/*! 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 39 Suppose Jack and Diane are each ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 39

Suppose Jack and Diane are each attempting to use simulation to describe the sampling distribution from a population that is skewed left with mean 50 and standard deviation \(10 .\) Jack obtains 1000 random samples of size \(n=3\) from the population, finds the mean of the 1000 samples, draws a histogram of the means, finds the mean of the means, and determines the standard deviation of the means. Diane does the same simulation, but obtains 1000 random samples of size \(n=30\) from the population. (a) Describe the shape you expect for Jack's distribution of sample means. Describe the shape you expect for Diane's distribution of sample means. (b) What do you expect the mean of Jack's distribution to be? What do you expect the mean of Diane's distribution to be? (c) What do you expect the standard deviation of Jack's distribution to be? What do you expect the standard deviation of Diane's distribution to be?

Short Answer

Expert verified
(a) Jack: skewed left; Diane: nearly normal. (b) Both: mean = 50. (c) Jack: std dev 鈮� 5.77; Diane: std dev 鈮� 1.83.

Step by step solution

01

Understanding the Central Limit Theorem

The Central Limit Theorem (CLT) states that, for a large enough sample size, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution. This will be important to determine the shapes of Jack's and Diane's distributions.
02

Predicting the Shape of Jack's Distribution

Since Jack's sample size is small (=3), the distribution of sample means is less likely to be normally distributed and more likely to retain some skewness from the original population. Therefore, Jack's distribution of sample means is expected to be skewed left, similar to the original population.
03

Predicting the Shape of Diane's Distribution

Since Diane's sample size is much larger (=30), according to the CLT, the distribution of sample means is expected to be approximately normal, even though the population itself is skewed left. Thus, Diane's distribution of sample means will be nearly normally distributed.
04

Expected Mean of Jack's and Diane's Distributions

The mean of the sampling distribution (i.e., the mean of the sample means) is equal to the mean of the population, regardless of the sample size. Therefore, the mean of Jack's distribution and the mean of Diane's distribution are both expected to be 50.
05

Expected Standard Deviation of Jack's and Diane's Distributions

The standard deviation of the sampling distribution (standard error) is given by \(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation, and \(n\) is the sample size. For Jack: \ \sigma_{\bar{X}_{Jack}} = \frac{10}{\sqrt{3}} \approx 5.77\. For Diane: \ \sigma_{\bar{X}_{Diane}} = \frac{10}{\sqrt{30}} \approx 1.83\.

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.

Sampling Distribution
The sampling distribution is a critical concept in statistics. It refers to the probability distribution of a given statistic based on a random sample. For Jack and Diane, the statistic of interest is the sample mean.
When Jack and Diane draw multiple samples, they each calculate the sample mean for those samples. These sample means form their own distributions. This is what we call the 'sampling distribution of the sample mean' (or sometimes just 'sampling distribution' for brevity).
For Jack, who has a sample size of 3, his sampling distribution will likely reflect the skewness present in the population because smaller sample sizes tend to retain more characteristics of the original population. Hence, Jack鈥檚 sampling distribution is expected to be skewed left, similar to the original skewed-left population distribution.
In contrast, Diane鈥檚 larger sample size of 30 means that, according to the Central Limit Theorem, her sampling distribution will be closer to a normal distribution, even if the original population is skewed. The larger sample size helps in averaging out the skewness, leading to a more symmetric, bell-shaped distribution.
Mean of Sample Means
The mean of the sample means (also known as the expected value) is a fascinating property. No matter how skewed the population is, the mean of the sample means is always equal to the population mean, as long as the samples are random and unbiased.
For both Jack and Diane, irrespective of their sample sizes, the mean of the sample means will always be the same as the population mean. Since the population mean is given as 50, we can confidently say that the mean of Jack鈥檚 distribution of sample means and Diane鈥檚 distribution of sample means will both be 50.
Standard Error
The standard error (SE) is the term we use for the standard deviation of the sampling distribution. It gives us an idea of how much the sample means are expected to vary around the population mean.
The formula for the standard error of the sample mean is: \ \( \text{SE} = \frac{\text{Population Standard Deviation}}{\text{Square Root of Sample Size}} \)
For Jack, with a sample size of 3:
\ \( \text{SE}_{Jack} = \frac{10}{\text{sqrt}(3)} \approx 5.77 \)
For Diane, with a sample size of 30:
\ \( \text{SE}_{Diane} = \frac{10}{\text{sqrt}(30)} \approx 1.83 \)
Notice how Diane鈥檚 standard error is much smaller. This is because a larger sample size reduces the variability of the sample means. Hence, Diane鈥檚 sample means will be clustered more tightly around the population mean, while Jack鈥檚 will be more spread out.

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

True or False: The mean of the sampling distribution of \(\hat{p}\) is \(p\).

