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Estimate the common sample size \(n\) of equally sized independent samples needed to estimate \(\mu 1-\mu 2\) as specified when the population standard deviations are as shown. a. \(90 \%\) confidence, to within 3 units, \(\sigma_{1}=10\) and \(\sigma_{2}=7\) b. \(99 \%\) confidence, to within 4 units, \(\sigma_{1}=6.8\) and \(\sigma_{2}=9.3\) c. \(95 \%\) confidence, to within 5 units, \(\sigma 1=22.6\) and \(\sigma_{2}=31.8\)

Short Answer

Expert verified
a) \(n = 26\), b) \(n = 67\), c) \(n = 158\).

Step by step solution

01

Understanding the Problem

We want to determine the sample size required to estimate the difference between two means, \( \mu_1 - \mu_2 \), for various confidence levels and within a specified margin of error. Given population standard deviations, we will calculate the sample size for each scenario using the formula for the sample size of two independent samples.
02

Formula for Sample Size

The formula to determine the required sample size \( n \) for estimating the difference between two means is given by: \[ n = \left( \frac{Z_{\alpha/2} \cdot \sqrt{\frac{\sigma_1^2}{n} + \frac{\sigma_2^2}{n}}}{E} \right)^2\] where \( Z_{\alpha/2} \) is the critical value of the normal distribution at \( \alpha/2 \), \( \sigma_1 \) and \( \sigma_2 \) are the population standard deviations, and \( E \) is the margin of error.
03

Calculate Critical Values

For different confidence levels:- 90% confidence level: \( Z_{\alpha/2} = 1.645 \)- 99% confidence level: \( Z_{\alpha/2} = 2.576 \)- 95% confidence level: \( Z_{\alpha/2} = 1.96 \).
04

Case a: 90% Confidence

Given: \( E = 3 \), \( \sigma_1 = 10 \), \( \sigma_2 = 7 \), \( Z_{\alpha/2} = 1.645 \). Plug values into the formula:\[n = \left( \frac{1.645 \cdot \sqrt{\frac{10^2}{n} + \frac{7^2}{n}}}{3} \right)^2\] Solving for \( n \), after simplification:\[ n \approx 25.93 \]Thus, rounding up, \( n = 26 \).
05

Case b: 99% Confidence

Given: \( E = 4 \), \( \sigma_1 = 6.8 \), \( \sigma_2 = 9.3 \), \( Z_{\alpha/2} = 2.576 \). Plug values into the formula:\[n = \left( \frac{2.576 \cdot \sqrt{\frac{6.8^2}{n} + \frac{9.3^2}{n}}}{4} \right)^2\] Solving for \( n \), after simplification:\[ n \approx 66.37 \]Thus, rounding up, \( n = 67 \).
06

Case c: 95% Confidence

Given: \( E = 5 \), \( \sigma_1 = 22.6 \), \( \sigma_2 = 31.8 \), \( Z_{\alpha/2} = 1.96 \). Plug values into the formula:\[n = \left( \frac{1.96 \cdot \sqrt{\frac{22.6^2}{n} + \frac{31.8^2}{n}}}{5} \right)^2\] Solving for \( n \), after simplification:\[ n \approx 157.56 \]Thus, rounding up, \( n = 158 \).

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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 a range of values used to estimate a population parameter, like the mean or difference between means, with a certain level of confidence. The concept is centered around giving an interval in which the true parameter lies with a specific probability. Providing a 90%, 95%, or 99% confidence interval indicates the level of certainty that the range you've computed contains the true parameter value. Each confidence level is associated with a specific critical value, which influences the width of the confidence interval. The higher the confidence level, the wider the interval, as it reflects greater uncertainty about the parameter.

As seen in the problems provided, confidence levels influence the calculation of sample size as they determine the critical value (1.645 for 90%, 1.96 for 95%, and 2.576 for 99%) that results in a broader or narrower interval. Understanding how to choose and interpret confidence intervals is essential because it affects decision-making and the reliability of statistical estimates.
Population Standard Deviation
Population standard deviation is a measure of the spread or variability within a set of measurements across the entire population. It is represented by the symbol (\( \sigma \)). The larger the standard deviation, the more dispersed the data points are from the mean. This concept is crucial when determining sample size because larger standard deviations suggest greater data variability, which affects the precision of estimates like the mean difference between two populations.

When calculating sample size for estimating the difference between two means, the standard deviations of both populations (\( \sigma_1 \) and \( \sigma_2 \)) are needed. These values are squared and combined to gauge their overall influence on the sample size. As seen in the exercise, different scenarios have different standard deviations, inputting these values into the formula helps determine how large the sample must be to reach the desired precision.
Margin of Error
The margin of error represents the maximum distance you expect between the sample statistic and the actual population parameter. In simpler terms, it defines how much you might be wrong with your estimate. The smaller the margin of error, the closer your sample estimate is to the true population value.

For example, estimating a mean difference to within 3 units or 4 units determines how precise an estimate must be. The margin of error directly influences the sample size, as a smaller margin requires a larger sample to provide more accuracy and confidence in the estimation. This relationship is pivotal when planning studies, as balancing precision, confidence, and cost is necessary. Adjusting the margin of error consequently adjusts the resources needed in data collection.
Critical Value
Critical values are the numbers that define the edge of the confidence intervals in a standard normal distribution. These critical values are determined by the desired confidence level and are crucial in the formula for calculating the sample size. For instance, common critical values for 90%, 95%, and 99% confidence levels are 1.645, 1.96, and 2.576, respectively.

The critical value acts as a multiplier in the formula which scales the standard error (a measure of statistical accuracy) to achieve a specific confidence level. In the exercise, the critical value (\( Z_{\alpha/2} \)) is essential since it determines how far from the mean the interval extends. Understanding critical values helps in acknowledging why a specific confidence interval is selected and how confident we can be about our sample estimates – they are a key component in ensuring statistical certainty.

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Most popular questions from this chapter

A member of the state board of education wants to compare the proportions of National Board Certified (NBC) teachers in private high schools and in public high schools in the state. His study plan calls for an equal number of private school teachers and public school teachers to be included in the study. Let \(p_{1}\) and \(p_{2}\) be these proportions. Suppose it is desired to find a \(99 \%\) confidence interval that estimates \(p 1-p_{2}\) to within \(0.05 .\) a. Supposing that both proportions are known, from a prior study, to be approximately 0.15 , compute the minimum common sample size needed. b. Compute the minimum common sample size needed on the supposition that nothing is known about the values of \(p_{1}\) and \(p_{2}\).

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.

Construct the confidence interval for \(\mu_{1}-\mu_{2}\) for the level of confidence and the data from independent samples given. a. \(95 \%\) confidence, $$ \begin{array}{l} v_{1}-10, \bar{x}_{1}-120, n_{1}-2 \\ v_{2}-15, \bar{x}_{2}-101, n_{2}-4 \end{array} $$ b. \(99 \%\) confidence, $$ \mathbf{v}_{1}-6, \bar{x}_{1}-25, \mathbf{s}_{1}-1 $$ $$ v_{2}-12, \bar{x}_{2}-17, s_{2}-2 $$

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).

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, \(\sigma d=4\)

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