1. 1 Matched Samples The paired t test

2. 2 Sometimes in a statistical setting we will have information about the same person at different points in time. A classical example of this is the pretest/posttest design (or the before and after test). The idea of this design is that we measure something (knowledge, weight, blood pressure, how much coke is enjoyed), then we do something to the person (educate them, have them do something, etc). After this intervention the subject is measured again. The point then is did the intervention somehow change the value of the measurement. (another application here is to have the same people do two things and measure after each and see if there is a difference.) It seems we have two populations to compare, the population before and the population after the intervention. In the context of having the same people involved at each point in time means we have a special design. We can do the matched sample inference methods.

3. 3 Image you have data on people before and after some program. If for each person you calculate the value before minus after you will have a number for each person. In this sense we are right back to having a single variable. The inference procedures are right back to the population mean inference procedures for one variable. So, when we have matched samples two groups really collapses back to one group and the one variable is a difference in value variable. Another context would be that you do not have the same people, but the people in two groups have enough characteristics in common that you can consider them a matched sample. Then, again you calculate differences in the values and treat the differences as a single variable (instead of people you might have stores, or states or other units of analysis). In the context here we use a t statistic with df = n – 1. (It’s probably true the population standard deviation is unknown here.)

4. Problem 10.20 page 403 4 A1A2A1 - A2A2 - A1 5053-33 2527-22 3038-88 5055-55 60611 8085-55 45 00 30311 6572-77 7078-88 t-Test: Paired Two Sample for Means A1A2 Mean50.554.5 Variance341.3888889389.3888889 Observations10 Pearson Correlation0.98966949 Hypothesized Mean Difference0 df9 t Stat-4.190581775 P(T<=t) one-tail0.00116955 t Critical one-tail1.833112933 P(T<=t) two-tail0.0023391 t Critical two-tail2.262157163 t-Test: Paired Two Sample for Means A2A1 Mean54.550.5 Variance389.3888889341.3888889 Observations10 Pearson Correlation0.98966949 Hypothesized Mean Difference0 df9 t Stat4.190581775 P(T<=t) one-tail0.00116955 t Critical one-tail1.833112933 P(T<=t) two-tail0.0023391 t Critical two-tail2.262157163 You will refer back to this information.

5. 5 In this example there is information on people about their attitude before and after being part of a study. Presumably being part of the study changed the person. Moreover, here we think that being in the study increased their attitude. (Their attitude could be some state of happiness.) Thus, if you took pre minus post attitude you would think the values would be negative. Another way to say this is that you expect the post minus the pre to be positive. So, you could have 1)pre minus post < 0 or 2)post minus pre > 0. Either way the post would be higher. The way the null and alternative hypotheses are stated depends on how you construct the difference.

6. example 6 If we have pre minus post then we are expecting differences to be negative. In that case the null would be Ho: μD ≥ 0, against the alternative H1: μD < 0 If we have post minus pre we would reverse the signs on the null and alternative above. Notice on slide 4 the Excel information is virtually the same. On the bottom left I had pre minus post and thus the t stat is negative. Since we have a less than sign in the alternative hypothesis we do a one tailed test. The p-value of.0012 means we reject the null at the.1,.05, and.01 levels, but not at the.001 level of significance. We have very strong evidence of a difference.

7. example 7 If we make a confidence interval – at the 95% level – with a mean difference equal to a minus 4 we need to get the t value from the t table with 9 degrees of freedom (the value is 2.262) and we need the sample standard deviation of the differences (here = to 3.02, which you could find by calculating the sample standard deviation for the numbers in the column A1 – A2.). The interval would be -4 – and + 2.262(3.02)/square root of 10, or -4 – and + 2.16, and so the interval is [-6.16, -1.84]. The minimum difference is then -1.84 at the 95% level of confidence.