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A researcher wants to determine a \(99 \%\) confidence interval for the mean number of hours that adults spend per week doing community service. How large a sample should the researcher select so that the estimate is within \(1.2\) hours of the population mean? Assume that the standard deviation for time spent per week doing community service by all adults is 3 hours.

Short Answer

Expert verified
The researcher should select a sample of size 30 to construct a 99% confidence interval for the mean number of hours adults spend per week doing community service.

Step by step solution

01

Understanding the information given

What we know from the problem: \n1. Confidence level is 99% which makes the significance level 0.1 (1-0.99 = 0.01).\n2. The corresponding z score for this confidence level from the z-table or calculator is approximately 2.576.\n3. Standard deviation, \( \sigma \), is given as 3 hours.\n4. Desired Margin of error (E) is 1.2 hours.
02

Apply the formula for sample size

The formula to calculate the sample size 'n' for a confidence interval is given by: \[ n = \left( \frac{z \cdot \sigma}{E} \right)^2 \] Here, z is the z score, \( \sigma \) is the standard deviation and E is the margin of error.
03

Substitute values into the formula

Let's substitute the calculated and given values into the formula: Hence, \[ n = \left( \frac{2.576 \cdot 3}{1.2} \right)^2 \] Now simplify that to find the sample size.
04

Calculate the Sample size

After performing the above calculation, the solution comes out to be approximately 29.24. However, we cannot have a fraction of a person, so you generally always round up to the next whole number to ensure the desired confidence level. So the required sample size is 30.

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Key Concepts

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

Sample Size Calculation
When planning a study or experiment, it is crucial to determine how large your sample should be to achieve a desired level of confidence in your results. This process is called sample size calculation. The sample size refers to the number of observations or data points that you need to collect from your population to ensure that the results of your study are representative and reliable.Sample size is essential in statistics because:
  • It impacts the precision of your estimates. Larger samples tend to be more accurate.
  • It helps control the costs and resources needed for the study, as larger samples require more time and expense.
To calculate the minimum sample size needed, we use the formula: \[ n = \left( \frac{z \cdot \sigma}{E} \right)^2 \]Here, \( n \) is the sample size, \( z \) is the Z-score, \( \sigma \) is the standard deviation, and \( E \) is the margin of error.By plugging in values into this formula, researchers can determine the number of subjects required to achieve a specified precision in their confidence interval.
Z-score
A Z-score in statistics is a measurement that describes a data point's relationship to the mean of a group of values. It is expressed in terms of standard deviations away from the mean. In simpler terms, the Z-score tells you how far and in what direction a data point deviates from the average.
  • A Z-score of 0 indicates that the data point is exactly at the mean.
  • A positive Z-score signifies that the data point is above the mean.
  • A negative Z-score shows it is below the mean.
In the context of confidence intervals, the Z-score is vital because it helps determine how confident we can be about our estimates. To find the appropriate Z-score for a desired confidence level, we refer to a Z-table or use statistical software. For instance, a 99% confidence interval corresponds to a Z-score of approximately 2.576. This Z-score is essential for calculating the sample size and ensuring that the results meet the desired level of accuracy.
Standard Deviation
Standard Deviation, represented by the symbol \( \sigma \), is a measure of the amount of variation or dispersion in a set of values. Simply put, it tells us how spread out the numbers are in a data set. A low standard deviation means that the values tend to be close to the mean of the data set, whereas a high standard deviation indicates that the values are more spread out over a wider range.
  • In our initial exercise, the standard deviation was given as 3 hours, indicating the variability in the time spent by adults on community service per week.
  • The standard deviation plays a critical role in confidence interval calculations, as it affects how precise your estimate of the population mean will be.
In formulas, such as the one used to calculate sample size, the standard deviation serves as a scale variable that directly influences the number of observations required.
Margin of Error
The margin of error is a statistical term that indicates the amount of error you can tolerate in your estimates. It's the "buffer zone" that surrounds your estimate, within which the true population parameter is likely to fall. Essentially, it's a way to quantify the uncertainty of your results. The margin of error affects the width of the confidence interval: the larger the margin of error, the less precise your estimate is.
  • In the given exercise, the desired margin of error is 1.2 hours. Thus, the estimate of the mean time allowance includes this range around it.
  • The margin of error is determined by the standard deviation and the Z-score, based on the desired confidence level.
The formula to compute the margin of error for a confidence interval is \[ E = \frac{z \cdot \sigma}{\sqrt{n}} \]Where \( z \) is the Z-score, \( \sigma \) is the standard deviation, and \( n \) is the sample size. Understanding the margin of error is crucial for interpreting the results of a statistical analysis and for making informed conclusions.

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Most popular questions from this chapter

A consumer agency that proposes that lawyers' rates are too high wanted to estimate the mean hourly rate for all lawyers in New York City. A sample of 70 lawyers taken from New York City showed that the mean hourly rate charged by them is \(\$ 420\). The population standard deviation of hourly charges for all lawyers in New York City is \(\$ 110\). a. Construct a \(99 \%\) confidence interval for the mean hourly charges for all lawyers in New York City. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

According to liposuction4you.com, the maximum amount of fat and fluid that can be removed safely during a liposuction procedure is 6 liters. Suppose that the following data represent the amount of fat and fluid removed during 12 randomly selected liposuction procedures. Assume that the population distribution of such amounts is normal. $$ \begin{array}{llllll} 1.84 & 2.66 & 2.96 & 2.42 & 2.88 & 2.86 \\ 3.66 & 3.65 & 2.33 & 2.66 & 3.20 & 2.24 \end{array} $$ a. What is the point estimate of the corresponding population mean? b. Construct a \(98 \%\) confidence interval for the corresponding population mean.

When one is attempting to determine the required sample size for estimating a population mean, and the information on the population standard deviation is not available, it may be feasible to take a small preliminary sample and use the sample standard deviation to estimate the required sample size, \(n .\) Suppose that we want to estimate \(\mu\), the mean commuting distance for students at a community college, to within 1 mile with a confidence level of \(95 \%\). A random sample of 20 students yields a standard deviation of \(4.1\) miles. Use this value of the sample standard deviation, \(s\), to estimate the required sample size, \(n\). Assume that the corresponding population has a normal distribution.

The mean time taken to design a house plan by 40 architects was found to be 23 hours with a standard deviation of \(3.75\) hours. a. Construct a \(98 \%\) confidence interval for the population mean \(\mu\). b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

Because of inadequate public school budgets and lack of money available to teachers for classroom materials, many teachers often use their own money to buy materials used in the classrooms. A random sample of 100 public school teachers selected from an eastern state showed that they spent an average of \(\$ 273\) on such materials during the 2009 school year. The population standard deviation was \$60. a. What is the point estimate of the mean of such expenses incurred during the 2009 school year by all public school teachers in this state? b. Make a \(95 \%\) confidence interval for the corresponding population mean.

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