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Chapter 8: Problem 85

It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 240 such companies showed that 96 of them provide such facilities on site. a. What is the point estimate of the percentage of all such companies that provide such facilities on site? b. Construct a \(97 \%\) confidence interval for the percentage of all such companies that provide such facilities on site. What is the margin of error for this estimate?

Short Answer

Expert verified
a. The point estimate is 40%. b. The 97% confidence interval is (33.18%, 46.82%), with a margin of error of 6.86%.

Step by step solution

01

Calculate the Point Estimate

The point estimate is calculated by dividing the number of successes (companies providing health facilities) by the total number of trials (total companies sampled). In this case, it involves \( \frac{96}{240} = 0.4 \). Therefore, the point estimate is 0.4 or 40%.
02

Determine the Appropriate Z-value

The z-value to construct a 97% confidence interval can be found using a standard z-table or calculator. The table value for a 97% interval, which leaves 3% in the tails (1.5% in each tail), corresponds to a z-value of approximately 2.17.
03

Calculate the standard error

The standard error is calculated using the formula: \( SE = \sqrt{p(1 - p) / n} \) where p is the sample proportion and n is the total sample size. So, \( SE = \sqrt{0.4(1 - 0.4) / 240} \approx 0.0316 \)
04

Constructing the Confidence Interval

We use the formula for constructing the confidence interval for population proportions which is \( p \pm Z * SE \) where p is the population proportion, Z is the z-value and SE is the standard error. Thus, the confidence interval is \( 0.4 \pm 2.17 * 0.0316 = (0.3318, 0.4682) \)
05

Calculate the Margin of Error

The margin of error here is the amount added and subtracted from the point estimate to create the confidence interval. This is \( Z * SE = 2.17 * 0.0316 = 0.0686 \)

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Most popular questions from this chapter

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. 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 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 normal.

Suppose, for a sample selected from a normally distributed population, \(\bar{x}=68.50\) and \(s=8.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=16\). b. Construct a \(90 \%\) confidence interval for \(\mu\) assuming \(n=16 .\) Is the width of the \(90 \%\) confidence interval smaller 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 \(n=25 .\) Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=25\) smaller than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=16\) calculated in part a? If so, why? Explain.

Determine the most conservative sample size for the estimation of the population proportion for the following. a. \(E=.025\), confidence level \(=95 \%\) b. \(E=.05, \quad\) confidence level \(=90 \%\) c. \(E=.015\), confidence level \(=99 \%\)

Refer to Exercise \(8.116\). Assume that a preliminary sample has shown that \(63 \%\) of the adults in this city favor legalized casino gambling. How large should the sample size be so that the \(95 \%\) confidence interval for the population proportion has a margin of error of \(.05 ?\)

A random sample of 300 female members of health clubs in Los Angeles showed that they spend, on average, \(4.5\) hours per week doing physical exercise with a standard deviation of \(.75\) hour. Find a \(98 \%\) confidence interval for the population mean.
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