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Chapter 8: Problem 47

A 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?

Short Answer

Expert verified
The point estimate of \(\mu\) is the sample mean from step 1. The 95% confidence interval for \(\mu\) is the result from step 4. The margin of error of the estimate for \(\mu\) is the result from step 5.

Step by step solution

01

Calculate the sample mean

First, add all the 11 data numbers together and then divide by 11 to find the mean of the sample. The sample mean will be the point estimate of \(\mu\).
02

Calculate the standard deviation

To calculate the standard deviation, subtract the mean from each data point, square the result, sum them up, divide the sum by the number of data points minus 1 (i.e., 10), and then take the square root.
03

Calculate the z-score

The z-score for a 95% confidence level in a normal distribution is approximately 1.96, determined from the z-table or from statistical software or calculators.
04

Calculate the confidence interval

To calculate the 95% confidence interval for \(\mu\), subtract and add the product of the z-score and the standard deviation divided by the square root of the sample size to the sample mean. The result will be the lower and upper boundaries of the confidence interval.
05

Determine the margin of error

The margin of error of the estimate for \(\mu\) in part b is half of the width of the confidence interval. Therefore, subtract the lower boundary of the confidence interval from the upper boundary and divide by 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point Estimate
The point estimate is a single value used to estimate a population parameter. In this exercise, we're interested in estimating the population mean, denoted by \( \mu \). To find the point estimate for \( \mu \), we use the sample mean. The sample mean is calculated by adding all the observations in the sample and dividing by the number of observations. This calculated mean serves as our point estimate for \( \mu \), providing a simple but powerful summary of the sample data.
Sample Mean
The sample mean is a foundational concept in statistics, representing the average value of a set of data points. To find it, add together all the values in your sample, and then divide by the number of values. This value is often symbolized by \( \bar{x} \).- **Formula**: \[ \bar{x} = \frac{\sum x_i}{n} \] where \( x_i \) represents each value in the sample and \( n \) is the number of values. The sample mean is used in various calculations, including forming the basis for the confidence interval and point estimate.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean of the set, while a high standard deviation means that the values are spread out.Here鈥檚 how you calculate the standard deviation:1. Subtract the sample mean from each data point.2. Square the result for each calculation. 3. Sum all the squared results.4. Divide by \( n - 1 \) (where \( n \) is the number of data points).5. Take the square root of the result to get the standard deviation.This tells us how much the individual data points deviate, on average, from the mean.
Margin of Error
The margin of error is a crucial component of a confidence interval. It provides a range that the true population parameter is likely to fall within.To find the margin of error, you need:- **Z-score**: For a 95% confidence level, this is approximately 1.96.- **Standard deviation**: Dispersion measure of your sample.- **Sample size**: The number of observations. The formula to calculate the margin of error is: \( \text{Margin of Error} = z \times \left(\frac{\text{standard deviation}}{\sqrt{n}}\right) \)It is half the width of the confidence interval, thereby indicating the uncertainty level in your point estimate for \( \mu \).

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Most popular questions from this chapter

A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a sample of 50 orders and found that 35 of them were mailed within 72 hours of the placement of the orders. a. Construct a \(98 \%\) confidence interval for the percentage of all orders that are mailed within 72 hours of their placement. 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 alternative is the best?

a. A sample of 1100 observations taken from a population produced a sample proportion of \(.32 .\) Make a \(90 \%\) confidence interval for \(p\). b. Another sample of 1100 observations taken from the same population produced a sample proportion of .36. Make a \(90 \%\) confidence interval for \(p\). c. A third sample of 1100 observations taken from the same population produced a sample proportion of .30. Make a \(90 \%\) confidence interval for \(p\). d. The true population proportion for this population is \(.34 .\) Which of the confidence intervals constructed in parts a through c cover this population proportion and which do not?

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?

A group of veterinarians wants to test a new canine vaccine for Lyme disease. (Lyme disease is transmitted by the bite of an infected deer tick.) In an area that has a high incidence of Lyme disease, 100 dogs are randomly selected (with their owners' permission) to receive the vaccine. Over a 12 -month period, these dogs are periodically examined by veterinarians for symptoms of Lyme disease. At the end of 12 months, 10 of these 100 dogs are diagnosed with the disease. During the same 12 -month period, \(18 \%\) of the unvaccinated dogs in the area have been found to have Lyme disease. Let \(p\) be the proportion of all potential vaccinated dogs who would contract Lyme disease in this area. a. Find a \(95 \%\) confidence interval for \(p\). b. Does \(18 \%\) lie within your confidence interval of part a? Does this suggest the vaccine might or might not be effective to some degree? c. Write a brief critique of this experiment, pointing out anything that may have distorted the results or conclusions.

An article in the Los Angeles Times (latimesblogs.latimes.com/pardonourdust/) quoted from the National Association of Realtors that "we now sell our homes and move an average of every six years." Suppose that the average time spent living in a house prior to selling it for a random sample of 400 recent home sellers was \(6.18\) years and the sample standard deviation was \(2.87\) years. a. What is the point estimate of the corresponding population mean? b. Construct a \(98 \%\) confidence interval for the average time spent living in a house prior to selling it for all home owners. What is the margin of error for this estimate?
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