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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

A computer company that recently developed a new software product wanted to estimate the mean time taken to learn how to use this software by people who are somewhat familiar with computers. A random sample of 12 such persons was selected. The following data give the times taken (in hours) by these persons to learn how to use this software. $$ \begin{array}{llllll} 1.75 & 2.25 & 2.40 & 1.90 & 1.50 & 2.75 \\ 2.15 & 2.25 & 1.80 & 2.20 & 3.25 & 2.60 \end{array} $$ Construct a \(95 \%\) confidence interval for the population mean. Assume that the times taken by all persons who are somewhat familiar with computers to learn how to use this software are approximately normally distributed.

a. How large a sample should be selected so that the margin of error of estimate for a \(99 \%\) confidence interval for \(p\) is \(.035\) when the value of the sample proportion obtained from a preliminary sample is \(.29\) ? b. Find the most conservative sample size that will produce the margin of error for a \(99 \%\) confidence interval for \(p\) equal to \(.035\).

A gas station attendant would like to estimate \(p\), the proportion of all households that own more than two vehicles. To obtain an estimate, the attendant decides to ask the next 200 gasoline customers how many vehicles their households own. To obtain an estimate of \(p\), the attendant counts the number of customers who say there are more than two vehicles in their households and then divides this number by 200. How would you critique this estimation procedure? Is there anything wrong with this procedure that would result in sampling and/or nonsampling errors? If so, can you suggest a procedure that would reduce this error?

\(8.104\) A random sample of 25 life insurance policyholders showed that the average premium they pay on their life insurance policies is \(\$ 685\) per year with a standard deviation of \(\$ 74\). Assuming that the life insurance policy premiums for all life insurance policyholders have a normal distribution, make a \(99 \%\) confidence interval for the population mean, \(\mu\).

What assumptions must hold true to use the \(t\) distribution to make a confidence interval for \(\mu ?\)
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