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Chapter 8: Problem 114

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
Therefore, the researcher needs to select a sample size of 37 in order to estimate the population mean within 1.2 hours with a confidence level of 99%.

Step by step solution

01

Identify Z-score

The first step is to identify the z-score associated with a 99% confidence interval. For a 99% confidence interval, the z-score value is 2.575
02

Determine Standard Deviation

The standard deviation indicates the variation of community service per week. This is given as 3 hours.
03

Determine Decimal Equivalent of Desired Precision

The margin error given in the problem is 1.2 hours. To use this in the calculations, convert it to a decimal. Therefore, this gives us 1.2.
04

Calculate Sample Size

Use the formula for finding sample size (\(n = (Z^2 * 蟽^2) / E^2\)), where Z is the Z-score, 蟽 is the standard deviation and E is the margin of error. So, \(n = (2.575^2 * 3^2) / 1.2^2\). When you calculate the given formula, it will yield 36.54. But the sample size is always a whole number, so round it up to 37.

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Most popular questions from this chapter

What is the point estimator of the population proportion, \(p ?\)

Determine the sample size for the estimate of \(\mu\) for the following. a. \(E=2.3, \quad \sigma=15.40\), confidence level \(=99 \%\) b. \(E=4.1, \quad \sigma=23.45\), confidence level \(=95 \%\) c. \(E=25.9, \quad \sigma=122.25\), confidence level \(=90 \%\)

For a population, the value of the standard deviation is 4.96. A sample of 32 observations taken from this population produced the following data. \(\begin{array}{llllllll}74 & 85 & 72 & 73 & 86 & 81 & 77 & 60 \\ 83 & 78 & 79 & 88 & 76 & 73 & 84 & 78 \\ 81 & 72 & 82 & 81 & 79 & 83 & 88 & 86 \\ 78 & 83 & 87 & 82 & 80 & 84 & 76 & 74\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(99 \%\) confidence interval for \(\mu .\) c. What is the margin of error of estimate for part b?

Suppose, for a sample selected from a population, \(\bar{x}=25.5\) and \(s=4.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=47\). b. Construct a \(99 \%\) confidence interval for \(\mu\) assuming \(n=47\). Is the width of the \(99 \%\) confidence interval larger than the width of the \(95 \%\) confidence interval calculated in part a? If yes, explain why. c. Find a \(95 \%\) confidence interval for \(\mu\) assuming \(n=32 .\) Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=32\) larger than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=47\) calculated in part a? If so, why? Explain.

Determine the sample size for the estimate of \(\mu\) for the following. a. \(E=.17, \quad \sigma=.90\), confidence level \(=99 \%\) b. \(E=1.45, \quad \sigma=5.82\) confidence level \(=95 \%\) c. \(E=5.65, \quad \sigma=18.20\), confidence level \(=90 \%\)
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