/*! 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 43 An anthropologist wishes to esti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 43

An anthropologist wishes to estimate the average height of men for a certain race of people. If the population standard deviation is assumed to be 2.5 inches and if she randomly samples 100 men, find the probability that the difference between the sample mean and the true population mean will not exceed .5 inch.

Short Answer

Expert verified
The probability that the difference does not exceed 0.5 inch is approximately 0.9545.

Step by step solution

01

Understand the Problem

We are given that the standard deviation of the population is known to be 2.5 inches. We have to find the probability that the sample mean of a sample of 100 men has a difference from the population mean that does not exceed 0.5 inches.
02

Determine the Standard Error

To calculate the probability, we need the standard error of the sample mean. The formula for the standard error (SE) is \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 2.5 \) is the population standard deviation, and \( n = 100 \) is the sample size. Substitute the values into the formula: \[ SE = \frac{2.5}{\sqrt{100}} = \frac{2.5}{10} = 0.25 \]
03

Calculate the Z-Score

The Z-score tells us how many standard errors away a particular value is from the mean. The formula for the Z-score is \( Z = \frac{X - \mu}{SE} \), where \( X = 0.5 \) (the maximum allowable difference) and \( \mu = 0 \) (since we are considering no difference as the baseline). Calculate this: \[ Z = \frac{0.5}{0.25} = 2 \]
04

Find the Probability Using the Z-Score

Using the standard normal distribution table, we find the probability corresponding to a Z-score of 2. This table value tells us the probability that a value is less than or equal to our calculated Z-score. For \( Z = 2 \), this probability is approximately 0.9772.
05

Calculate the Desired Probability

Since we want the probability that the difference is within 0.5 inches (not exceeding it), we are interested in the probability of being both above -0.5 and below 0.5. Thus, we need to consider the Z-scores of both +2 and -2. We find that the probability between these two Z-scores is: \[ 2 \times 0.9772 - 1 = 0.9545 \]

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 Error
The standard error (SE) is an essential concept in statistics, especially when dealing with sample data. It provides a measure of the variability or dispersion of a sample mean based on the population standard deviation. This makes it easier to gauge how much the sample mean can differ from the true population mean.

The formula to calculate the standard error is \[SE = \frac{\sigma}{\sqrt{n}}\]where:
  • \( \sigma \) is the population standard deviation
  • \( n \) is the sample size
In our exercise, the population standard deviation is 2.5 inches and the sample size is 100. Substituting these values into the formula, we get:\[SE = \frac{2.5}{\sqrt{100}} = \frac{2.5}{10} = 0.25\]This result indicates that the standard error of the sample mean is 0.25 inches. A smaller standard error suggests that the sample mean is likely close to the population mean.
Z-Score
Understanding Z-scores is crucial for comparing different data points within a distribution. A Z-score expresses how many standard deviations a specific data point is from the mean. It helps to standardize data by transforming it into a normal distribution.

The formula for calculating a Z-score is:\[Z = \frac{X - \mu}{SE}\]where:
  • \( X \) is the value of interest
  • \( \mu \) is the mean (which is 0 in this context since we're checking the difference)
  • \( SE \) is the standard error
In the problem, we want to know how far a difference of 0.5 inches is from the mean, so we calculate:\[Z = \frac{0.5}{0.25} = 2\]This Z-score of 2 means the difference of 0.5 inches is 2 standard errors away from the mean. It helps us use the standard normal distribution to find probabilities.
Normal Distribution
The normal distribution is a fundamental concept in statistics characterized by its bell-shaped curve. It is symmetric around the mean and is defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)). Many natural phenomena tend to follow a normal distribution, which makes it incredibly useful for probability estimation.

When dealing with sample means, as in this exercise, we assume the sample means are normally distributed if the population from which they're drawn is normal or if the sample size is large enough (usually \( n > 30 \) is considered acceptable due to the Central Limit Theorem). The mean of the sampling distribution of the sample mean is the population mean, and its standard deviation is the standard error.

Using the Z-scores, we can map our problem onto this distribution to find out the likelihood (probability) of a sample mean being within a certain range.
Probability Estimation
Probability estimation involves predicting the chance that a given event will occur. In this exercise, we aim to estimate the probability that the sample mean deviates from the population mean by no more than 0.5 inches in absolute terms.
  • Step 1: Convert the difference into a Z-score using the standard error.
  • Step 2: Use the Z-score in conjunction with the standard normal distribution table to find the corresponding probability.
After calculating a Z-score of 2, we look up this value in the standard normal distribution table. The value of 0.9772 suggests a high probability for a Z-score less than or equal to 2.

Since we need the probability that the difference is within 0.5 inches on either side of the mean, we consider both the probabilities for Z-scores of +2 and -2. Combining these gives us:\[2 \times 0.9772 - 1 = 0.9545\]This results in a probability estimation of approximately 95.45%, indicating a high likelihood that the sample mean will fall within 0.5 inches of the population mean.

