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Chapter 8: Problem 68

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 \(.03\) ?

Short Answer

Expert verified
The sample size should be at least 962.

Step by step solution

01

Consider the given population proportion and margin of error

The given population proportion is 0.76 (or 76%) and the desired margin of error (E) is 卤0.03.
02

Identify the Z-score associated with the given confidence level

The problem requires a 99% confidence interval. The Z value for a 99% confidence interval is 2.575, which you can find using a standard normal distribution table.
03

Implement the formula for sample size

The formula for finding sample size (n) when population proportion and margin of error are known is: \[ n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right) \] Here, Z = 2.575, p = 0.76 and E = 0.03. Substituting these values in to the formula,
04

Compute the sample size

After substituting Z, p and E into the formula, the next step is to compute the sample size. This calculation yields a sample size of approximately 962, which is normally rounded upwards. Therefore, the sample size should be 962.
05

Interpret the result

This means that at least 962 drivers should be in the survey sample in order for the margin of error of the survey's estimate to be within 3% at a 99% confidence level.

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Most popular questions from this chapter

A hospital administration wants to estimate the mean time spent by patients waiting for treatment at the emergency room. The waiting times (in minutes) recorded for a random sample of 35 such patients are given below. The population standard deviation is not known. \(\begin{array}{rrrrrrr}30 & 7 & 68 & 76 & 47 & 60 & 51 \\ 64 & 25 & 35 & 29 & 30 & 35 & 62 \\ 96 & 104 & 58 & 32 & 32 & 102 & 27 \\ 45 & 11 & 64 & 62 & 72 & 39 & 92 \\ 84 & 47 & 12 & 33 & 55 & 84 & 36\end{array}\) Construct a \(99 \%\) confidence interval for the corresponding population mean.

An economist wants to find a \(90 \%\) confidence interval for the mean sale price of houses in a state. How large a sample should she select so that the estimate is within \(\$ 3500\) of the population mean? Assume that the standard deviation for the sale prices of all houses in this state is \(\$ 31,500\).

A department store manager wants to estimate the mean amount spent by all customers at this store at a \(98 \%\) confidence level. The manager knows that the standard deviation of amounts spent by all customers at this store is \(\$ 31\). What minimum sample size should he choose so that the estimate is within \(\$ 3\) of the population mean?

Yunan Corporation produces bolts that are supplied to other companies. These bolts are supposed to be 4 inches long. The machine that makes these bolts does not produce each bolt exactly 4 inches long but the length of each bolt varies slightly. It is known that when the machine is working properly, the mean length of the bolts made on this machine is 4 inches. The standard deviation of the lengths of all bolts produced on this machine is always equal to \(.04\) inch. The quality control department takes a random sample of 20 such bolts every week, calculates the mean length of these bolts, and makes a \(98 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(4.02\) inches or the lower limit of this confidence interval is less than \(3.98\) inches, the machine is stopped and adjusted. A recent such sample of 20 bolts produced a mean length of \(3.99\) inches. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the population distribution is approximately normal.

Explain the meaning of a point estimate and an interval estimate.
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