/*! 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

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods vary slightly. It is known that when the machine is working properly, the mean length of the rods made on this machine is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to \(.10\) inch. The quality control department takes a sample of 20 such rods every week, calculates the mean length of these rods, and makes a \(99 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(36.05\) inches or the lower limit of this confidence interval is less than \(35.95\) inches, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of \(36.02\) inches. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the lengths of all such rods have a normal distribution.

a. A sample of 300 observations taken from a population produced a sample proportion of .63. Make a \(95 \%\) confidence interval for \(p\). b. Another sample of 300 observations taken from the same population produced a sample proportion of .59. Make a \(95 \%\) confidence interval for \(p\). c. A third sample of 300 observations taken from the same population produced a sample proportion of .67. Make a \(95 \%\) confidence interval for \(p\). d. The true population proportion for this population is .65. Which of the confidence intervals constructed in parts a through c cover this population proportion and which do not?

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?

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)

Check if the sample size is large enough to use the normal distribution to make a confidence interval for \(p\) for each of the following cases. a. \(n=80\) and \(\hat{p}=.85\) b. \(n=110\) and \(\hat{p}=.98\) c. \(n=35\) and \(\hat{p}=.40\) d. \(n=200\) and \(\hat{p}=.08\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

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.