/*! 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 6 A population consists of the fol... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A population consists of the following five values: 2,2 , \(4,4,\) and 8 a. List all samples of size \(2,\) and compute the mean of each sample. b. Compute the mean of the distribution of sample means and the population mean. Compare the two values. c. Compare the dispersion in the population with that of the sample means.

Short Answer

Expert verified
The mean of the sample means (4.11) approximates the population mean (4). The dispersion of the sample means (1.22) is lower than that of the population (2.19).

Step by step solution

01

List All Samples

List all possible samples of size 2 from the population \( \{2, 2, 4, 4, 8\} \). The samples are:- \((2, 2)\)- \((2, 4)\)- \((2, 4)\)- \((2, 8)\)- \((4, 2)\)- \((4, 4)\)- \((4, 8)\)- \((8, 2)\)- \((8, 4)\)
02

Calculate Sample Means

Calculate the mean for each sample:- For \((2, 2)\), Mean = \( \frac{2 + 2}{2} = 2\)- For \((2, 4)\), Mean = \( \frac{2 + 4}{2} = 3\)- For \((2, 4)\), Mean = \( \frac{2 + 4}{2} = 3\) (second occurrence)- For \((2, 8)\), Mean = \( \frac{2 + 8}{2} = 5\)- For \((4, 2)\), Mean = \( \frac{4 + 2}{2} = 3\)- For \((4, 4)\), Mean = \( \frac{4 + 4}{2} = 4\)- For \((4, 8)\), Mean = \( \frac{4 + 8}{2} = 6\)- For \((8, 2)\), Mean = \( \frac{8 + 2}{2} = 5\)- For \((8, 4)\), Mean = \( \frac{8 + 4}{2} = 6\)
03

Compute Mean of Sample Means

The means of all samples are \(2, 3, 3, 5, 3, 4, 6, 5, 6\). Calculate the mean of these sample means:Mean of Sample Means = \( \frac{2 + 3 + 3 + 5 + 3 + 4 + 6 + 5 + 6}{9} = \frac{37}{9} \approx 4.11 \).
04

Compute Population Mean

Compute the mean of the population \( \{2, 2, 4, 4, 8\} \):Population Mean = \( \frac{2 + 2 + 4 + 4 + 8}{5} = \frac{20}{5} = 4\).
05

Compare Means

The population mean is \(4\), while the mean of the sample means is approximately \(4.11\). These values are very close, as expected due to the law of large numbers.
06

Compute Population Dispersion

Calculate the standard deviation for the population. The variance is:\[ \frac{(2-4)^2 + (2-4)^2 + (4-4)^2 + (4-4)^2 + (8-4)^2}{5} = \frac{4 + 4 + 0 + 0 + 16}{5} = \frac{24}{5} = 4.8 \].Therefore, the population standard deviation is \( \sqrt{4.8} \approx 2.19 \).
07

Compute Dispersion of Sample Means

Calculate the standard deviation of the sample means \(2, 3, 3, 5, 3, 4, 6, 5, 6\). The variance is:\[ \frac{(2-4.11)^2 + (3-4.11)^2 + (3-4.11)^2 + (5-4.11)^2 + (3-4.11)^2 + (4-4.11)^2 + (6-4.11)^2 + (5-4.11)^2 + (6-4.11)^2}{9} \].After calculation, this results in a variance of approximately \(1.48\).Thus, the standard deviation is \( \sqrt{1.48} \approx 1.22 \).
08

Compare Dispersions

The population standard deviation is approximately \(2.19\), and the standard deviation of the sample means is approximately \(1.22\). The dispersion among the sample means is lower compared to the population dispersion, which is typical as sample means tend to cluster closer around the true mean.

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 foundational concept in statistics. It represents the average of all values within a set, or population. Calculating it involves summing all the values and dividing by the number of values.

In our example, the population set is \( \{2, 2, 4, 4, 8\} \). To find the population mean, you sum the values and then divide by the total number of values. So, you get \( \frac{2 + 2 + 4 + 4 + 8}{5} = 4 \).

  • Summing the values: \(2 + 2 + 4 + 4 + 8 = 20\)
  • Number of values (N): 5
  • Population Mean = \( \frac{20}{5} = 4 \)
The result reflects the central tendency of the data, showing where most values cluster.
Sample Dispersion
Sample dispersion indicates how much variation exists in the sample data. One common measure of dispersion is standard deviation, which tells us how much individual sample means tend to deviate from the true mean.

