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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 500 from population \(A\) with proportion 0.58 and samples of size 200 from population \(B\) with proportion 0.49

Short Answer

Expert verified
The mean of the difference in sample proportions is 0.09 and the standard error is 0.04264. The Central Limit Theorem can be applied in this case since the sample sizes were large enough, leading to the approximation of the sampling distribution as a normal distribution.

Step by step solution

01

Find the mean of difference in sample proportions

The mean of the difference in sample proportions is given by \(\mu_{\hat{p}_{A}-\hat{p}_{B}} = p_{A} - p_{B}\). Substitute \(p_{A}\) = 0.58 and \(p_{B}\) = 0.49 into the equation to get \(\mu_{\hat{p}_{A}-\hat{p}_{B}} = 0.58 - 0.49 = 0.09\).
02

Find the standard error of difference in sample proportions

The standard error of the difference in sample proportions is given by \(\sigma_{\hat{p}_{A}-\hat{p}_{B}} = \sqrt{\frac{p_{A}(1-p_{A})}{n_{A}} + \frac{p_{B}(1-p_{B})}{n_{B}}}\). Substitute \(p_{A}\) = 0.58, \(p_{B}\) = 0.49, \(n_{A}\) = 500 and \(n_{B}\) = 200 into the equation to get \(\sigma_{\hat{p}_{A}-\hat{p}_{B}} = \sqrt{\frac{0.58(1-0.58)}{500} + \frac{0.49(1-0.49)}{200}} = 0.04264 \)
03

Apply Central Limit Theorem and draw the curve

The Central Limit Theorem tells us that if samples sizes are large enough (generally if \(np > 5\) and \(n(1-p) > 5\)), the sampling distribution of the difference in sample proportions will be approximately normal. We used samples of size 500 and 200 which are large enough, so the Theorem can be applied. The curve will be a normal curve centered at 0.09 (the mean), with standard deviation 0.04264, density varying as we get further from the mean. The three values on the horizontal axis could be \(0.09 - 0.04264\), 0.09 and \(0.09 + 0.04264\).

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Key Concepts

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

sampling distribution
The concept of a sampling distribution is crucial in statistics. When we talk about a sampling distribution, we're looking at the distribution of a particular statistic across many samples from the same population. In this context, we are interested in the sampling distribution of the difference in sample proportions, \( \hat{p}_{A} - \hat{p}_{B} \).

Imagine you repeatedly take samples from a population, calculate the difference between two sample proportions, and record that difference. If you do this many times, the recorded differences will form a distribution — this is your sampling distribution.

Thanks to the Central Limit Theorem, we know that when sample sizes are large enough, the shape of this sampling distribution will become approximately normal.
  • The center of this distribution is the true difference in population proportions.
  • Its spread (or variability) is determined by the standard error.
This normal approximation is a powerful tool. It allows us to make inferences about a population without directly studying all its members.
mean difference
Understanding the mean difference is fundamental when comparing two populations. The mean of the difference in sample proportions is straightforward to compute. Represented as \( \mu_{\hat{p}_{A} - \hat{p}_{B}} \), it's simply the difference between the two population proportions: \( p_{A} - p_{B} \).

This tells us on average how much the proportions differ across different samples. In our problem, \( p_{A} = 0.58 \) and \( p_{B} = 0.49 \), resulting in \( \mu_{\hat{p}_{A} - \hat{p}_{B}} = 0.09 \).

This mean difference is the center of our sampling distribution. It tells us the expected direction and magnitude of the disparity in proportions when we randomly sample from these populations repeatedly.
standard error
The standard error (SE) is like a magnifying glass, highlighting the variability in a statistic from sample to sample. In the context of our exercise, the standard error of the difference in sample proportions is pivotal.

Calculated as \( \sigma_{\hat{p}_{A} - \hat{p}_{B}} = \sqrt{\frac{p_{A}(1-p_{A})}{n_{A}} + \frac{p_{B}(1-p_{B})}{n_{B}}} \), this measure reflects how much the sample proportions differ naturally due to random sampling.

Plugging in values: \( p_{A} = 0.58, p_{B} = 0.49, n_{A} = 500, \) and \( n_{B} = 200 \), we find \( \sigma_{\hat{p}_{A} - \hat{p}_{B}} = 0.04264 \).

The smaller this standard error, the closer our sampling proportion is likely to be to the true population proportion difference. Thus, it helps in constructing confidence intervals and conducting hypothesis tests. The SE is crucial for understanding the precision of our estimate — the smaller the SE, the more precise our estimate is.

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Most popular questions from this chapter

According to the 2006 Australia Census, \(^{43} 25.5 \%\) of Australian women over the age of 25 had a college degree, while the percentage for Australian men was \(21.4 \% .\) Suppose we select random samples of 200 women and 200 men from this population and look at the differences in proportions with college degrees, \(\hat{p}_{f}-\hat{p}_{m}\), in those samples. (a) Describe the distribution (center, spread,shape) for the difference in sample proportions. Include a rough sketch of the distribution with values labeled on the horizontal axis. (b) What is the chance that the proportion with college degrees in the men's sample is actually more than the proportion in the women's sample? (Hint: Think about what must be true about \(\hat{p}_{f}-\hat{p}_{m}\) when this happens.)

What Percent of Houses Are Owned vs }\end{array}\( Rented? The 2010 US Census \)^{4}\( reports that, of all the nation's occupied housing units, \)65.1 \%\( are owned by the occupants and \)34.9 \%$ are rented. If we take random samples of 50 occupied housing units and compute the sample proportion that are owned for each sample, what will be the mean and standard deviation of the distribution of sample proportions?

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 lie detector trials in which the technology misses a lie, with \(n=48\) and \(\hat{p}=0.354\)

Impact of the Population Proportion on SE Compute the standard error for sample proportions from a population with proportions \(p=0.8, p=0.5\), \(p=0.3,\) and \(p=0.1\) using a sample size of \(n=100\). Comment on what you see. For which proportion is the standard error the greatest? For which is it the smallest?

Who Is More Trusting: Internet Users or Non-users? In a randomly selected sample of 2237 US adults, 1754 identified themselves as people who use the Internet regularly while the other 483 indicated that they do not. In addition to Internet use, participants were asked if they agree with the statement "most people can be trusted." The results show that 807 of the Internet users agree with this statement, while 130 of the non-users agree. \(^{54}\) (a) Which group is more trusting in the sample (in the sense of having a larger percentage who agree): Internet users or people who don't use the Internet? (b) Can we generalize the result from the sample? In other words, does the sample provide evidence that the level of trust is different between the two groups in the broader population? (c) Can we conclude that Internet use causes people to be more trusting? (d) Studies show that formal education makes people more trusting and also more likely to use the Internet. Might this be a confounding factor in this case?

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