Raising the bar

Chapter 8. Estimation: One or Two Populations

Unit 8.4.2. Independent samples. Case I

Example 8.4. When efficiency of a new medicine is tested, usually its effect on a group of patients is compared to the health of members of a control group, who are not administered any medication. In this case we have two distinct samples of different sizes $latex n_X,\, n_Y$. They come independently from two different populations.

Assumptions. The samples $latex x = (x_1, \ldots, x_{n_X}),\, y = (y_1, \ldots, y_{n_Y})$ are independent normal. The population means are unknown and the population variances are known.

By properties of means and variances

$latex E(\bar X - \bar Y) = E\bar X - E\bar Y = \mu_X - \mu_Y,$

$latex V(\bar X - \bar Y) = V(\bar X) + V(\bar Y) = \frac{\sigma_X^2}{n_X} + \frac{\sigma_Y^2}{n_Y}. \qquad (\textrm{Eq. 8.8})$

Step 1. Find the sample means.

Step 2. Choose the level of significance $latex \alpha$ and find the critical value of the $latex z$ statistic from $latex P(z_{n-1} > z_{\alpha/2}) = \alpha/2$.

Step 3. Then the confidence interval for $latex \mu_X - \mu_Y$ is

$latex P(\bar X - \bar Y - ME < \mu_X - \mu_Y < \bar X - \bar Y + ME) = 1 - \alpha, \quad ME = z_{\alpha/2} \sqrt{\frac{\sigma_X^2}{n_X} + \frac{\sigma_Y^2}{n_Y}}.$

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