/*! 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 153 In Exercises 6.153 to \(6.158,\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
The mean is 0.1 and the standard error is 0.086. The Central Limit Theorem applies, so the sampling distribution of the difference in proportions is approximately normal with a curve centered at 0.1.

Step by step solution

01

Calculate the Mean

Substitute the given values into the formula for the mean of the distribution of differences between the two sample proportions. Here, \(p_A = 0.7\) and \(p_B = 0.6\), so \(\mu_{\hat{p}_A - \hat{p}_B} = p_A - p_B = 0.7 - 0.6 = 0.1\).
02

Compute the Variances and Standard Error

First, compute the variances of the two populations using the formula \(\sigma^2 = p*(1-p)\). For population \(A\), \({\sigma_A}^2 = 0.7 * (1 - 0.7) = 0.21\). For population \(B\), \({\sigma_B}^2 = 0.6 * (1 - 0.6) = 0.24\). Secondly, plug these values into the formula for the standard error, getting \(\sigma_{\hat{p}_A - \hat{p}_B} = \sqrt{{\sigma_A}^2/n_A + {\sigma_B}^2/n_B} = \sqrt{0.21/50 + 0.24/75} = \sqrt{0.0042 + 0.0032} = \sqrt{0.0074} = 0.086\).
03

Check the Conditions for the Central Limit Theorem

Check the conditions for the CLT to apply. For population \(A\), \(n_A*p_A = 50*0.7 = 35 > 5\) and \(n_A*(1 - p_A) = 50*(1 - 0.7) = 15 > 5\). For population \(B\), \(n_B*p_B = 75*0.6 = 45 > 5\) and \(n_B*(1 - p_B) = 75*(1 - 0.6) = 30 > 5\). Therefore, the CLT applies to both populations, and the sampling distribution of the difference between proportions is approximately normally distributed.
04

Draw the Distribution Curve

Draw a normal curve centered at the mean of the difference in proportions, which is 0.1. Include at least three values on the horizontal axis such as -0.1, 0.1, and 0.3.

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 principle in statistics that describes the distribution of sample means. It tells us that regardless of the distribution of the original population, the distribution of the sample mean will approximate a normal distribution as the sample size becomes large enough. This is crucial when working with sampling distributions because it allows us to apply normal distribution techniques even if the population distribution is unknown.

For the CLT to hold true, the sample size should be sufficiently large. In practical terms, a threshold of 30 is often used, but even smaller sizes can work if the population distribution is not too skewed. The CLT becomes particularly important when dealing with the differences between sample proportions, as it ensures that the difference follows a normal distribution when the samples are large.

A key point is to check if the conditions for the CLT apply. This involves ensuring that both the number of successes and failures in your samples exceed 5, which was verified in the exercise. So when you draw a plot for these differences, it should resemble a bell-shaped curve. This curve will be centered around the mean of the difference, which helps us understand and infer population traits even with limited sample data.
Sample Proportions
Sample proportions represent the fraction of the sampled population that shares a particular characteristic. In statistical studies and surveys, they help provide a glimpse of the broader population by examining a smaller, manageable group. For instance, if you survey 50 people out of a city and 30 of them favor a local policy, the sample proportion is 30/50 = 0.6.

In the context of comparing two populations, each sample will have its own proportion. In our exercise, this was illustrated by having sample sizes of 50 and 75 from populations with proportions of 0.7 and 0.6, respectively. The difference in these sample proportions gives insight into how different the two populations are regarding the trait of interest.
  • To compute this difference, you subtract one sample proportion from the other, providing a measure of variability between populations.
  • The mean of this difference helps us understand the central tendency of the differences over multiple sampling iterations.
  • Using sample proportions enables robust predictions about population behavior based on a smaller segment.
Through random sampling and repeated measurements, sample proportions become powerful tools to estimate population parameters accurately.
Standard Error
The standard error (SE) is a critical concept in statistics, signifying the degree to which a sample statistic fluctuates from the population parameter. In essence, it is the measure of the standard deviation of the sample mean or sample proportion. A smaller SE signifies the sample estimate is more representative of the true population parameter.

When you deal with sample proportions, the SE assists in understanding the variability of these proportions. Calculating the SE involves examining each population's variance and sample sizes. This is done by using the formula for the standard error of the difference between two sample proportions: \[ \sigma_{\hat{p}_A - \hat{p}_B} = \sqrt{\frac{{\sigma_A}^2}{n_A} + \frac{{\sigma_B}^2}{n_B}} \]where \(\sigma_A^2\) and \(\sigma_B^2\) are the variances of populations A and B, and \(n_A\) and \(n_B\) are the sample sizes.

This formula captures the combined variability from both sample components, providing us the SE of the differences in proportions. In practical applications, SE tells us how tight or wide our confidence intervals should be; narrower intervals suggest more precise estimates. By pinpointing this variance in sample results, researchers can draw more valid conclusions about the underlying population demographics.

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

Difference in mean commuting distance (in miles) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=18.16,\) and \(s_{1}=13.80\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=14.16,\) and \(s_{2}=10.75\) for St. Louis.

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 body temperature, in \({ }^{\circ} \mathrm{F},\) using the data in BodyTemp50 with \(n=50, \bar{x}=98.26,\) and \(s=0.765\)

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is 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 300 from a population with proportion 0.08

Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting cancer of any kind? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The distribution of sample means \(\bar{x}_{N}-\bar{x}_{E},\) where \(\bar{x}_{N}\) represents the mean Mathematics score for a sample of 100 people for whom the native language is not English and \(\bar{x}_{E}\) represents the mean Mathematics score for a sample of 100 people whose native language is English, is centered at 10 with a standard deviation of 17.41 . Give notation and define the quantity we are estimating with these sample differences. In the population of all students taking the test, who scored higher on average, non-native English speakers or native English speakers? Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.224 and \(6.225,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in 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 deviations as estimates of the population standard deviations.

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.