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Estimate the number \(n\) of pairs that must be sampled in order to estimate \(\mu_{d}=\mu 1-\mu\) as specified when the standard deviation \(s_{d}\) of the population of differences is as shown. a. \(80 \%\) confidence, to within 6 units, \(\sigma d=26.5\) b. \(95 \%\) confidence, to within 4 units, \(\sigma d=12\) c. \(90 \%\) confidence, to within 5.2 units, \(\sigma d=11.3\)

Short Answer

Expert verified
a. 33 pairs, b. 35 pairs, c. 13 pairs.

Step by step solution

01

Understanding the formula

To calculate the required sample size \( n \), we use the formula: \[n = \left(\frac{Z \cdot \sigma_d}{E}\right)^2\] where \( Z \) is the z-score corresponding to the desired confidence level, \( \sigma_d \) is the given population standard deviation, and \( E \) is the acceptable error.
02

Find Z-value for 80% confidence

For 80% confidence, the z-value is approximately 1.28.
03

Calculate n for 80% confidence

Using \( Z = 1.28 \), \( \sigma_d = 26.5 \), and \( E = 6 \), plug the values into the formula: \[n = \left(\frac{1.28 \times 26.5}{6}\right)^2 = \left(\frac{33.92}{6}\right)^2 \approx 32.05\] Rounding up, we get \( n = 33 \).
04

Find Z-value for 95% confidence

For 95% confidence, the z-value is approximately 1.96.
05

Calculate n for 95% confidence

Using \( Z = 1.96 \), \( \sigma_d = 12 \), and \( E = 4 \), plug the values into the formula: \[n = \left(\frac{1.96 \times 12}{4}\right)^2 = \left(\frac{23.52}{4}\right)^2 \approx 34.57\]Rounding up, we get \( n = 35 \).
06

Find Z-value for 90% confidence

For 90% confidence, the z-value is approximately 1.645.
07

Calculate n for 90% confidence

Using \( Z = 1.645 \), \( \sigma_d = 11.3 \), and \( E = 5.2 \), plug the values into the formula: \[n = \left(\frac{1.645 \times 11.3}{5.2}\right)^2 = \left(\frac{18.5885}{5.2}\right)^2 \approx 12.76\]Rounding up, we get \( n = 13 \).

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values, derived from sample data, that is likely to contain the true value of an unknown population parameter. When constructing a confidence interval, you are essentially specifying a degree of certainty that the interval reflects the true parameter. For example, an 80% confidence interval suggests that if you repeated the sampling process numerous times, around 80% of those intervals would contain the actual population parameter.

Confidence intervals are typically expressed in terms of a percentage and depend on the desired level of precision and certainty. The wider the confidence interval, the more likely it includes the true population parameter. However, this also means that the level of precision decreases. When performing a sample size estimation, the confidence interval's width will be determined by variables like the z-score, population standard deviation, and the acceptable margin of error. Understanding these concepts is vital to ensure that your interval reflects the level of confidence you desire.
Population Standard Deviation
The population standard deviation, often denoted as \( \sigma \), is a measure of the amount of variation or dispersion of a set of values in a population. It's an essential element in statistics because it indicates how much individual data points differ from the mean value of the data set. In the context of sample size estimation, the population standard deviation allows us to gauge the variability we might expect in different samples from that population.

When calculating the necessary sample size for a given confidence interval, the population standard deviation provides insight into the distribution of data points. A larger \( \sigma \) suggests more variability, which in turn may require a larger sample size to achieve the same level of confidence and precision compared to a population with a smaller \( \sigma \). When dealing with such statistical exercises, if the population standard deviation isn’t known, it is common to use a sample standard deviation as an estimate.
Z-Score
The z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. It is expressed in terms of standard deviations. In more basic terms, the z-score tells us how many standard deviations away a particular data point is from the mean. This concept becomes critical when working with confidence intervals and sample size calculations.

To determine the sample size for estimation, you need to reference a z-score associated with the desired confidence level. Common confidence levels and their corresponding z-scores include:
  • 80% confidence, z-score approximately 1.28
  • 90% confidence, z-score approximately 1.645
  • 95% confidence, z-score approximately 1.96
The z-score contributes to the calculation formula: \( n = \left(\frac{Z \cdot \sigma_d}{E}\right)^2 \), where \( Z \) signifies the z-score for the targeted confidence level, ensuring that estimates made from sample data are as accurate as possible.

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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 0.05 (five percentage points) 1\. When no prior knowledge of \(p_{1}\) or \(p_{2}\) is available 2\. When prior studies indicate that \(p 1 \approx 0.20\) and \(p 2 \approx 0.65\) b. \(90 \%\) confidence, to within 0.02 (two percentage points) 1\. When no prior knowledge of \(p_{1}\) or \(p_{2}\) is available 2\. When prior studies indicate that \(p 1 \approx 0.75\) and \(p 2 \approx 0.63\) c. \(95 \%\) confidence, to within 0.10 (ten percentage points) 1\. When no prior knowledge of \(p_{1}\) or \(p_{2}\) is available 2\. When prior studies indicate that \(p 1 \approx 0.11\) and \(p 2 \approx 0.37\)

The owner of a professional football team believes that the league has become more offense oriented since five years ago. To check his belief, 32 randomly selected games from one year's schedule were compared to 32 randomly selected games from the schedule five years later. Since more offense produces more offensive yards per game, the owner analyzed the following information on offensive yards per game (oypg).

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.

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.

In order to cut costs a wine producer is considering using duo or \(1+1\) corks in place of full natural wood corks, but is concerned that it could affect buyers's perception of the quality of the wine. The wine producer shipped eight pairs of bottles of its best young wines to eight wine experts. Each pair includes one bottle with a natural wood cork and one with a duo cork. The experts are asked to rate the wines on a one to ten scale, higher numbers corresponding to higher quality. The results are: \(\begin{array}{|l|c|c|} \hline \text { Wine Expert } & \text { Duo Cork } & \text { Wood Cork } \\ \hline 1 & 8.5 & 8.5 \\ \hline 2 & 8.0 & 8.5 \\ \hline 3 & 6.5 & 8.0 \\ \hline 4 & 7.5 & 8.5 \\ \hline 5 & 8.0 & 7.5 \\ \hline 6 & 8.0 & 8.0 \\ \hline 7 & 9.0 & 9.0 \\ \hline \end{array}\) \(\begin{array}{|l|c|c|} \hline \text { Wine Expert } & \text { Duo Cork } & \text { Wood Cork } \\ \hline 8 & 7.0 & 7.5 \\ \hline \end{array}\) a. Give a point estimate for the difference between the mean ratings of the wine when bottled are sealed with different kinds of corks. b. Construct the \(90 \%\) confidence interval based on these data for the difference. c. Test, at the \(10 \%\) level of significance, the hypothesis that on the average duo corks decrease the rating of the wine.

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