/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 115 An economist wants to find a \(9... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 115

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 \(\$ 31,500\).

Short Answer

Expert verified
The economist needs to select a sample size of 51 houses to get within \(\$3500\) of the population mean, with a confidence level of \(90\%\).

Step by step solution

01

Identify the known values

In this problem, we know the confidence level (which is \(90\%\), so the significance level is \(1-0.90 = 0.10\) or \(10\% \)), the standard deviation (\(\$31,500\)), and the error margin (\(\$3500\)). We can use these known values to calculate the critical value Z.
02

Find the critical value Z

Since, the confidence level is \(90\%\), the significance level \(\alpha\) is \(0.10\). Thus half of the significance level \(\(\alpha/2\) is \(0.05\). Looking this up in the Z-table, we find that the critical value \(Z_{\alpha/2}\) is \(1.645\).
03

Apply the formula for sample size

Now, it's time to insert these values into the formula for sample size when estimating means: \(n = \left( \frac{Z_{\alpha/2} . \sigma}{E} \right)^{2}\). Replacing \(Z_{\alpha/2}\) by \(1.645\), \(\sigma\) by \(\$31,500\), and \(E\) by \(\$3500\), the equation is: \(n = \left( \frac{1.645 * 31500}{3500} \right)^{2}\).
04

Calculate the sample size

This gives us: \(n = \left( \frac{1.645 * 31500}{3500} \right)^{2} \approx 50.8\). We must then round to the nearest whole number since we can't select a fraction of a house, and the sample size needed is \(n = 51\).

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Most popular questions from this chapter

A dentist wants to find the average time taken by one of her hygienists to take X-rays and clean teeth for patients. She recorded the time to serve 24 randomly selected patients by this hygienist. The data (in minutes) are as follows: \(\begin{array}{llllllllllll}36.80 & 39.80 & 38.60 & 38.30 & 34.30 & 32.60 & 38.70 & 34.50 & 37.00 & 36.80 & 40.90 & 33.80 \\ 37.10 & 33.00 & 35.10 & 38.20 & 36.60 & 38.80 & 39.60 & 39.70 & 35.10 & 38.20 & 32.70 & 39.50\end{array}\) Assume that such times for this hygienist for all patients are approximately normal. a. What is the point estimate of the corresponding population mean. b. Construct a \(99 \%\) confidence interval for the average time taken by this hygienist to take X-rays and to clean teeth for all patients.

An entertainment company is in the planning stages of producing a new computer-animated movie for national release, so they need to determine the production time (labor-hours necessary) to produce the movie. The mean production time for a random sample of 14 big-screen computer-animated movies is found to be 53,550 labor-hours. Suppose that the population standard deviation is known to be 7462 laborhours and the distribution of production times is normal. a. Construct a \(98 \%\) confidence interval for the mean production time to produce a big-screen computer-animated movie. b. Explain why we need to make the confidence interval. Why is it not correct to say that the average production time needed to produce all big-screen computer-animated movies is 53,550 labor-hours?

In a Time/Money Magazine poll of Americans of age 18 years and older, \(65 \%\) agreed with the statement, "We are less sure our children will achieve the American Dream" (Time, October 10,2011 ). Assume that this poll was based on a random sample of 1600 Americans. a. Construct a \(95 \%\) confidence interval for the proportion of all Americans of age 18 years and older who will agree with the aforementioned statement. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(65 \%\) of all Americans of age 18 years and older agree with the aforementioned statement?

A random sample of 25 life insurance policyholders showed that the average premium they pay on their life insurance policies is \(\$ 685\) per year with a standard deviation of \(\$ 74\). Assuming that the life insurance policy premiums for all life insurance policyholders have a normal distribution, make a \(99 \%\) confidence interval for the population mean, \(\mu\).

Almost all employees working for financial companies in New York City receive large bonuses at the end of the year. A sample of 65 employees selected from financial companies in New York City showed that they received an average bonus of \(\$ 55,000\) last year with a standard deviation of \(\$ 18,000\). Construct a \(95 \%\) confidence interval for the average bonus that all employees working for financial companies in New York City received last year.
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