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Chapter 5: Problem 21

Discuss the problem of finding a confidence interval for the difference \(\mu_{1}-\mu_{2}\) between the two means of two normal distributions if the variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\) are known but not necessarily equal.

Short Answer

Expert verified
To find the confidence interval for the difference between two means of two normal distributions when the variances are known but not necessarily equal, you can use the formula \( CI = (\mu_{1}-\mu_{2}) 卤 Z* \sqrt{(\sigma_{1}^{2}/n_{1}) + (\sigma_{2}^{2}/n_{2})} \) where \( CI \) represents the confidence interval, \( \mu_{1}-\mu_{2} \) is the difference of the population means, \( Z* \) is the z-score that corresponds to the desired confidence level, \( \sigma_{1}^{2} \) and \( \sigma_{2}^{2} \) are the variances, and \( n_{1} \) and \( n_{2} \) are the sample sizes. Simply plug in the appropriate values into this formula to compute the confidence interval.

Step by step solution

01

Understand Confidence Interval Formula

The general formula for a confidence interval for the difference between two means is \( CI = (\mu_{1}-\mu_{2}) 卤 Z* \sqrt{(\sigma_{1}^{2}/n_{1}) + (\sigma_{2}^{2}/n_{2})} \). Here, \( n_{1} \) and \( n_{2} \) represent the sample sizes, \( Z* \) is the z-score that corresponds to your chosen level of confidence, \( \mu_{1}-\mu_{2} \) is the difference of the means, and \( \sigma_{1}^{2} \) and \( \sigma_{2}^{2} \) are the variances of the two populations.
02

Apply Known Variances

Next, you apply the known variances \( \sigma_{1}^{2} \) and \( \sigma_{2}^{2} \) into the formula. This becomes \( CI = (\mu_{1}-\mu_{2}) 卤 Z* \sqrt{(\sigma_{1}^{2}/n_{1}) + (\sigma_{2}^{2}/n_{2})} \).
03

Calculate Confidence Interval

Lastly, to calculate the confidence interval, you will need to know or estimate other factors in the equation such as the sample sizes \( n_{1} \) and \( n_{2} \), the z-score for your desired confidence level, and the difference between the two means \( \mu_{1}-\mu_{2} \). Once these are known or estimated, put them in the equation to compute the confidence interval.

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Most popular questions from this chapter

Let \(X\) and \(Y\) denote independent random variables with respective probability density functions \(f(x)=2 x, 0

Let \(Y_{1}1\) is accepted if the observed \(Y_{4} \geq c\) (a) Find the constant \(c\) so that the significance level is \(\alpha=0.05\). (b) Determine the power function of the test.

Let \(X_{1}, X_{2}, \ldots, X_{9}\) be a random sample of size 9 from a distribution that is \(N\left(\mu, \sigma^{2}\right)\) (a) If \(\sigma\) is known, find the length of a 95 percent confidence interval for \(\mu\) if this interval is based on the random variable \(\sqrt{9}(\bar{X}-\mu) / \sigma\). (b) If \(\sigma\) is unknown, find the expected value of the length of a 95 percent confidence interval for \(\mu\) if this interval is based on the random variable \(\sqrt{9}(\bar{X}-\mu) / S\). Hint: \(\quad\) Write \(E(S)=(\sigma / \sqrt{n-1}) E\left[\left((n-1) S^{2} / \sigma^{2}\right)^{1 / 2}\right]\). (c) Compare these two answers.

In Exercise \(5.4 .25\), in finding a confidence interval for the ratio of the variances of two normal distributions, we used a statistic \(S_{1}^{2} / S_{2}^{2}\), which has an \(F\) distribution when those two variances are equal. If we denote that statistic by \(F\), we can test \(H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}\) against \(H_{1}: \sigma_{1}^{2}>\sigma_{2}^{2}\) using the critical region \(F \geq c\). If \(n=13, m=11\), and \(\alpha=0.05\), find \(c .\)

Assume a binomial model for a certain random variable. If we desire a 90 percent confidence interval for \(p\) that is at most \(0.02\) in length, find \(n\). Hint: Note that \(\sqrt{(y / n)(1-y / n)} \leq \sqrt{\left(\frac{1}{2}\right)\left(1-\frac{1}{2}\right)}\).
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