 units, \(\sigma d=4\)

Short Answer

Expert verified
Sample sizes are 39, 33, and 33 for 90%, 95%, and 99% confidence levels, respectively.

Step by step solution

01

Identifying the formula for sample size estimation

To determine the sample size needed to estimate the difference between two population means, we use the formula: \[ n = \left(\frac{Z \cdot \sigma_{d}}{E}\right)^2 \] where \( n \) is the sample size, \( Z \) is the Z-value corresponding to the desired confidence level, \( \sigma_{d} \) is the population's standard deviation of differences, and \( E \) is the margin of error.
02

Calculating sample size for 90% confidence level

For 90% confidence, the Z-value is 1.645. We are given \( \sigma_d = 75.5 \) and the margin of error \( E = 20 \). Substitute these values into the formula:\[ n = \left(\frac{1.645 \times 75.5}{20}\right)^2 \approx 38.69 \]Round up to the nearest whole number, so the required sample size is 39.
03

Calculating sample size for 95% confidence level

For 95% confidence, the Z-value is 1.96. We are given \( \sigma_d = 31.4 \) and the margin of error \( E = 11 \). Substitute these values into the formula:\[ n = \left(\frac{1.96 \times 31.4}{11}\right)^2 \approx 32.22 \]Round up to the nearest whole number, so the required sample size is 33.
04

Calculating sample size for 99% confidence level

For 99% confidence, the Z-value is 2.576. We are given \( \sigma_d = 4 \) and the margin of error \( E = 1.8 \). Substitute these values into the formula:\[ n = \left(\frac{2.576 \times 4}{1.8}\right)^2 \approx 32.68 \]Round up to the nearest whole number, so the required sample size is 33.

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Key Concepts

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

Confidence Level
When conducting a statistical study, the confidence level helps us understand how certain we can be about our estimates. The confidence level is the probability that the value of a parameter falls within a specified range of values.
  • Commonly used confidence levels are 90%, 95%, and 99%.
  • A 90% confidence level suggests that if the study were repeated many times, 90% of the calculated intervals would contain the true mean difference.
  • The higher the confidence level, the broader the range, and thus, the greater the sample size needed.
Confidence levels are represented by Z-values in statistical calculations. These Z-values correspond to the area under the normal distribution curve:
  • For 90%, the Z-value is 1.645.
  • For 95%, it's 1.96.
  • For 99%, it is 2.576.
Understanding these values can help you determine how precise your statistical estimates need to be for your study.
Margin of Error
The margin of error, also known as the error bound, indicates the range within which we expect the true population parameter to fall. It represents the maximum expected difference between the true population parameter and a sample estimate.
  • The smaller the margin of error, the more precise the estimate.
  • Reducing the margin of error typically requires a larger sample size.
  • It is closely tied to the confidence level you choose.
In the formula for sample size estimation, the margin of error ( $E$ ) represents this aspect of accuracy. For example, being "within 20 units" means the margin of error is 20. Understanding the margin of error helps in planning how precise and reliable your statistical analysis intends to be.
Standard Deviation of Differences
The standard deviation of the differences, symbolized as \( \sigma_d \) , measures the variability or dispersion of the differences between paired observations in a dataset.
  • It helps assess how much individual differences deviate from the average difference.
  • Higher standard deviations indicate greater variability in the differences.
  • The standard deviation is crucial in determining the sample size needed for reliable estimates.
In the context of sample size estimation, \( \sigma_d \) influences the size of the sample because:
  • Larger standard deviations require larger samples to achieve the same degree of precision.
  • It plays a vital role in calculating the confidence intervals for population means.
By understanding the standard deviation of differences, you can gain insights into the consistency and reliability of the datasets you're working with. This helps in deciding the necessary sample size to achieve the desired accuracy in your study.

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Most popular questions from this chapter

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).

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).

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.

A neighborhood home owners association suspects that the recent appraisal values of the houses in the neighborhood conducted by the county government for taxation purposes is too high. It hired a private company to appraise the values of ten houses in the neighborhood. The results, in thousands of dollars, are \(\begin{array}{|l|c|c|} \hline \text { House } & \text { County Government } & \text { Private Company } \\ \hline 1 & 217 & 219 \\ \hline 2 & 350 & 338 \\ \hline 3 & 296 & 291 \\ \hline 4 & 237 & 237 \\ \hline \end{array}\) \(\begin{array}{|l|c|c|} \hline \text { House } & \text { County Government } & \text { Private Company } \\ \hline 5 & 237 & 235 \\ \hline 6 & 272 & 269 \\ \hline 7 & 257 & 239 \\ \hline 8 & 277 & 275 \\ \hline 9 & 312 & 320 \\ \hline 10 & 335 & 335 \\ \hline \end{array}\) a. Give a point estimate for the difference between the mean private appraisal of all such homes and the government appraisal of all such homes. b. Construct the \(99 \%\) confidence interval based on these data for the difference. c. Test, at the \(1 \%\) level of significance, the hypothesis that appraised values by the county government of all such houses is greater than the appraised values by the private appraisal company.

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 \(p_{2} \approx 0.21\)
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