/*! 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 17 Standard Error from a Formula an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Standard Error from a Formula and Simulation In Exercises 6.15 to \(6.18,\) find the mean and standard error of the sample proportions two ways: (a) Use StatKey or other technology to simulate at least 1000 sample proportions. Give the mean and standard error and comment on whether the distribution appears to be normal. (b) Use the formulas in the Central Limit Theorem to compute the mean and standard error. Are the results similar to those found in part (a)? Sample proportions of sample size \(n=40\) from a population with \(p=0.5\)

Short Answer

Expert verified
In both simulation and the application of the Central Limit Theorem, the mean should be close to \(p = 0.5\). The standard error from both methods should also be close. The distribution of the simulated sample proportions is expected to be approximately normal given the large number of samples used.

Step by step solution

01

Simulating Sample Proportions

Use StatKey or appropriate technology to generate 1000 random samples, each of size 40, from a population where the proportion \(p=0.5\). Calculate the sample proportion for each of these samples to get 1000 sample proportions. Record these proportions as you'll need them for further computation.
02

Calculate Mean & Standard Error for Simulation

After generating the 1000 sample proportions, calculate the mean and standard error of these proportions. The mean is the sum of all sample proportions divided by the number of samples - 1000 in this case. The standard error is the standard deviation of these 1000 sample proportions. Using visual tools available in StatKey or other technology to plot these sample proportions can help provide an idea if the distribution appears normal.
03

Applying Central Limit Theorem Formulas

Remarkably, the Central Limit Theorem provides us with formulas to directly calculate the mean and standard error of sample proportions. According to this theorem, the mean of sample proportions is equal to the population proportion, \(p\), and the standard error is calculated by the formula \(\sqrt{(p * (1-p)) / n}\), where \(n\) is the sample size, which is 40 in this case.
04

Comparing Results

Compare the mean and standard error computed using the formulas from the Central Limit Theorem with those calculated in step 2 from the simulation. The two results should be similar given the large number of samples (1000 sample proportions) used in the simulation.

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.

Standard Error
The standard error of a sample proportion is a crucial concept when working with statistical data. In simple terms, the standard error measures the variability or spread of the sample proportions around the actual population proportion. It helps us understand how much the sample proportion will differ from the true population proportion due to random sampling.

You can think of it as a "margin of error" for the proportion, giving us an idea of how much the sample results might vary if we were to take multiple samples. In mathematical terms, when calculating the standard error of sample proportions, we use the formula:
  • Standard Error (SE) = \( \sqrt{\frac{p(1-p)}{n}} \)
where \(p\) is the population proportion, and \(n\) is the sample size.

A smaller standard error indicates that the sample proportion is a more accurate estimate of the population proportion. Conversely, a larger standard error means more variability and less reliability. Understanding the standard error helps researchers make informed decisions based on the data they have collected.
Sample Proportions
Sample proportions are estimates of the actual proportion in a population, derived from a sample. These proportions are very useful, especially when dealing with large populations where it's impractical to measure everyone.

For example, if a company wants to know the percentage of customers who like a new product, it's easier to survey a sample of customers rather than the entire customer base. The proportion of those who like the product in the sample gives us an estimate of the overall population's preference.

When we talk about sampling, we're often interested in the sample proportion \( \hat{p} \), which is calculated using the formula:
  • Sample Proportion (\( \hat{p} \)) = \( \frac{x}{n} \)
where \(x\) is the number of successful outcomes, and \(n\) is the sample size.

Because samples don't perfectly represent the population every time, this is where sample proportions become valuable. They allow us to make inferences about the population, knowing that there will always be some level of approximation and uncertainty involved.
Simulation in Statistics
Simulation in statistics is a powerful tool that helps us understand complex probability distributions and models that are not easily solved analytically. Through simulation, we can generate a large number of sample data points to mimic or simulate a real-world process or experiment.

In the context of our original exercise, simulation is used to generate 1000 random samples of proportions to gain a visual and quantitative understanding of the distribution. This practical approach helps verify theoretical results predicted by the Central Limit Theorem. It allows us to visualize whether the distribution of sample proportions is approximately normal by looking at histograms or other plots.

Simulations are particularly useful when you want to:
  • Test hypotheses in scenarios where theoretical distributions are complex.
  • Validate analytical results by seeing how they hold in "simulated reality."
  • Explore the potential variability and behavior of sample estimates under different conditions.
Overall, using simulation in statistics provides more evidence and confidence in the results, especially when dealing with large datasets or when theoretical assumptions need practical verification.

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

Use StatKey or other technology to generate a bootstrap distribution of sample means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviation as an estimate of the population standard deviation. Mean price of used Mustang cars online (in \(\$ 1000\) s) using the data in MustangPrice with \(n=25,\) \(\bar{x}=15.98,\) and \(s=11.11\)

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (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 1 with mean 87 and standard deviation 12 and samples of size 80 from Population 2 with mean 81 and standard deviation 15

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The average score on the Mathematics part of the SAT exam for males is 534 with a standard deviation of 118 , while the average score for females is 500 with a standard deviation of 112 . (a) If random samples are taken with 40 males and 60 females, find the mean and standard deviation of the distribution of differences in sample means, \(\bar{x}_{m}-\bar{x}_{f},\) where \(\bar{x}_{m}\) represents the sample mean for the males and \(\bar{x}_{f}\) represents the sample mean for the females. (b) Repeat part (a) if the random samples contain 400 males and 600 females. (c) What effect do the different sample sizes have on center and spread of the distribution?

In Exercises 6.153 to \(6.158,\) 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 50 from population \(A\) with proportion 0.70 and samples of size 75 from population \(B\) with proportion 0.60

We saw in Exercise 6.260 on page 425 that drinking tea appears to offer a strong boost to the immune system. In a study extending the results of the study described in that exercise, \(^{70}\) blood samples were taken on five participants before and after one week of drinking about five cups of tea a day (the participants did not drink tea before the study started). The before and after blood samples were exposed to e.coli bacteria, and production of interferon gamma, a molecule that fights bacteria, viruses, and tumors, was measured. Mean production went from 155 \(\mathrm{pg} / \mathrm{mL}\) before tea drinking to \(448 \mathrm{pg} / \mathrm{mL}\) after tea drinking. The mean difference for the five subjects is \(293 \mathrm{pg} / \mathrm{mL}\) with a standard deviation in the differences of 242 . The paper implies that the use of the t-distribution is appropriate. (a) Why is it appropriate to use paired data in this analysis? (b) Find and interpret a \(90 \%\) confidence interval for the mean increase in production of interferon gamma after drinking tea for one week.

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.