Raising the bar

Chapter 8. Estimation: One or Two Populations

Unit 8.4.1. Matched pairs

As we move into a more applied area, problem statements, their solutions and conditions for validity of the solutions will become more complex. The previous material should provide many enough stepping stones. The choices to be studied in this and the next units are:

Figure 8.3: Chart of different possibilities to be studied.

Example 8.3. A group of students takes a course and their performance in the subject is measured before and after taking it. The sample sizes are equal and there is a high dependence between the grade $latex x_i$ of the $latex i$th student before taking the course and the grade $latex y_i$ after taking it. Instead of writing the samples as separate vectors $latex x = (x_1, \ldots, x_n),\, y = (y_1, \ldots, y_n)$ it's logical to write the observations in pairs $latex (x_1, y_1), \ldots, (x_n, y_n)$.

Assumptions. Suppose $latex n$ matched pairs $latex (x_1, y_1), \ldots, (x_n, y_n)$ of observations from populations with means $latex \mu_X,\, \mu_Y$ are given and suppose that the differences $latex d_i = x_i - y_i$ are i.i.d. normal.

Step 1. Find the sample mean $latex \bar d$ and standard deviation $latex s_d$. ☞ $latex \mu_d = \mu_X - \mu_Y$.

Step 2. Choose the level of significance $latex \alpha$. Since we use sample standard deviation, the right statistic is $latex t$. Find its critical value for the two-tail confidence interval from the condition $latex P(t_{n-1} > t_{n-1, \alpha/2}) = \alpha/2$.

Step 3. By what we know from Exercise 8.6 the confidence interval is

$latex P(\bar d - ME < \mu_d < \bar d + ME) = 1 - \alpha, \quad ME = t_{n-1, \alpha/2} \frac{s_d}{\sqrt n}.$

It remains to replace $latex \mu_d$ by $latex \mu_X - \mu_Y$.

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