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Chapter 8: Problem 77

A survey of 500 randomly selected adult men showed that the mean time they spend per week watching sports on television is \(9.75\) hours with a standard deviation of \(2.2\) hours. Construct a \(90 \%\) confidence interval for the population mean, \(\mu\).

Short Answer

Expert verified
The 90% confidence interval for the population mean time adult men spend per week watching sports on television is between 9.588 and 9.912 hours.

Step by step solution

01

Identify given values

From the exercise, it can be identified that the sample mean, \(\bar{x}\), is 9.75 hours, the standard deviation for the sample, \(s\), is 2.2 hours and the size of the sample, \(n\), is 500.
02

Determine the z-score

The z-score associated with a 90% confidence interval can be looked up on a Z-table, found in most statistics textbooks, or calculated using statistical software. The chosen level of confidence is 90%, which leaves 5% in each tail of the distribution. Thus, a z-score for a 90% confidence level is \(Z = 1.645\).
03

Apply the confidence interval formula

You should now insert the given values into the confidence interval formula: \(CI = \bar{x} \pm z\times \frac{s}{\sqrt{n}} = 9.75 \pm 1.645\times \frac{2.2}{\sqrt{500}}\). Calculating the expressions in the formula then gives the interval.
04

Calculate the confidence interval

Performing the operations gives the confidence interval \(CI = 9.75 \pm 0.162\), which can be expressed as \((9.588, 9.912)\). This means with 90% confidence, the average time adult men spend watching sports on television is between \(9.588\) and \(9.912\) hours per week.

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Most popular questions from this chapter

A large city with chronic economic problems is considering legalizing casino gambling. The city council wants to estimate the proportion of all adults in the city who favor legalized casino gambling. What is the most conservative estimate of the minimum sample size that would limit the margin of error to be within \(.05\) of the population proportion for a \(95 \%\) confidence interval?

You are working for a bank. The bank manager wants to know the mean waiting time for all customers who visit this bank. She has asked you to estimate this mean by taking a sample. Briefly explain how you will conduct this study. Collect data on the waiting times for 45 customers who visit a bank. Then estimate the population mean. Choose your own confidence level.

A random sample of 11 observations taken from a normally distributed population produced the following data: $$ \begin{array}{lllllllllll} -7.1 & 10.3 & 8.7 & -3.6 & -6.0 & -7.5 & 5.2 & 3.7 & 9.8 & -4.4 & 6.4 \end{array} $$ a. What is the point estimate of \(\mu\) ? b. Make a \(95 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for \(\mu\) in part b?

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

A city planner wants to estimate the average monthly residential water usage in the city at a \(97 \%\) confidence level. Based on earlier data, the population standard deviation of the monthly residential water usage in this city is \(389.60\) gallons. How large a sample should be selected so that the estimate for the average monthly residential water usage in this city is within 100 gallons of the population mean?
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