Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

If random samples of the given sizes are drawn from populations with the given proportions: (a) Find the standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) Determine whether the sample sizes are large enough for the Central Limit Theorem to apply. Samples of size 100 from population \(A\) with proportion 0.20 and samples of size 50 from population \(B\) with proportion 0.30

Short Answer

Expert verified
The standard error of the distribution of differences in sample proportions is approximately 0.07. Yes, the sample sizes are large enough for the Central Limit Theorem to apply.

Step by step solution

01

Finding the Standard Error

To start, the formula for the standard error of two proportions is formulated as follows: \(\sigma_{\hat{p}_{A}-\hat{p}_{B}}=\sqrt{\frac{{p_{A}(1-p_{A})}}{{n_{A}}}} + \frac{{p_{B}(1-p_{B})}}{{n_{B}}}}\). Here, \(p_{A}\) and \(p_{B}\) are the proportions in the two populations and \(n_{A}\) and \(n_{B}\) are the sizes of the samples drawn from the two populations. Substituting the given values into this formula will yield the standard error.
02

Calculation

Substituting given values, we get \(\sigma_{\hat{p}_{A}-\hat{p}_{B}}=\sqrt{\frac{{0.20(1-0.20)}}{{100}} + \frac{{0.30(1-0.30)}}{{50}}}\). This simplifies to \(\sigma_{\hat{p}_{A}-\hat{p}_{B}}=\sqrt{\frac{{0.16}}{{100}} + \frac{{0.21}}{{50}}}\), and further simplification gives the answer for the standard error.
03

Apply the Central Limit Theorem

The Central Limit Theorem can be applied if both \(n_{A}p_{A}\) and \(n_{B}p_{B}\) are greater than or equal to 5. In this case, \(n_{A}p_{A}=100*0.20=20\) and \(n_{B}p_{B}=50*0.30=15\). Both results are greater than 5, hence, the conditions for the Central Limit Theorem are satisfied, and it can be applied here.

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Error
When dealing with statistics, one crucial concept to grasp is standard error. It measures the amount of variability or dispersion in a sampling distribution. Specifically, in the context of differences between sample proportions, the standard error quantifies the expected distribution of these differences if you were to repeat the sampling process many times. Think of it as a gauge of how much the statistic you calculate from a sample, like a sample mean or proportion, would differ from the true population value.

To calculate the standard error for the difference between two sample proportions, denoted as \(\sigma_{\hat{p}_{A}-\hat{p}_{B}}\), you virtually blend the formulas for the standard error of both samples. The formula is \(\sigma_{\hat{p}_{A}-\hat{p}_{B}}=\sqrt{\frac{{p_{A}(1-p_{A})}}{{n_{A}}} + \frac{{p_{B}(1-p_{B})}}{{n_{B}}}}\), where \(p_{A}\) and \(p_{B}\) represent the known proportions in the populations, and \(n_{A}\) and \(n_{B}\) are the sample sizes. The square root encapsulates both variability and sample size, reflecting how they influence the precision of the sample proportions.
Distribution of Sample Proportions
The distribution of sample proportions describes how the proportion values are spread out for all possible random samples of a certain size from a population. It's like creating a map of all the outcomes you might expect if you took many, many samples. This distribution tends to form a shape that is approximately normal (bell-shaped) when the sample size is large enough, which is particularly handy as it allows statisticians to use the normal distribution's properties to make inferences about the population.

In real-world scenarios, you rarely have the luxury of polling an entire population. Therefore, when you select a sample and calculate its proportion, this acts as an estimate for the population proportion. The central limit theorem reassures us that with a big enough sample size, these estimates become more reliable. The distribution's standard deviation, or its standard error, helps determine how much the sample proportion will likely differ from the actual population proportion.
Sampling Distributions
The foundational bedrock for many statistical inference techniques is the concept of sampling distributions. This concept imagines what would happen if you took every possible sample of a particular size from the population and then mapped out the frequency of the sample statistics. The sampling distribution represents the probability distribution of a given statistic based on a random sample and it is crucial in the field of inferential statistics.

The beauty of the sampling distribution lies in its predictable nature, as dictated by the central limit theorem. This theorem states that no matter what the shape of the original population distribution is, the sampling distribution of the sample mean (or proportion) will approximate a normal distribution if the sample size is large enough. This property holds true as long as the samples are independent and identically distributed (i.i.d.). A practical use of the sampling distribution is it allows the calculation of confidence intervals, which are ranges within which the true population parameter is likely to lie. With larger samples, the wider range of possible sample means (or proportions) narrows down, making your estimate more precise.

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

Split the Bill? Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The data in SplitBill includes the cost of what each person ordered (in Israeli shekels) and the payment method (Individual or Split). Some summary statistics are provided in Table 6.20 and both distributions are reasonably bell-shaped. Use this information to test (at a \(5 \%\) level ) if there is evidence that the mean cost is higher when people split the bill. You may have done this test using randomizations in Exercise 4.118 on page 302 .

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=15.7, s_{d}=12.2\) \(n_{d}=25 .\)

Autism and Maternal Antidepressant Use A recent study \(^{41}\) compared 298 children with Autism Spectrum Disorder to 1507 randomly selected control children without the disorder. Of the children with autism, 20 of the mothers had used antidepressant drugs during the year before pregnancy or the first trimester of pregnancy. Of the control children, 50 of the mothers had used the drugs. (a) Is there a significant association between prenatal exposure to antidepressant medicine and the risk of autism? Test whether the results are significant at the \(5 \%\) level. (b) Can we conclude that prenatal exposure to antidepressant medicine increases the risk of autism in the child? Why or why not? (c) The article describing the study contains the sentence "No increase in risk was found for mothers with a history of mental health treatment in the absence of prenatal exposure to selective serotonin reuptake inhibitors [antidepressants]." Why did the researchers conduct this extra analysis?

Use the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal. Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the sample results \(\bar{x}_{1}=56, s_{1}=8.2\) with \(n_{1}=30\) and \(\bar{x}_{2}=51, s_{2}=6.9\) with \(n_{2}=40\).

Use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of survey respondents who say exercise is important, with \(n=1000\) and \(\hat{p}=0.753\)

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.

Sign-up for free