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91Ó°ÊÓ

An online survey by ShareBuilder, a retirement plan provider, and Harris Interactive reported that \(60 \%\) of female business owners are not confident they are saving enough for retirement (SmallBiz, Winter 2006). Suppose we would like to do a follow-up study to determine how much female business owners are saving each year toward retirement and want to use \(\$ 100\) as the desired margin of error for an interval estimate of the population mean. Use \(\$ 1100\) as a planning value for the standard deviation and recommend a sample size for each of the following situations. a. \(\quad\) A \(90 \%\) confidence interval is desired for the mean amount saved. b. \(\quad\) A \(95 \%\) confidence interval is desired for the mean amount saved. c. \(\quad\) A \(99 \%\) confidence interval is desired for the mean amount saved.

Short Answer

Expert verified
a) 328, b) 465, c) 803.

Step by step solution

01

Identify Known Values and Formula

To determine the sample size required for estimating the population mean with a given margin of error, we use the formula: \[ n = \left( \frac{z \times \sigma}{E} \right)^2 \] where \(z\) is the z-value corresponding to the desired confidence level, \(\sigma\) is the standard deviation, and \(E\) is the margin of error. The problem states \(\sigma = 1100\) and \(E = 100\). We need to determine the appropriate \(z\) value for each confidence level.
02

Determine z-value for 90% Confidence Level

For a 90% confidence interval, the z-value is approximately 1.645. This value comes from the standard normal distribution table and represents the critical value for a two-tailed distribution with 5% in each tail.
03

Calculate Sample Size for 90% Confidence

Substitute the known values into the formula:\[ n = \left( \frac{1.645 \times 1100}{100} \right)^2 \approx \left( 18.095 \right)^2 \approx 327.43 \]Rounding up, the required sample size is 328 females.
04

Determine z-value for 95% Confidence Level

For a 95% confidence interval, the z-value is approximately 1.96. This is based on the standard normal distribution for a two-tailed test with 2.5% in each tail.
05

Calculate Sample Size for 95% Confidence

Substitute the known values into the formula:\[ n = \left( \frac{1.96 \times 1100}{100} \right)^2 \approx \left( 21.56 \right)^2 \approx 464.17 \]Rounding up, the required sample size is 465 females.
06

Determine z-value for 99% Confidence Level

For a 99% confidence interval, the z-value is approximately 2.576. This value represents the critical value for a two-tailed distribution with 0.5% in each tail.
07

Calculate Sample Size for 99% Confidence

Substitute the known values into the formula:\[ n = \left( \frac{2.576 \times 1100}{100} \right)^2 \approx \left( 28.336 \right)^2 \approx 802.92 \]Rounding up, the required sample size is 803 females.

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Key Concepts

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

Confidence Interval
When conducting a statistical study, it's important to determine how reliable your results are. This is where confidence intervals come into play. They help provide a range of values that likely contain the population parameter, such as a population mean. To have a good estimate of what the true value in a population might be, statisticians calculate a confidence interval. It's usually expressed with a percentage, such as 90%, 95%, or 99%, which reflects how confident we are that the interval contains the true mean.
  • 90% Confidence Interval: Suggests there's a 90% chance that the true population parameter lies within the calculated interval.
  • 95% Confidence Interval: Most common in research, reflects 95% confidence.
  • 99% Confidence Interval: Provides high assurance that the interval includes the population parameter.
To find the correct interval, a z-value is used according to the desired level of confidence, derived from the standard normal distribution.
Standard Deviation
Standard deviation (\(\sigma\)) is a key concept in statistics that tells us how much variation or "spread" there is in a set of values. A smaller standard deviation means that the data points tend to be very close to the mean, while a larger standard deviation indicates that the data points are spread out over a wider range of values. In our study context, we used \(\sigma = 1100\), meaning there is a relatively high variance in retirement savings among female business owners. It's crucial because it affects the width of the confidence interval. The larger the standard deviation, the larger the sample size needed to achieve a desired margin of error. Calculating the potential spread helps in understanding how representative a sample might be of the overall population.
Margin of Error
Margin of error is essentially the maximum expected difference between the true population parameter and a sample estimate. It provides a way of expressing uncertainty in an estimate. In research, it's important to set a margin of error that balances precision and practicality. The formula to calculate sample sizes incorporates the margin of error (\(E\)), which in this exercise is \(100\) dollars.
  • Larger margins result in smaller necessary sample sizes because they tolerate more variability.
  • Smaller margins demand larger sample sizes for more precise estimates.
Deciding the margin affects how narrow or wide the confidence interval will be, which can impact decision-making processes in research.
Population Mean
The population mean is an average that represents the true center of the population distribution. In the context of our study, it's the average amount that female business owners save for retirement, which we want to estimate. By calculating a confidence interval around the estimated sample mean, the goal is to get a good approximation of this population mean.
  • Sample mean is used to estimate population mean since it's impractical to survey every individual.
  • Achieving a representative sample size helps ensure the sample mean is close to the population mean.
In the exercise, sample size calculations are mathematically derived to ensure confidence in the sample mean, providing an interval where the true mean likely falls.

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Most popular questions from this chapter

In a survey, 200 people were asked to identify their major source of news information; 110 stated that their major source was television news. a. Construct a \(95 \%\) confidence interval for the proportion of people in the population who consider television their major source of news information. b. How large a sample would be necessary to estimate the population proportion with a margin of error of .05 at \(95 \%\) confidence?

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