Question

To illustrate Exercise 4.2.24, let \(X_{1}, X_{2}, \ldots, X_{9}\) and \(Y_{1}, Y_{2}, \ldots, Y_{12}\) represent two independent random samples from the respective normal distributions \(N\left(\mu_{1}, \sigma_{1}^{2}\right)\) and \(N\left(\mu_{2}, \sigma_{2}^{2}\right) .\) It is given that \(\sigma_{1}^{2}=3 \sigma_{2}^{2}\), but \(\sigma_{2}^{2}\) is unknown. Define a random variable that has a \(t\) -distribution that can be used to find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

Short answer

The required random variable, having a \(t\)-distribution, is defined as \(t = \frac{\bar{X} - \bar{Y}}{\sqrt{s_{p}^{2} \left(\frac{1}{n} + \frac{1}{m}\right)}}\) with \(n + m - 2 = 19\) degrees of freedom. The 95% confidence interval for \(\mu_{1}-\mu_{2}\) is \((\bar{X} - \bar{Y}) \pm t_{0.025, 19} \sqrt{s_{p}^{2} \left(\frac{1}{n} + \frac{1}{m}\right)}\).

Step-by-step solution

Compute sample means

Given the two independent samples \(X_{1}, X_{2}, \ldots, X_{9}\) and \(Y_{1}, Y_{2}, \ldots, Y_{12}\), calculate their respective sample means: \[\bar{X} = \frac{1}{9}(X_{1} + X_{2} + \ldots + X_{9})\] \[\bar{Y} = \frac{1}{12}(Y_{1} + Y_{2} + \ldots + Y_{12})\] These represent the sample estimates for \(\mu_1\) and \(\mu_2\) respectively.

Estimate variances

Use the relationship between the variances, \(\sigma_1^2 = 3\sigma_2^2\), and compute the pooled variance. Given there is no direct information about \(\sigma_2^2\), replace \(\sigma_1^2\) with \(s_1^2\) and \(\sigma_2^2\) with \(s_2^2\) in the equation. The estimates for the population variances \(\sigma_1^2\) and \(\sigma_2^2\) are \(s_1^2\) and \(s_2^2\) respectively. So, \(s_1^2 = 3s_2^2\). This equation can be used in the formula for pooled variance.

Compute the t-statistic

The t-statistic can now be computed as follows: \[t = \frac{\bar{X} - \bar{Y}}{\sqrt{s_{p}^{2} \left(\frac{1}{n} + \frac{1}{m}\right)}}\] where \(n=9\) and \(m=12\) are the respective sample sizes and \(s_{p}^{2}\) is the pooled variance. This will have a t-distribution with \(n + m - 2 = 9 + 12 - 2 = 19\) degrees of freedom.

Define the Confidence Interval

The 95% confidence interval for \(\mu_{1}-\mu_{2}\) can be defined as follows: \[(\bar{X} - \bar{Y}) \pm t_{0.025, 19} \sqrt{s_{p}^{2} \left(\frac{1}{n} + \frac{1}{m}\right)}\] where \(t_{0.025, 19}\) is the t-score corresponding to a 95% confidence level with 19 degrees of freedom.