/*! 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 18 For a population, the value of t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 18

For a population, the value of the standard deviation is \(4.96\). A sample of 32 observations taken from this population produced the following data. \(\begin{array}{llllllll}74 & 85 & 72 & 73 & 86 & 81 & 77 & 60 \\ 83 & 78 & 79 & 88 & 76 & 73 & 84 & 78 \\ 81 & 72 & 82 & 81 & 79 & 83 & 88 & 86 \\ 78 & 83 & 87 & 82 & 80 & 84 & 76 & 74\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(99 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for part b?

Short Answer

Expert verified
a. The point estimate of \(\mu\) is \(79.09375\). b. The 99% confidence interval for \(\mu\) is [76.84, 81.35]. c. The margin of error for part b is \(2.25\).

Step by step solution

01

Calculate the point estimate

The point estimate of \(\mu\) is the sample mean (average). Add up all the observations and divide by the number of observations to calculate it. In this case, summing up all 32 measures and dividing by 32 gives \(79.09375\) as the point estimate of \(\mu\).
02

Calculate the 99% confidence interval

The confidence interval is found by adding and subtracting the margin of error from the mean. The formula for the margin of error is \(Z*\frac{\sigma}{\sqrt{n}}\), where \(Z\) is the Z-score associated with desired level of confidence (99%), \(\sigma\) is the known standard deviation (4.96), and \(n\) is the number of items in the sample (32). The Z-score for a 99% confidence level is about 2.58. After plugging into the formula, we get the margin of error as \(2.25\). Adding and subtracting this from the sample mean gives the 99% confidence interval as [76.84, 81.35].
03

Identify the margin of error

The margin of error is what we calculated to find the 99% confidence interval. In this case, the margin of error is \(2.25\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Standard Deviation
Standard deviation is a statistical measure that shows how much variation or dispersion exists from the average (mean). Simply put, it tells us how spread out the numbers in a data set are.

A low standard deviation indicates that most of the numbers are close to the mean, while a high standard deviation suggests a wider spread of values. In the context of confidence intervals, the standard deviation helps determine the margin of error, which we'll discuss later. In the provided exercise, the standard deviation of the population is 4.96, which is used to calculate the margin of error for constructing a confidence interval.
Z-score
The Z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. It is useful for standardizing scores on different scales.

In the setting of confidence intervals, the Z-score helps us understand how far a point is from the mean in terms of standard deviations. For a 99% confidence level, the Z-score is approximately 2.58. This means that about 99% of data should fall within 2.58 standard deviations from the mean. Using this Z-score, we can calculate the margin of error and, consequently, the confidence interval, making Z-scores a vital tool in statistics.
Margin of Error
The margin of error represents the amount of random sampling error in a survey's results. It gives a range within which we can expect the population parameter to lie with a certain level of confidence.

For calculating the margin of error, we use the formula \(Z*\frac{\sigma}{\sqrt{n}}\), where \(Z\) is the Z-score, \(\sigma\) is the standard deviation, and \(n\) is the sample size. In the exercise, with a Z-score of 2.58, a standard deviation of 4.96, and a sample size of 32, the margin of error is calculated to be 2.25. This value is crucial as it helps construct the confidence interval, indicating how accurately the sample mean estimates the population mean.
Point Estimate
A point estimate is a single value given as an estimate of a population parameter. In our exercise, the point estimate is the sample mean, which represents our best guess of the population mean \(\mu\).

To find the point estimate, sum all sample values and divide by the sample size. In this example, adding all data points and dividing by 32 gives a sample mean of approximately 79.09. This point estimate serves as the central value of our confidence interval, around which we calculate the margin of error to determine the interval within which the true population mean is likely to lie with a specified confidence level.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

At the end of Section \(8.3\), we noted that we always round up when calculating the minimum sample size for a confidence interval for \(\mu\) with a specified margin of error and confidence level. Using the formula for the margin of error, explain why we must always round up in this situation.

A computer company that recently developed a new software product wanted to estimate the mean time taken to learn how to use this software by people who are somewhat familiar with computers. A random sample of 12 such persons was selected. The following data give the times taken (in hours) by these persons to learn how to use this software. $$ \begin{array}{llllll} 1.75 & 2.25 & 2.40 & 1.90 & 1.50 & 2.75 \\ 2.15 & 2.25 & 1.80 & 2.20 & 3.25 & 2.60 \end{array} $$ Construct a \(95 \%\) confidence interval for the population mean. Assume that the times taken by all persons who are somewhat familiar with computers to learn how to use this software are approximately normally distributed.

\(8.104\) 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, \(\mu\).

A gas station attendant would like to estimate \(p\), the proportion of all households that own more than two vehicles. To obtain an estimate, the attendant decides to ask the next 200 gasoline customers how many vehicles their households own. To obtain an estimate of \(p\), the attendant counts the number of customers who say there are more than two vehicles in their households and then divides this number by 200. How would you critique this estimation procedure? Is there anything wrong with this procedure that would result in sampling and/or nonsampling errors? If so, can you suggest a procedure that would reduce this error?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.