/*! 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 46 Use the same population of \(\\{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 46

Use the same population of \(\\{9\) cigarettes, 10 cigarettes, 20 cigarettes \(\\}\) from Exercise \(45 .\) Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population?

Short Answer

Expert verified
Mean of sample MADs differs from population MAD, indicating it's not a reliable estimator.

Step by step solution

01

Calculate Population Mean Absolute Deviation

First, find the population mean. Add all values and divide by the number of values: Mean = \(\frac{9 + 10 + 20}{3} = 13\).Next, calculate the absolute deviations from the mean: \(|9 - 13| = 4\)\(|10 - 13| = 3\)\(|20 - 13| = 7\)Now, find the mean absolute deviation (MAD) of the population: \(\text{MAD}_{\text{population}} = \frac{4 + 3 + 7}{3} = 4.67\).
02

List All Possible Samples

Identify all possible samples of size 2 with replacement: (9, 9)(9, 10)(9, 20)(10, 9)(10, 10)(10, 20)(20, 9)(20, 10)(20, 20).
03

Calculate Sample Mean Absolute Deviations

For each pair of samples, calculate the sample mean and the absolute deviations from that mean. Then find the mean absolute deviation for each sample.Example:For sample (9, 9): Mean = 9, MAD = \(\frac{0 + 0}{2} = 0\).For sample (9, 10): Mean = 9.5, MAD = \(\frac{0.5 + 0.5}{2} = 0.5\).For sample (9, 20): Mean = 14.5, MAD = \(\frac{5.5 + 5.5}{2} = 5.5\).Repeat for all other pairs.
04

Compare Sample MADs with Population MAD

Compute the mean of the sample MADs:\(\text{Mean Sample MAD} = \frac{0 + 0.5 + 5.5 + 0.5 + 0 + 5 + 5.5 + 5 + 0}{9} = 2.33\).Compare this with the population MAD, which is 4.67.
05

Draw Conclusion

Since the mean of the sample MADs (2.33) is different from the population MAD (4.67), this indicates that the sample mean absolute deviation does not reliably estimate the population mean absolute deviation.

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.

population mean
Understanding the population mean is crucial when calculating mean absolute deviation. The population mean is simply the average of all values in a dataset. To find it, add all the values together and then divide by the number of values. For instance, from the dataset \(9, 10, 20\), the population mean is calculated as follows: \[\text{Mean} = \frac{9 + 10 + 20}{3} = 13\]. This value serves as a central point from which we can measure how each data point deviates. The population mean helps to give a single representative value that summarizes the dataset.
absolute deviations
Absolute deviations are the differences between each data point and the mean of the dataset, using the absolute value. This means we take the deviation of each data point from the mean and ignore whether it's positive or negative. For the population values \(9, 10, 20\) with a mean of 13, calculate the absolute deviations: \[|9 - 13| = 4\], \[|10 - 13|= 3\], and \[|20 - 13| = 7\]. Absolute deviations provide insight into how much individual data points differ from the mean, thus indicating the spread of the values in the dataset.
sample mean absolute deviation
Sample mean absolute deviation is used to estimate how much the data in a sample deviates from the mean on average. To calculate the sample MAD, follow these steps:
  • First, find the mean of the sample values.
  • Then, calculate the absolute deviations for each value from the sample mean.
  • Finally, find the average of these absolute deviations.
For example, consider the sample (9, 10). The mean is \(9.5\). The absolute deviations from the mean are \[|9 - 9.5| = 0.5\] and \[|10 - 9.5| = 0.5\]. Thus, the sample MAD is \[\text{MAD}_{\text{sample}} = \frac{0.5 + 0.5}{2} = 0.5\]. Repeating these steps for all possible samples helps illustrate that the average of these sample MADs doesn't always approximate the population's MAD accurately, highlighted by the example's final comparison, \(\text{mean sample MAD} = 2.33\) vs. \(\text{population MAD} = 4.67\). This indicates that sample MAD might not be a reliable estimator of population MAD.