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 does the sampling distribution of the sample mean look like if samples are taken from an approximately normal distribution? Use the applet Sampling Distribution of the Mean (at https://college.cengage.com/nextbook/statistics/wackerly 966371/student/html/index.html) to complete the following. The population to be sampled is approximately normally distributed with \(\mu=16.50\) and \(\sigma=6.03\) (these values are given above the population histogram and denoted \(M\) and S, respectively). a. Use the button "Next Obs" to select a single value from the approximately normal population. Click the button four more times to complete a sample of size \(5 .\) What value did you obtain for the mean of this sample? Locate this value on the bottom histogram (the histogram for the values of \(Y\) ). b. Click the button "Reset" to clear the middle graph. Click the button "Next Obs" five more times to obtain another sample of size 5 from the population. What value did you obtain for the mean of this new sample? Is the value that you obtained equal to the value you obtained in part (a)? Why or why not? c. Use the button "1 Sample" eight more times to obtain a total of ten values of the sample mean. Look at the histogram of these ten means. i. What do you observe? ii. How does the mean of these \(10 \bar{y}\) -values compare to the population mean \(\mu\) ? d. Use the button "1 Sample" until you have obtained and plotted 25 realized values for the sample mean \(\bar{Y}\), each based on a sample of size 5 . i. What do you observe about the shape of the histogram of the 25 values of \(\bar{y}_{i}\) $$ i=1,2, \ldots, 25 ? $$ ii. How does the value of the standard deviation of the \(25 \bar{y}\) values compare with the theoretical value for \(\sigma_{Y}\) obtained in Example 5.27 where we showed that, if \(Y\) is computed based on a sample of size \(n,\) then \(V(\bar{Y})=\sigma^{2} / n ?\) e. Click the button "1000 Samples" a few times, observing changes to the histogram as you generate more and more realized values of the sample mean. What do you observe about the shape of the resulting histogram for the simulated sampling distribution of \(Y\) ? f. Click the button "Toggle Normal" to overlay (in green) the normal distribution with the same mean and standard deviation as the set of values of \(Y\) that you previously generated. Does this normal distribution appear to be a good approximation to the sampling distribution of \(Y ?\)

Let \(Y_{1}, Y_{2}, \ldots, Y_{5}\) be a random sample of size 5 from a normal population with mean 0 and variance 1 and let \(\bar{Y}=(1 / 5) \sum_{i=1}^{5} Y_{i} .\) Let \(Y_{6}\) be another independent observation from the same population. What is the distribution of a. \(W=\sum_{i=1}^{5} Y_{i}^{2} ?\) Why? b. \(U=\sum_{i=1}^{5}\left(Y_{i}-\bar{Y}\right)^{2} ?\) Why? c. \(\sum_{i=1}^{5}\left(Y_{i}-\bar{Y}\right)^{2}+Y_{6}^{2} ?\) Why?

Suppose that independent samples (of sizes \(n_{i}\) ) are taken from each of \(k\) populations and that population \(i\) is normally distributed with mean \(\mu_{i}\) and variance \(\sigma^{2}, i=1,2, \ldots, k\). That is, all populations are normally distributed with the same variance but with (possibly) different means. Let \(\bar{X}_{i}\) and \(S_{i}^{2}, i=1,2, \ldots, k\) be the respective sample means and variances. Let \(\theta=c_{1} \mu_{1}+c_{2} \mu_{2}+\cdots+c_{k} \mu_{k},\) where \(c_{1}, c_{2}, \ldots, c_{k}\) are given constants. a. Give the distribution of \(\hat{\theta}=c_{1} \bar{X}_{1}+c_{2} \bar{X}_{2}+\cdots+c_{k} \bar{X}_{k}\). Provide reasons for any claims that you make. b. Give the distribution of $$\frac{\mathrm{SSE}}{\sigma^{2}}, \quad \text { whereSSE }=\sum_{i=1}^{k}\left(n_{i}-1\right) S_{i}^{2}$$. Provide reasons for any claims that you make. c. Give the distribution of $$\frac{\hat{\theta}-\theta}{\sqrt{\left(\frac{c_{1}^{2}}{n_{1}}+\frac{c_{2}^{2}}{n_{2}}+\cdots+\frac{c_{k}^{2}}{n_{k}}\right) \mathrm{MSE}}}, \quad \text { whereMSE }=\frac{\mathrm{SSE}}{n_{1}+n_{2}+\cdots+n_{k}-k}$$ Provide reasons for any claims that you make.

An airline finds that \(5 \%\) of the persons who make reservations on a certain flight do not show up for the flight. If the airline sells 160 tickets for a flight with only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?

Suppose that \(T\) is a \(t\) -distributed random variable. a. If \(T\) has 5 df, use Table 5 , Appendix 3 , to find \(t_{10}\), the value such that \(P\left(T>t_{10}\right)=.10\). Find \(t_{10}\) using the applet Student's \(t\) Probabilities and Quantiles. b. Refer to part (a). What quantile does \(t_{.10}\) correspond to? Which percentile? c. Use the applet Student's t Probabilities and Quantiles to find the value of \(t_{.10}\) for \(t\) distributions with 30, 60, and 120 df. d. When \(Z\) has a standard normal distribution, \(P(Z>1.282)=.10\) and \(z_{.10}=1.282 .\) What property of the \(t\) distribution (when compared to the standard normal distribution) explains the fact that all of the values obtained in part (c) are larger than \(z_{10}=1.282 ?\) e. What do you observe about the relative sizes of the values of \(t_{.10}\) for \(t\) distributions with 30,60 and \(120 \mathrm{df} ?\) Guess what \(t_{.10}\) "converges to" as the number of degrees of freedom gets large. [Hint: Look at the row labeled \(\infty \text { in Table } 5 \text { , Appendix } 3 .\)]
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Geometry

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.