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Chapter 4: Problem 26

Let \(\bar{X}\) and \(\bar{Y}\) be the means of two independent random samples, each of size \(n\), from the respective distributions \(N\left(\mu_{1}, \sigma^{2}\right)\) and \(N\left(\mu_{2}, \sigma^{2}\right)\), where the common variance is known. Find \(n\) such that $$ P\left(\bar{X}-\bar{Y}-\sigma / 5<\mu_{1}-\mu_{2}<\bar{X}-\bar{Y}+\sigma / 5\right)=0.90 $$

Short Answer

Expert verified
The required sample size is \(n = 50\).

Step by step solution

01

Identify the distribution

Understand that \(\bar{X}\) and \(\bar{Y}\) are the sample means from two independent normal distributions, and their difference follows another normal distribution with mean \( \mu_1 - \mu_2 \) and variance \( 2 \sigma^{2}/n \).
02

Calculate the z-score

Know that the z-score is given by the difference between the sample mean and the population mean divided by the standard deviation of the sample mean. Here the expression \( \sigma / 5 \) represents the standard deviation of the difference of the sample means, which implies that \( \sqrt{2 \sigma^{2} / n} = \sigma / 5 \). Simplifying this equation gives \( n = 50 \).
03

Check the associated probability

Realize that you are looking for a 90% probability, which corresponds to a z-score of approximately 卤1.645 for a two-sided test. Since the z-score calculated in the previous step is within this range, \( n = 50 \) is the right solution.

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