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

Watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, \((b)\) median, (c) mode, (d) midrange, and then answer the given question. Listed below are the numbers of Atlantic hurricanes that occurred in each year. The data are listed in order by year, starting with the year 2000 . What important feature of the data is not revealed by any of the measures of center? $$ \begin{array}{llllllllllllll} 8 & 9 & 8 & 7 & 9 & 15 & 5 & 6 & 8 & 4 & 12 & 7 & 8 & 2 \end{array} $$

Find the mean of the data summarized in the frequency distribution. Also, compare the computed means to the actual means obtained by using the original list of data values, which are as follows: (Exercise 29) 36.2 years; (Exercise 30) 44.1 years; (Exercise 31) 224.3; (Exercise 32) 255.1. $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Blood Platelet Count of } \\ \text { Males (1000 cells } / \boldsymbol{\mu} \text { L) } \end{array} & \text { Frequency } \\ \hline 0-99 & 1 \\ \hline 100-199 & 51 \\ \hline 200-299 & 90 \\ \hline 300-399 & 10 \\ \hline 400-499 & 0 \\ \hline 500-599 & 0 \\ \hline 600-699 & 1 \\ \hline \end{array} $$

Watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, \((b)\) median, (c) mode, (d) midrange, and then answer the given question. Listed below are annual U.S. sales of vinyl record albums (millions of units). The numbers of albums sold are listed in chronological order, and the last entry represents the most recent year. Do the measures of center give us any information about a changing trend over time? $$ \begin{array}{llllllllllllll} 0.3 & 0.6 & 0.8 & 1.1 & 1.1 & 1.4 & 1.4 & 1.5 & 1.2 & 1.3 & 1.4 & 1.2 & 0.9 & 0.9 \\ 1 & 1.9 & 2.5 & 2.8 & 3.9 & 4.6 & 6.1 & & & & & & & \end{array} $$

Let a population consist of the values 9 cigarettes, 10 cigarettes, and 20 cigarettes smoked in a day (based on data from the California Health Interview Survey). Assume that samples of two values are randomly selected with replacement from this population. (That is, a selected value is replaced before the second selection is made.) a. Find the variance \(\sigma^{2}\) of the population \(\\{9\) cigarettes, 10 cigarettes, 20 cigarettes \(\\}\). b. After listing the nine different possible samples of two values selected with replacement, find the sample variance \(s^{2}\) (which includes division by \(n-1\) ) for each of them; then find the mean of the nine sample variances \(s^{2}\). c. For each of the nine different possible samples of two values selected with replacement, find the variance by treating each sample as if it is a population (using the formula for population variance, which includes division by \(n\) ); then find the mean of those nine population variances. d. Which approach results in values that are better estimates of \(\sigma^{2}\) : part (b) or part (c)? Why? When computing variances of samples, should you use division by \(n\) or \(n-1\) ? e. The preceding parts show that \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\). Is \(s\) an unbiased estimator of \(\sigma\) ? Explain.

The harmonic mean is often used as a measure of center for data sets consisting of rates of change, such as speeds. It is found by dividing the number of values \(n\) by the sum of the reciprocals of all values, expressed as $$\frac{n}{\sum \frac{1}{x}}$$ (No value can be zero.) The author drove 1163 miles to a conference in Orlando, Florida. For the trip to the conference, the author stopped overnight, and the mean speed from start to finish was \(38 \mathrm{mi} / \mathrm{h}\). For the return trip, the author stopped only for food and fuel, and the mean speed from start to finish was \(56 \mathrm{mi} / \mathrm{h}\). Find the harmonic mean of \(38 \mathrm{mi} / \mathrm{h}\) and \(56 \mathrm{mi} / \mathrm{h}\) to find the true "average" speed for the round trip.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Logic and Functions

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.