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Chapter 8: Problem 107

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.

Short Answer

Expert verified
The 98% confidence interval for the population mean amount of physical exercise done per week by female members of health clubs in Los Angeles is between 4.399 and 4.601 hours.

Step by step solution

01

Identify Given Values

The random sample size (n) is 300, the sample mean (\(\bar{x}\)) is 4.5 hours, and the standard deviation (s) is 0.75 hours.
02

Determine the Z-Score for Confidence Level

A 98% confidence level corresponds to alpha level of 0.02 (since 1 - 0.98 = 0.02), split between two tails of the distribution (since a confidence interval is two-tailed by nature; \(\alpha/2 = 0.01\)). Interpreting from Z-table, this corresponds to a Z score of 2.33.
03

Use the Confidence Interval Formula

Now, it's time to plug into the formula: \(\bar{x} \pm Z(\frac{s}{\sqrt{n}})\). Substituting the values, we get: \(4.5 \pm 2.33*\frac{0.75}{\sqrt{300}}\)
04

Compute Confidence Interval

Compute the calculations to find the confidence interval. This results in a confidence interval of \(4.5 \pm 0.101\), thus the interval is \(4.399, 4.601\)

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Most popular questions from this chapter

The management of a health insurance company wants to know the percentage of its policyholders who have tried alternative treatments (such as acupuncture, herbal therapy, etc.). A random sample of 24 of the company's policyholders were asked whether or not they have ever tried such treatments. The following are their responses. \(\begin{array}{llllllll}\text { Yes } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { Yes } & \text { No } & \text { No } \\ \text { No } & \text { Yes } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { Yes } & \text { No } \\ \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } & \text { No }\end{array}\) a. What is the point estimate of the corresponding population proportion? b. Construct a \(99 \%\) confidence interval for the percentage of this company's policyholders who have tried alternative treatments.

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\).

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.

A sample selected from a population gave a sample proportion equal to . 73 . a. Make a \(99 \%\) confidence interval for \(p\) assuming \(n=100\). b. Construct a \(99 \%\) confidence interval for \(p\) assuming \(n=600\) c. Make a \(99 \%\) confidence interval for \(p\) assuming \(n=1500\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) decrease as the sample size increases? If yes, explain why.

The principal of a large high school is concerned about the amount of time that his students spend on jobs to pay for their cars, to buy clothes, and so on. He would like to estimate the mean number of hours worked per week by these students. He knows that the standard deviation of the times spent per week on such jobs by all students is \(2.5\) hours. What sample size should he choose so that the estimate is within \(.75\) hour of the population mean? The principal wants to use a \(98 \%\) confidence level.
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