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Chapter 8: Problem 30

A marketing researcher wants to find a \(95 \%\) confidence interval for the mean amount that visitors to a theme park spend per person per day. She knows that the standard deviation of the amounts spent per person per day by all visitors to this park is \(\$ 11\). How large a sample should the researcher select so that the estimate will be within \(\$ 2\) of the population mean?

Short Answer

Expert verified
To determine how large a sample the researcher should select so that the estimate will be within $2 of the population mean with 95% confidence, she should use formula to calculate sample size and replace all known values in. After calculation, the sample size will be \(n\), the result of the formula, rounded up to the nearest whole number.

Step by step solution

01

Identify the correct formula

For a confidence interval with known standard deviation, we use this formula to calculate the necessary sample size: \(n = (\frac{Z_{\frac{\alpha}{2}} * \sigma}{E})^2\). Here, \(Z_{\frac{\alpha}{2}}\) is the z-value associated with the desired confidence level, \(\sigma\) is the population standard deviation, and \(E\) is the desired margin of error (how close we would like to come to the population mean).
02

Find the Z-value for the 95% Confidence Level

The z-value for a 95% confidence interval is 1.96 (based on the z-table or using a statistical calculator or software). This is because 95% of the data lies within 1.96 standard deviations from the mean in a normal distribution. Therefore, \(Z_{\frac{\alpha}{2}} = 1.96\).
03

Substituting values into formula

The standard deviation \(\sigma\) is $11 and the desired margin of error \(E\) is $2. Substitute these values and the Z-value from step 2 into the sample size formula: \(n = (\frac{1.96 * 11}{2})^2\).
04

Calculating the sample size

Performing the calculation will yield the minimum sample size necessary to ensure that our estimate is within $2 of the population mean with 95% confidence. Remember to always round up to the nearest whole number, because you can't have a fraction of a person in the sample.

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Most popular questions from this chapter

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)

The high cost of health care is a matter of major concern for a large number of families. A random sample of 25 families selected from an area showed that they spend an average of \(\$ 253\) per month on health care with a standard deviation of \(\$ 47 .\) Make a \(98 \%\) confidence interval for the mean health care expenditure per month incurred by all families in this area. Assume that the monthly health care expenditures of all families in this area have a normal distribution.

A bank manager wants to know the mean amount of mortgage paid per month by homeowners in an area. A random sample of 120 homeowners selected from this area showed that they pay an average of \(\$ 1575\) per month for their mortgages. The population standard deviation of such mortgages is \(\$ 215\) a. Find a \(97 \%\) confidence interval for the mean amount of mortgage paid per month by all homeowners in this area. 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?

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=2.467\) and \(d f=28\) b. \(t=-1.672\) and \(d f=58\) c. \(t=-2.670\) and \(n=55\) d. \(t=2.383\) and \(n=23\)

A consumer agency wants to estimate the proportion of all drivers who wear seat belts while driving. Assume that a preliminary study has shown that \(76 \%\) of drivers wear seat belts while driving. How large should the sample size be so that the \(99 \%\) confidence interval for the population proportion has a margin of error of \(.03\) ?
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