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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 300 from Population 1 with mean 75 and standard deviation 18 and samples of size 500 from Population 2 with mean 83 and standard deviation 22

Short Answer

Expert verified
The mean of the differences is -8 and the standard error of the differences is approximately 1.43. The distribution can be approximated by a normal distribution with these parameters due to the Central Limit Theorem.

Step by step solution

01

Calculate the Mean of the Differences

Mean difference \(\mu_{\bar{x1} - \bar{x2}} = \mu_{x1} - \mu_{x2} = 75 - 83 = -8\). The mean difference is -8.
02

Calculate the Standard Error of the Differences

Standard error \(\sigma_{\bar{x1} - \bar{x2}} = \sqrt{\frac{{\sigma_{x1}^2}}{n1} + \frac{{\sigma_{x2}^2}}{n2}} = \sqrt{\frac{{18^2}}{300} + \frac{{22^2}}{500}} = \sqrt{\frac{324}{300} + \frac{484}{500}} = \sqrt{1.08 + 0.968} = \sqrt{2.048}\). The standard error is approximately 1.43.
03

Draw the Curve of the Sampling Distribution

Using the Central Limit Theorem, a curve illustrating the sampling distribution could be drawn. As the sample sizes here are large enough for the theorem to apply, the distribution is assumed to be normally distributed with a mean of -8 (from step 1) and a standard error of 1.43 (from step 2). On the horizontal axis, plot at least three values: the mean (-8), and two values equalling to the mean plus/minus the standard error (-8 + 1.43, -8 - 1.43).

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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 provides a bridge between the world of probability and the practical world of data analysis. It states that when an adequately large number of random samples are taken from any population with a finite level of variance, the sampling distribution of the sample means will approximate a normal distribution (also known as a Gaussian distribution), regardless of the population's original distribution shape.

The beauty of the CLT lies in its ability to simplify complex distributions. By asserting that the sampling distribution converges to a normal distribution as sample sizes increase, it grants us permission to use statistical methods that are predicated on normality, even when the population itself is not normally distributed.

To illustrate with the given exercise, when we take large samples of size 300 and 500 from our two populations, the distribution of the differences in sample means \(\bar{x}_{1}-\bar{x}_{2}\) would be expected to form a bell-shaped curve, typical of a normal distribution, due to the CLT. This allows us to make inferences about the population based on the sample data.
Standard Error
Understanding the standard error is essential when you're dealing with sample statistics. The standard error quantifies the variability or uncertainty around a sample statistic, such as the sample mean. Specifically, it measures how much the sample mean is expected to fluctuate from the true population mean were we to take multiple samples.

In practical terms, a smaller standard error indicates that the sample mean is likely a good approximation of the population mean, because there is less variation from sample to sample. Conversely, a larger standard error suggests more variability, indicating that our sample mean may be less representative of the population mean.

Using the formula presented in the exercise's solution, \(\sigma_{\bar{x1} - \bar{x2}} = \sqrt{\frac{{\sigma_{x1}^2}}{n1} + \frac{{\sigma_{x2}^2}}{n2}}\), we calculated the standard error for the difference in sample means. This particular formula allows us to consider the variability in both samples and not just within a single sample, adding depth to our assessment of how the sample means might differ from each other.
Mean Difference
In the context of our exercise, the mean difference is concerned with the difference between the means of the two populations as estimated by the difference between the means of our two samples. In simpler terms, it is the average amount by which one sample mean exceeds the other.

The mean difference plays a pivotal role when we're comparing two sets of data. It is the basis for many statistical tests that examine whether there's a significant difference between two groups, such as the independent samples t-test. The formula for the mean difference between two independent samples is \(\mu_{\bar{x1} - \bar{x2}} = \mu_{x1} - \mu_{x2}\), where \(\mu_{x1}\) and \(\mu_{x2}\) are the population means.

In the provided exercise, the mean difference is calculated as -8, signifying that, on average, sample means taken from Population 1 are 8 units less than those from Population 2. It is important to note that mean differences could be both negative and positive; the sign reflects the direction of the difference between the group means.

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Most popular questions from this chapter

IQ tests scale the scores so that the mean IQ score is \(\mu=100\) and standard deviation is \(\sigma=15\). Suppose that 30 fourth graders in one class are given such an IQ test that is appropriate for their grade level. If the students are really a random sample of all fourth graders, what is the chance that the average IQ score for the class is above \(105 ?\)

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