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Chapter 11: Problem 1

Consider two populations for which \(\mu_{1}=30, \sigma_{1}=2\), \(\mu_{2}=25\), and \(\sigma_{2}=3\). Suppose that two independent random samples of sizes \(n_{1}=40\) and \(n_{2}=50\) are selected. Describe the approximate sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) (center, spread, and shape).

Short Answer

Expert verified
The center (expected mean) of the sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) is 5, the spread (standard deviation) is approximately 0.48, and the shape is approximately normal.

Step by step solution

01

Determine the expected value of \(\bar{x}_{1}-\bar{x}_{2}\)

This can be found by subtracting the population means \(E[\bar{x}_{1} - \bar{x}_{2}] = \mu_{1} - \mu_{2}\). Substituting the given values we get, \(E[\bar{x}_{1}-\bar{x}_{2}] = 30 - 25 = 5\).
02

Determine the standard deviation of \(\bar{x}_{1}-\bar{x}_{2}\)

The standard deviation \(\sigma_{\bar{x}_{1} - \bar{x}_{2}}\) is given by the formula: \(\sqrt{\sigma_{1}^{2}/n_{1} + \sigma_{2}^{2}/n_{2}}\). Replacing our known values into the formula: \(\sqrt{2^2/40 + 3^2/50} = \sqrt{0.05 + 0.18} = \sqrt{0.23} 鈮� 0.48\). The standard deviation, therefore, is about 0.48.
03

Determine the shape of the distribution

The Central Limit Theorem (CLT) states that if the sample size is large enough (>30), the distribution of the sample means will approximate a normal distribution, regardless of the shape of the population distribution. Since both sample sizes, \(n_{1}=40\) and \(n_{2}=50\), exceed 30, the distribution of \(\bar{x}_{1}-\bar{x}_{2}\) will be approximately normal.

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