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91Ó°ÊÓ

In developing patient appointment schedules, a medical center wants to estimate the mean time that a staff member spends with each patient. How large a sample should be taken if the desired margin of error is two minutes at a \(95 \%\) level of confidence? How large a sample should be taken for a \(99 \%\) level of confidence? Use a planning value for the population standard deviation of eight minutes.

Short Answer

Expert verified
For 95% confidence, a sample size of 62 is needed; for 99% confidence, a sample size of 107 is required.

Step by step solution

01

Identify the known values

We have a desired margin of error (E) of 2 minutes, population standard deviation (\( \sigma \)) of 8 minutes, and we need to determine sample sizes for confidence levels of 95% and 99%. The corresponding Z-scores are 1.96 for 95% confidence and 2.576 for 99% confidence.
02

Write the formula for sample size estimation

The formula to estimate the required sample size \( n \) is given by: \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] where \( Z \) is the Z-score corresponding to the confidence level, \( \sigma \) is the standard deviation, and \( E \) is the margin of error.
03

Calculate sample size for 95% confidence level

Substitute the values for the 95% confidence level: \( Z = 1.96 \), \( \sigma = 8 \), and \( E = 2 \) into the formula: \[ n = \left( \frac{1.96 \times 8}{2} \right)^2 \] Calculate this to get: \[ n = \left( 7.84 \right)^2 = 61.47 \] Round up to the nearest whole number to find that the sample size required is 62.
04

Calculate sample size for 99% confidence level

Substitute the values for the 99% confidence level: \( Z = 2.576 \), \( \sigma = 8 \), and \( E = 2 \) into the formula: \[ n = \left( \frac{2.576 \times 8}{2} \right)^2 \] Calculate this to get: \[ n = \left( 10.304 \right)^2 = 106.18 \] Again, round up to the nearest whole number to find that the sample size required is 107.

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Key Concepts

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

Confidence Level
In statistics, when we talk about a confidence level, we are referring to the degree of certainty that a parameter lies within a specific range. It's an important concept in sample size estimation because it tells us how confident we are that our sample reflects the population.
The most common confidence levels are 95% and 99%.
  • A 95% confidence level means that if we were to take 100 different samples and compute an interval estimate for each sample, we would expect about 95 of those intervals to contain the population parameter.
  • Similarly, a 99% confidence level indicates a greater certainty, meaning only 1 out of 100 intervals would not contain the population parameter.
Confidence levels are closely tied to the idea of Z-scores, which we will discuss later. The higher the confidence level, the larger the sample size you'll generally need to get a reliable estimate.
Margin of Error
The margin of error represents the range of values above and below the sample statistic in a confidence interval. It is a crucial component in sample size estimation because it helps us understand the precision of our estimate.
In simpler terms, the margin of error shows how much our sample results might vary from the actual population parameter.
  • For example, if you have a margin of error of 2 minutes, it means the true mean could be 2 minutes more or less than the sample mean.
  • A smaller margin of error provides a more precise estimate, but it often requires a larger sample size.
Achieving a desired margin of error involves balancing three factors: the confidence level, the population standard deviation, and the sample size. The formula used in the original exercise directly reflects this triad, combining them to calculate the needed sample size for specific margins of error.
Z-score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It plays a crucial role in determining the size of the sample needed for estimating a population parameter.
In the context of sample size estimation, a Z-score is directly linked to the confidence level. It essentially translates the percentage of confidence into a standard score that can be used in statistical formulas.
  • For a 95% confidence level, the corresponding Z-score is 1.96.
  • For a 99% confidence level, this Z-score increases to 2.576.
By using these Z-scores in the sample size calculation formula, we ensure that our sample's estimated interval will indeed reflect the chosen confidence level. This means that as the confidence level grows, so does the Z-score, which in turn increases the required sample size.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of sample size estimation, knowing the population standard deviation allows us to assess how spread out the data points in the population are.
This measurement is pivotal because it affects the size of the sample needed to achieve a desired margin of error at a specific confidence level. In general:
  • A higher standard deviation indicates more variation in the data, implying a larger sample size is needed to accurately estimate the population mean.
  • A lower standard deviation suggests less variation, potentially requiring a smaller sample size.
Using the standard deviation in conjunction with confidence levels and the margin of error gives a clear picture of the sample size required to make reliable and precise estimates. It's the reason why the original problem used a planning value for the standard deviation of eight minutes to estimate the sample sizes needed with different confidence levels.

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Most popular questions from this chapter

Playbill magazine reported that the mean annual household income of its readers is \(\$ 119,155\) (Playbill, January 2006 ). Assume this estimate of the mean annual household income is based on a sample of 80 households, and based on past studies, the population standard deviation is known to be \(\sigma=\$ 30,000\) a. Develop a \(90 \%\) confidence interval estimate of the population mean. b. Develop a \(95 \%\) confidence interval estimate of the population mean. c. Develop a \(99 \%\) confidence interval estimate of the population mean. d. Discuss what happens to the width of the confidence interval as the confidence level is increased. Does this result seem reasonable? Explain.

A survey of small businesses with Web sites found that the average amount spent on a site was \(\$ 11,500\) per year (Fortune, March 5, 2001). Given a sample of 60 businesses and a population standard deviation of \(\sigma=\$ 4000\), what is the margin of error? Use \(95 \%\) confidence. What would you recommend if the study required a margin of error of \(\$ 500 ?\)

The National Center for Education Statistics reported that \(47 \%\) of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. a. Provide a \(95 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. b. Provide a \(99 \%\) confidence interval for the population proportion of college students who work to pay for tuition and living expenses. c. What happens to the margin of error as the confidence is increased from \(95 \%\) to \(99 \% ?\)

A simple random sample of 50 items from a population with \(\sigma=6\) resulted in a sample mean of 32 a. Provide a \(90 \%\) confidence interval for the population mean. b. Provide a \(95 \%\) confidence interval for the population mean. c. Provide a \(99 \%\) confidence interval for the population mean.

For a \(t\) distribution with 16 degrees of freedom, find the area, or probability, in each region. a. To the right of 2.120 b. To the left of 1.337 c. To the left of -1.746 d. To the right of 2.583 e. Between -2.120 and 2.120 f. Between -1.746 and 1.746

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