Section 3: Two dependent means - matched pairs

First a definition:

 Two samples are independent if the sample selected from one population is not related to the sample selected from the other population. If one sample is related to the other, the samples are dependent. When we get samples from the same subject or one value from each of two samples sharing the same characteristic they are called paired or matched samples

 In many cases these are "before and after" situations. Look at this situation

 Example 1: The "new grapefruit diet" has been advertised to be a remarkable way to lose weight. Ten males subjects are put on the diet and here are the results after one month

Before we begin the hypothesis test on this example, we need to define some terms/concepts

 1. Two dependent samples must be selected from two populations in a way that is random

 2. Each of the two populations must be normally distributed

 Notation:

  = mean value of the differences d for the population of paired data

   = mean value of the differences d for the paired sample data

  = standard deviation of the differences for the paired sample data

 n = number of pairs of data

 To do a hypothesis test on paired data, we find the differences between the two sets of data (usually after - before) and then test it against 0. Here is the example given above:

 If the diet "works" we would expect the average difference to be significantly less than 0.  Since we are doing after minus before we would want the after weight to be smaller in this situation. So we make this our claim:

 Lets use the 0.05 significance level

 Now we need a test statistic - here it is:

  , where n-1 = degrees of freedom - note this is a t test often called the paired t test - use this if you have 30 or fewer pairs, use the normal distribution if you have greater than 30 pairs

 Now lets solve the problem like we would any hypothesis test. First calculate the mean and standard deviation of the differences, in our case

 For 9 degrees of freedom the critical value is t = -1.833. Here is the picture

 So we reject the null hypothesis. The sample evidence supports the claim that the diet works

 We can also do a confidence interval - Use the formula:

 In our case , so the confidence interval is

Notice the confidence interval does not contain 0, so we can say we are 95% confident that the actual mean difference is between –20.4615 lbs and –2.3385 lbs. So it is unlikely that the mean difference is 0 (diet has no affect) or greater than zero.

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