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91Ó°ÊÓ

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

Short Answer

Expert verified
The mean of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\), is 0.06. The standard error can be calculated using the formula provided in step 2. The sampling distribution will be approximately normal, and the three values on the horizontal axis should include 0.06 - SE, 0.06, and 0.06 + SE.

Step by step solution

01

Find the Mean

The mean of the difference in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\), is obtained by subtracting the population proportions. It can be calculated as: \(\mu_{\hat{p}_{A}-\hat{p}_{B}} = P_{A} - P_{B}\). Substituting \(P_{A} = 0.30\) and \(P_{B} = 0.24\) into the equation gives \(\mu_{\hat{p}_{A}-\hat{p}_{B}} = 0.30 - 0.24 = 0.06\)
02

Find the Standard Error

The standard error of the difference in sample proportions can be calculated using the formula: \(SE_{\hat{p}_{A}-\hat{p}_{B}} = \sqrt{ \frac{{P_{A} * (1-P_{A})}}{{n_{A}}} + \frac{{P_{B} * (1-P_{B})}}{{n_{B}}} }\). Substituting the given values (\(P_{A} = 0.30\), \(P_{B} = 0.24\), \(n_{A} = 40\), and \(n_{B} = 30\)) into the equation, calculating the result gives the standard error.
03

Draw the Sampling Distribution

The sampling distribution will be almost normal if the sample sizes are large enough according to the Central Limit Theorem. Our mean is 0.06 from step 1, and let's denote the standard error as SE. Then, for the normal distribution, the values on the horizontal axis should at least include \(\mu_{\hat{p}_{A}-\hat{p}_{B}} - 1 * SE_{\hat{p}_{A}-\hat{p}_{B}} = 0.06 - SE\), \(\mu_{\hat{p}_{A}-\hat{p}_{B}} = 0.06\), and \(\mu_{\hat{p}_{A}-\hat{p}_{B}} + 1 * SE_{\hat{p}_{A}-\hat{p}_{B}} = 0.06 + SE\).

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Key Concepts

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

Difference in Sample Proportions
Understanding the difference in sample proportions is crucial for statistical analysis when we want to compare two groups. For example, a researcher might want to compare the proportion of voters who support a particular policy in different regions. In practice, we estimate these proportions by taking random samples from each population.

To find the mean difference in sample proportions (\( \bar{p}_A-\bar{p}_B \)), we simply subtract the population proportions of group B from group A. It provides us with an expected value of the difference, assuming that our samples accurately reflect the populations they're drawn from. However, we also need to consider variability in this estimate, which involves calculating the standard error of the difference in sample proportions. This is essential for constructing confidence intervals and conducting hypothesis tests.
Standard Error
When we talk about the standard error, we're referring to a measure of the variation or spread of the sampling distribution of a statistic – typically the mean. In the context of differences in sample proportions, the standard error helps us grasp the variability of the difference between two sample proportions from their true population value.

The formula for the standard error of the difference between two independent sample proportions, as shown in the step-by-step solution, is based on the individual proportions and sample sizes. Understanding and calculating this value is fundamental because it plays a pivotal role in the precision of our estimates. A smaller standard error indicates that the sample statistic is likely to be closer to the population parameter. This is often the sought-after scenario in research, as it implies more confidence in the sample results.
Central Limit Theorem
At the heart of inferential statistics lies the Central Limit Theorem (CLT), a powerful concept that aids in understanding the distribution of sample means. This theorem states that, given a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution.

This notion is incredibly useful when dealing with sample proportions. Even if the original population does not follow a normal distribution, if our samples are large enough, we can assume that the sampling distribution of the difference in sample proportions is normal. This allows statisticians to use the normal probability model to make inferences about population parameters from sample statistics, even when the underlying population distribution is unknown or not normally distributed. By applying the CLT, we can apply z-scores and predict probabilities for our sample estimates, which are essential for further statistical analysis like hypothesis testing.

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Most popular questions from this chapter

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the area in a t-distribution above 2.1 if the samples have sizes \(n_{1}=12\) and \(n_{2}=12\).

We see in the AllCountries dataset that the percent of the population that is elderly (over 65 years old) is 17.0 in Austria and 15.9 in Denmark. Suppose we take random samples of size 200 from each of these countries and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{D}\), where \(\hat{p}_{A}\) represents the proportion of the sample that is elderly in Austria and \(\hat{p}_{D}\) represents the proportion of the sample that is elderly in Denmark. Find the mean and standard deviation of the differences in sample proportions.

In Exercises 6.166 and \(6.167,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section (and using the sample proportions to estimate the population proportions). Sample A has a count of 30 successes with \(n=100\) and Sample \(\mathrm{B}\) has a count of 50 successes with \(n=250\).

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(2.5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=15\) and \(n_{2}=25\)

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) 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 \(\hat{p}=0.753\)

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