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Chapter 8: Problem 3

What is the point estimator of the population mean, \(\mu\) ? How would you calculate the margin of error for an estimate of \(\mu\) ?

Short Answer

Expert verified
The point estimator of the population mean, \(\mu\), is the sample mean, \(\bar{x}\). The margin of error of the estimate can be calcualted by the formula \( E=Z \frac{\sigma}{\sqrt{n}} \).

Step by step solution

01

Point Estimator Calculation

The point estimator of the population mean, \(\mu\), is the sample mean, \(\bar{x}\). It is calculated by adding all the data values and dividing by the number of data values.
02

Margin of Error Calculation

The margin of error for an estimate of \(\mu\) is given by the formula \( E=Z \frac{\sigma}{\sqrt{n}} \), where \(E\) is the margin of error, \(Z\) is the z-value (which depends on the confidence level), \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size. This formula assumes that the sampling distribution of the mean is approximately normally distributed (Central Limit Theorem), and that the population standard deviation is known.

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Most popular questions from this chapter

In a random sample of 50 homeowners selected from a large suburban area, 19 said that they had serious problems with excessive noise from their neighbors. a. Make a \(99 \%\) confidence interval for the percentage of all homeowners in this suburban area who have such problems. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which option is best?

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. Find the value of \(t\) from the \(t\) distribution table for a sample size of 22 and a confidence level of \(95 \%\). b. Find the value of \(t\) from the \(t\) distribution table for 60 degrees of freedom and a \(90 \%\) confidence level. c. Find the value of \(t\) from the \(t\) distribution table for a sample size of 24 and a confidence level of \(99 \%\).

According to a Pew Research Center nationwide telephone survey of adults conducted March 15 to April 24, 2011, 55\% of college graduates said that their college education prepared them for a job (Time, May 30,2011 ). Suppose that this survey included 1450 college graduates. a. What is the point estimate of the corresponding population proportion? b. Construct a \(98 \%\) confidence interval for the proportion of all college graduates who will say that their college education prepared them for a job. What is the margin of error for this estimate?

Refer to Exercise \(8.24\). A city planner wants to estimate, with a \(97 \%\) confidence level, the average monthly residential water usage in the city. 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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