/*! 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 164 We see } in the AllCountries dat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

We see } in the AllCountries dataset that the percent of the population living in rural areas is 8.0 in Argentina and 34.4 in Bolivia. Suppose we take random samples of size 200 from each country, and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{B},\) where \(\hat{p}_{A}\) represents the sample proportion living in rural areas in Argentina and \(\hat{p}_{B}\) represents the proportion of the sample that lives in rural areas in Bolivia. (a) Find the mean and standard deviation of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. (c) Using the graph drawn in part (b), are we likely to see a difference in sample proportions as large in magnitude as -0.4 ? As large as \(-0.3 ?\) Explain.

Short Answer

Expert verified
The mean of the difference in proportions is -0.264 with a standard deviation of 0.0169. A difference of -0.4 or less is highly unlikely given it lies beyond 8 standard deviations from the mean. However, a difference of -0.3 or less is likely, since it is within 2 standard deviations from the mean.

Step by step solution

01

Calculating Mean and Standard Deviation

Begin by finding the mean and standard deviation for each countries sample proportion. The mean in this context is simply the subtraction of the populations proportions, so \(\mu_{\hat{p}_{A}-\hat{p}_{B}} = P_{A}-P_{B}\) where \(P_{A}=0.08\) and \(P_{B}=0.344\). Therefore, the expected value is \(\mu_{\hat{p}_{A}-\hat{p}_{B}} = 0.08 - 0.344 = -0.264\). The variance is the sum of the variances for each country's proportion (which is their sample proportion multiplied by 1 minus the sample proportion, divided by the sample size) i.e. \(Variance_{\hat{p}_{A}-\hat{p}_{B}} = \frac{P_{A}(1-P_{A})}{n_{A}} + \frac{P_{B}(1-P_{B})}{n_{B}} = \frac{0.08(1-0.08)}{200} + \frac{0.344(1-0.344)}{200} = 0.000286\). The standard deviation is the square root of this variance and equals 0.0169.
02

Graph Creation Using Central Limit Theorem

Here we're considering Central Limit Theorem, which says that if the sample sizes are large enough, the sampling distribution of the sample proportion will be approximately normally distributed. In this case, the sample sizes are large enough so the Central Limit Theorem applies. From here we can graph this normal distribution with mean -0.264 and standard deviation 0.0169. On the horizontal axis, include the mean, and two points one standard deviation to each side of the mean, i.e., -0.264, -0.2471 and -0.2809.
03

Deciding on Difference Magnitude

We are asked whether we are likely to see a difference in sample proportions as large in magnitude as -0.4 or -0.3. Observing our graph and considering the calculated mean of -0.264 and standard deviation of 0.0169, we conclude that getting a difference as large as -0.4 is quite unlikely, since it is more than 8 standard deviations beneath the mean. However, getting a difference as large as -0.3 is probable, since it is within 2 standard deviations from the 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.

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental statistical principle that explains why many distributions tend to appear normal or bell-shaped, especially when dealing with large datasets. In its essence, the CLT states that the sampling distribution of the sample mean (or proportion) will approximate a normal distribution as the sample size becomes larger, regardless of the shape of the population distribution.

This theorem is particularly useful when working with sample proportions like the scenario described in the exercise, where we're comparing the proportion of rural populations in Argentina and Bolivia. Once the sample sizes are large enough (usually a rule of thumb is n > 30), we can assume that the differences in sample proportions will also be normally distributed. This allows us to make predictions about the likelihood of observing certain outcomes and to calculate probabilities associated with these outcomes.

When applying the CLT to sample proportions, it's crucial to ensure that the samples are independent and randomly selected, as any bias could skew the results and render the normal approximation invalid.
Sample Proportions
Sample proportions represent the percentage of items in a sample that have a particular feature. For example, in the exercise, the sample proportion \(\hat{p}_{A}\) refers to the proportion of individuals living in rural areas in a sample taken from Argentina's population.

When dealing with sample proportions, we look to calculate the mean and the variance of the distribution of these proportions to understand their behavior. The mean of the sample proportion is the expected value that we predict to find if we were to repeat our sample many times. In the exercise, the mean difference in sample proportions \(\mu_{\hat{p}_{A}-\hat{p}_{B}}\) was calculated using the population proportions from both countries.

Variance, on the other hand, measures the spread of the sample proportion values around the mean. It is the average of the squared deviations from the mean. In sampling, it's important since it affects how precise our estimates are likely to be - smaller variance usually means higher precision.
Variance and Standard Deviation
Variance and standard deviation are two core concepts of statistics that measure variability or dispersion within a data set. Variance (\(\sigma^2\)) calculates the average of the squared differences from the mean, offering a numeric value that describes the degree to which each data point differs from the average. The standard deviation (\(\sigma\)) is the square root of the variance, providing a measurement that is on the same scale as the data itself, which makes it more interpretable.

In the context of sampling distributions, the standard deviation is critical because it helps describe the spread of the data points around the mean. A smaller standard deviation implies that the data points are generally closer to the mean. On the graph of the sampling distribution, it would appear as a steeper curve, indicating that there is less variability in sample outcomes. Conversely, a larger standard deviation suggests more spread out data points and a flatter curve.

To calculate the variance of the difference in sample proportions, as described in the exercise, we add the variances of individual sample proportions since they are independent. This calculation gives us an insight into how much variability to expect when we take repeated samples from the populations of Argentina and Bolivia to compare the proportions living in rural areas.

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

If random samples of the given sizes are drawn from populations with the given proportions: (a) Find the mean and standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 100 from population \(A\) with proportion 0.20 and samples of size 50 from population \(B\) with proportion 0.30

If random samples of the given sizes are drawn from populations with the given proportions: (a) Find the mean and standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 40 from population \(A\) with proportion 0.30 and samples of size 30 from population \(B\) with proportion 0.24

To study the effect of sitting with a laptop computer on one's lap on scrotal temperature, 29 men have their scrotal temperature tested before and then after sitting with a laptop for one hour.

NBA Free Throws In Exercise 6.10, we learn that the percent of free throws made in the \(\mathrm{NBA}\) (National Basketball Association) has been about \(75 \%\) for the last 50 years. If we take random samples of free throws in the NBA and compute the proportion of free throws made, what percent of samples of size \(n=200\) will have a sample proportion greater than \(80 \%\) ? Use the fact that the sample proportions are normally distributed and compute the mean and standard deviation of the distribution.

How big is the home field advantage in the National Football League (NFL)? In Exercise 6.240 on page 419 , we examine a difference in means between home and away teams using two separate samples of 80 games from each group. However, many factors impact individual games, such as weather conditions and the scoring of the opponent. It makes more sense to investigate this question using a matched pairs design, using scores for home and away teams matched for the same game. The data in NFLScores2011 include the points scored by the home and away team in 256 regular season games in \(2011 .\) We will treat these games as a sample of all NFL games. Estimate average home field scoring advantage and find a \(90 \%\) confidence interval for the mean difference.

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.