In this exercise, we compute it for the list of sample means: \(2, 3, 3, 5, 3, 4, 6, 5, 6\). The formula for variance, a component of standard deviation, is computed as follows:
  • Variance = \( \frac{\sum{(X_i - \bar{X})^2}}{n} \)
  • Where \( X_i \) are individual sample means, \( \bar{X} \) is the mean of sample means, and \( n \) is the number of sample means.
The variance calculated results in approximately \(1.48\). Thus, the standard deviation is \(\sqrt{1.48} \approx 1.22\). Lower dispersion among sample means indicates that they tend to be close to the true mean of the population.
Law of Large Numbers
The law of large numbers is a statistical principle that explains why the average of a large number of samples can be expected to approximate the population mean. Basically, as you increase the number of samples taken from a population, the average of the sample means will get closer and closer to the population mean.

In the exercise, the mean of the sample means (approximately \(4.11\)) is indeed very close to the population mean (\(4\)). Although it isn't exactly equal, it illustrates how sample averages converge to the population mean with larger sample sizes.

This principle helps us understand why in research, larger sample sizes generally produce more reliable estimates of the population parameters, since the effects of random variation diminish as sample size grows.
Standard Deviation
Standard deviation is a measure of how spread out numbers are. It is crucial in understanding dispersion in any set of data, whether it's a sample or an entire population.
In simpler terms, standard deviation gives insight into how much the data values differ from the average value.

For a population, you calculate the standard deviation by first finding the variance and then taking the square root. When calculating from the original dataset \( \{2, 2, 4, 4, 8\} \), the variance is \( \frac{24}{5} = 4.8 \), leading to a standard deviation of approximately \(\sqrt{4.8} \approx 2.19\).
  • Variance shows how much the numbers differ either above or below the mean.
  • Higher standard deviation indicates more variation in values.
In conclusion, standard deviation provides valuable information about data variability and reliability.

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

The mean amount purchased by a typical customer at Churchill's Grocery Store is \(\$ 23.50\) with a standard deviation of \(\$ 5.00 .\) Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions. a. What is the likelihood the sample mean is at least \(\$ 25.00 ?\) b. What is the likelihood the sample mean is greater than \(\$ 22.50\) but less than \(\$ 25.00 ?\) c. Within what limits will 90 percent of the sample means occur?

Scrapper Elevator Company has 20 sales representatives who sell its product throughout the United States and Canada. The number of units sold last month by each representative is listed below. Assume these sales figures to be the population values. a. Draw a graph showing the population distribution. b. Compute the mean of the population. c. Select five random samples of 5 each. Compute the mean of each sample. Use the methods described in this chapter and \(\underline{\text { Appendix }} \mathrm{B} .6\) to determine the items to be included in the sample. d. Compare the mean of the sampling distribution of the sample means to the population mean. Would you expect the two values to be about the same? e. Draw a histogram of the sample means. Do you notice a difference in the shape of the distribution of sample means compared to the shape of the population distribution?

A normal population has a mean of 60 and a standard deviation of \(12 .\) You select a random sample of \(9 .\) Compute the probability the sample mean is: a. Greater than \(63 .\) b. Less than 56 . c. Between 56 and \(63 .\)

CRA CDs, Inc., wants the mean lengths of the "cuts" on a CD to be 135 seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have plenty of time for commercials within each 10 -minute segment. Assume the distribution of the length of the cuts follows the normal distribution with a population standard deviation of 8 seconds. Suppose we select a sample of 16 cuts from various CDs sold by CRA CDs, Inc. a. What can we say about the shape of the distribution of the sample mean? b. What is the standard error of the mean? c. What percent of the sample means will be greater than 140 seconds? d. What percent of the sample means will be greater than 128 seconds? e. What percent of the sample means will be greater than 128 but less than 140 seconds?

A population consists of the following five values: 0,0 , 1,3,6 a. List all samples of size \(3,\) and compute the mean of each sample. b. Compute the mean of the distribution of sample means and the population mean. Compare the two values. c. Compare the dispersion in the population with that of the sample means.

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.