/*! 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 14 Determine the point estimate of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the point estimate of the population mean and margin of error for each confidence interval. Lower bound: \(20,\) upper bound: 30

Short Answer

Expert verified
Point Estimate: 25, Margin of Error: 5

Step by step solution

01

Understand the Problem

The given problem requires determining the point estimate of the population mean and the margin of error from the provided confidence interval bounds. The lower bound is 20 and the upper bound is 30.
02

Calculate the Point Estimate

The point estimate of the population mean can be found by calculating the average of the lower and upper bounds of the confidence interval. Use the formula:\[ \text{Point Estimate} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \]Substitute the given values:\[ \text{Point Estimate} = \frac{20 + 30}{2} = 25 \]
03

Calculate the Margin of Error

The margin of error can be determined by finding the difference between the upper bound and the lower bound of the confidence interval, and then divide by 2. Use the formula:\[ \text{Margin of Error} = \frac{\text{Upper Bound} - \text{Lower Bound}}{2} \]Substitute the given values:\[ \text{Margin of Error} = \frac{30 - 20}{2} = 5 \]

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 91Ó°ÊÓ!

Key Concepts

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

population mean
The population mean is a crucial concept in statistics. It represents the average value of a particular characteristic in the whole population. For example, if you calculate the average height of all students in a school, this average height is the population mean. Calculating the population mean is often impractical due to the size of the population, so we estimate it using a sample mean from selected data.

This sample mean helps us to make assumptions about the entire population. When we estimate the population mean, we use techniques like confidence intervals to understand how accurate our estimate might be. The true population mean lies within this interval, giving us a range where the actual mean is likely to be. By understanding the population mean, you can make more informed decisions and predictions based on data.
point estimate
A point estimate is a single value that approximates a population parameter. It's like picking a single point in a dart game to hit the target. In the context of population mean, the point estimate would be the average of the sample data, aimed to represent the population mean.

For instance, in our given exercise, to determine the point estimate of the population mean, we used the formula: \( \text{Point Estimate} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \). By substituting the given values (lower bound: 20 and upper bound: 30), we get: \( \text{Point Estimate} = \frac{20 + 30}{2} = 25 \).

This value of 25 is our point estimate, offering an approximation of the actual population mean from our sample data. Remember, a point estimate doesn't provide any information about the accuracy or uncertainty of the estimate. That's why we use confidence intervals along with it.
margin of error
The margin of error measures the range of uncertainty around a point estimate. It tells us how much our point estimate might vary due to sampling variability or inherent statistical errors. It is an essential part of the confidence interval, helping clarify the reliability of the estimate.

In the given exercise, we calculated the margin of error using the formula: \( \text{Margin of Error} = \frac{\text{Upper Bound} - \text{Lower Bound}}{2} \). By substituting the values (upper bound: 30 and lower bound: 20), we have: \( \text{Margin of Error} = \frac{30 - 20}{2} = 5 \).

This means our point estimate (25) could be off by 5 units in either direction. So, the true population mean is likely to be between 20 (25-5) and 30 (25+5).

That’s it! By understanding the margin of error, you can better gauge how close your sample's point estimate might be to the actual population parameter. This helps you make more informed decisions and reduces the risk of incorrect conclusions based on the data.

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

Alan wants to estimate the proportion of adults who walk to work. In a survey of 10 adults, he finds 1 who walk to work. Explain why a \(95 \%\) confidence interval using the normal model yields silly results. Then compute and interpret a \(95 \%\) confidence interval for the proportion of adults who walk to work using Agresti and Coull's method.

A Tootsie Pop is a sucker with a candy center. A famous commercial for Tootsie Pops once asked, "How many licks to the center of a Tootsie Pop?" In an attempt to answer this question, Cory Heid of Siena Heights University asked 92 volunteers to count the number of licks required before reaching the chocolate center. The mean number of licks required was 356.1 with a standard deviation of \(185.7 .\) Find and interpret a \(95 \%\) confidence interval for the number of licks required to reach the candy center of a Tootsie Pop.

What type of variable is required to construct a confidence interval for a population proportion?

The following small data set represents a simple random sample from a population whose mean is \(50 .\) $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline \end{array} $$ (a) A normal probability plot indicates that the data could come from a population that is normally distributed with no outliers. Compute a \(95 \%\) confidence interval for this data set. (b) Suppose that the observation, 41 , is inadvertently entered into the computer as 14 . Verify that this observation is an outlier (c) Construct a \(95 \%\) confidence interval on the data set with the outlier. What effect does the outlier have on the confidence interval? (d) Consider the following data set, which represents a simple random sample of size 36 from a population whose mean is 50\. Verify that the sample mean for the large data set is the same as the sample mean for the small data set from part (a). $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline 47 & 65 & 56 & 58 & 41 & 52 \\ \hline 49 & 56 & 57 & 50 & 38 & 42 \\ \hline 59 & 54 & 57 & 41 & 63 & 37 \\ \hline 46 & 54 & 42 & 48 & 53 & 41 \\ \hline \end{array} $$ (e) Compute a \(95 \%\) confidence interval for the large data set. Compare the results to part (a). What effect does increasing the sample size have on the confidence interval? (f) Suppose that the last observation, 41 , is inadvertently entered as 14 . Verify that this observation is an outlier. (g) Compute a \(95 \%\) confidence interval for the large data set with the outlier. Compare the results to part (e). What effect does an outlier have on a confidence interval when the data set is large?

A simple random sample of size \(n\) is drawn. The sample mean, \(\bar{x},\) is found to be \(35.1,\) and the sample standard deviation, \(s,\) is found to be \(8.7 .\) (a) Construct a \(90 \%\) confidence interval for \(\mu\) if the sample size, \(n,\) is \(40 .\) (b) Construct a \(90 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is \(100 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(98 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is \(40 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of error, \(E ?\) (d) If the sample size is \(n=18\), what conditions must be satisfied to compute the confidence interval?

See all solutions

Recommended explanations on Math Textbooks

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.