Election Prediction Exit polling is a popular technique used to determine the outcome of an election prior to results being tallied. Suppose a referendum to increase funding for education is on the ballot in a large town (voting population over 100,000 ). An exit poll of 310 voters finds that 164 voted for the referendum. How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is 0.49 ? Based on your result, comment on the dangers of using exit polling to call elections. Include a discussion of the potential nonsampling error that could disrupt your findings.

Fumbles The New England Patriots made headlines prior to the 2015 Super Bowl for allegedly playing with underinflated footballs. An underinflated ball is easier to grip, and therefore, less likely to be fumbled. What does the data say? The following data represent the number of plays a team has per fumble. For example, the Chicago Bears run 48 offensive plays for every fumble. $$ \begin{array}{llc|llc} \text { Team } & \text { Dome } & \text { Plays } & \text { Team } & \text { Dome } & \text { Plays } \\ \hline \text { Atlanta Falcons } & \text { Yes } & 83 & \text { Saint Louis Rams } & \text { Yes } & 50 \\ \hline \text { New Orleans Saints } & \text { Yes } & 80 & \text { Seattle Seahawks } & \text { No } & 50 \\ \hline \text { New England Patriots } & \text { No } & 78 & \text { Detroit Lions } & \text { Yes } & 49 \\ \hline \text { Houston Texans } & \text { Yes } & 61 & \text { Chicago Bears } & \text { No } & 48 \\ \hline \text { Minnesota Vikings } & \text { Yes } & 59 & \text { San Francisco 49ers } & \text { No } & 47 \\ \hline \text { Baltimore Ravens } & \text { No } & 58 & \text { New York Jets } & \text { No } & 46 \\ \hline \text { Carolina Panthers } & \text { No } & 57 & \text { Denver Broncos } & \text { No } & 46 \\ \hline \text { San Diego Chargers } & \text { No } & 57 & \text { Tampa Bay Buccaneers } & \text { No } & 45 \\ \hline \text { Indianapolis Colts } & \text { Yes } & 56 & \text { Dallas Cowboys } & \text { Yes } & 45 \\ \hline \text { Green Bay Packers } & \text { No } & 55 & \text { Tennessee Titans } & \text { No } & 45 \\ \hline \text { New York Giants } & \text { No } & 55 & \text { Oakland Raiders } & \text { No } & 44 \\ \hline \text { Cincinnati Bengals } & \text { No } & 53 & \text { Miami Dolphins } & \text { No } & 44 \\ \hline \text { Pittsburgh Steelers } & \text { No } & 52 & \text { Buffalo Bills } & \text { No } & 43 \\ \hline \text { Kansas City Chiefs } & \text { No } & 52 & \text { Arizona Cardinals } & \text { Yes } & 43 \\ \hline \text { Jacksonville Jaguars } & \text { No } & 51 & \text { Philadelphia Eagles } & \text { No } & 42 \\ \hline \text { Cleveland Browns } & \text { No } & 50 & \text { Washington Redskins } & \text { No } & 37 \\ \hline \end{array} $$ (a) Draw a boxplot of plays per fumble for all teams in the National Football League. Describe the shape of the distribution. Are there any outliers? If so, which team(s)? (b) Playing in a dome (inside) removes the effect of weather (such as rain) on the game. Draw a boxplot of plays per fumble for teams who do not play in a dome. Are there any outliers? If so, which team(s)?

Finite Population Correction Factor In this section, we assumed that the sample size was less than \(5 \%\) of the size of the population. When sampling without replacement from a finite population in which \(n>0.05 N,\) the standard deviation of the distribution of \(\hat{p}\) is given by $$ \sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n-1} \cdot\left(\frac{N-n}{N}\right)} $$ where \(N\) is the size of the population. A survey is conducted at a college having an enrollment of 6502 students. The student council wants to estimate the percentage of students in favor of establishing a student union. In a random sample of 500 students, it was determined that 410 were in favor of establishing a student union. (a) Obtain the sample proportion, \(\hat{p},\) of students surveyed who favor establishing a student union. (b) Calculate the standard deviation of the sampling distribution of \(\hat{p}\) using \(\hat{p}\) as an estimate of \(p\).

The amount of time Americans spend watching television is closely monitored by firms such as \(\mathrm{AC}\) Nielsen because this helps determine advertising pricing for commercials. (a) Do you think the variable "weekly time spent watching television" would be normally distributed? If not, what shape would you expect the variable to have? (b) According to the American Time Use Survey, adult Americans spend 2.35 hours per day watching television on a weekday. Assume that the standard deviation for "time spent watching television on a weekday" is 1.93 hours. If a random sample of 40 adult Americans is obtained, describe the sampling distribution of \(\bar{x},\) the mean amount of time spent watching television on a weekday. (c) Determine the probability that a random sample of 40 adult Americans results in a mean time watching television on a weekday of between 2 and 3 hours. (d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 35 individuals who consider themselves to be avid Internet users results in a mean time of 1.89 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.89 hours or less from a population whose mean is presumed to be 2.35 hours. Based on the result obtained, do you think avid Internet users watch less television?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

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.