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The Matched-Pairs Design for Comparing Two Treatment Means Study Guide (page 2)

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Updated on Oct 5, 2011

Example

A researcher wants to compare the quality of cooking roasts using two methods—open pan and bag. Four ovens are available for the study. Eight roasts of equal quality have been allocated for the study.

  1. Describe how to conduct the study using a matched-pairs design.
  2. Describe how to conduct the study using a twogroup design.
  3. Which of the two designs would you use for this study? Explain.

Solution

  1. For a matched-pairs design, two roasts would be cooked in each oven, one in an open pan and the other in a bag. The location of each roast within the oven would be randomly determined.
  2. For a two-group design, we randomly select two of the ovens to cook a roast using the open-pan method; the other two ovens would each be used to cook a roast in a bag.
  3. The matched-pairs design would be the best for this study. Ovens often vary in their ability to hold temperature at a specified level. By having both treatments in each oven, differences between ovens can be accounted for in the analysis. As described, we have used half as many roasts in the two-group design. We could put two roasts in each oven and cook using the same method. This gives us information on the differences within an oven and allows us to more precisely estimate the quality of the roasts cooked in a specific oven. Cooking two roasts in the same oven does not double the number of experimental units in the study. An oven would be the experimental unit because the cooking methods were randomly assigned to the ovens.

Matched-Pairs Design

Once we have decided to conduct an experiment using matched pairs, how do we actually go about conducting the study? First, the study units need to be obtained. As we learned in Lesson 2, if the study units are randomly selected from some population, conclusions can be made for that population at the end of the study; otherwise, conclusions apply only to the units in the study. In the shoe-shine study, children were randomly selected. The group from which these children were randomly selected is the population for which inferences can be made.

Next, the study units need to be paired. Individuals could be matched according to a characteristic that could explain some of the difference in the response variable. In the cholesterol study, individuals were matched by initial cholesterol level. Sometimes, both treatments can be sequentially applied to the same individual. This form of matched pairs is often very strong, but may require more time than is available for the study.

Once the pairs are formed, one treatment is randomly assigned to one unit in the pair; the other unit receives the second treatment. Notice that a separate randomization is used for each pair. For the shoe-shine study, it would not be sufficient to flip a coin and randomly assign the first treatment to all right shoes and the other treatment to all left shoes. Children are right or left-footed just as they are right- or left-handed. It is possible that one shoe, say the right shoe, tends to get the most wear because most children are right footed. If this is the case, then the treatment assigned to all right shoes would be at a disadvantage in the study. To avoid this and other biases of which we may not even be aware, we randomly assign treatments within each pair.

Once the study is complete, we record the response variable for each unit. Let X1i be the observed response from the first treatment in pair i, i = 1, 2, . . ., n, where there are n pairs. Similarly, let X2i be the observed response from the second treatment in pair i,i = 1, 2, . . . , n. Then Di = X1iX2i, i = 1, 2, . . . , n, is the observed difference in the two treatments for the ith pair. There is a conceptual population of Di's comprised of the differences in all possible pairs that could have been used in this study. This population has μD and standard deviation σD.

The sample mean difference in the two treatments, , is an estimate of the difference in the treatment means, μ1– μ2= μD, the mean of the population of paired treatment differences. The sample variance of the pairwise differences provides an estimate of σD2 and is . The sample standard deviation is sD = √. Notice that and sD are, respectively, the sample mean and sample standard deviation of the differences. This would lead us to speculate that the standard error of is . This is, in fact, the case! The analysis of a paired study is based on these quantities. We will consider this further in the next two lessons.

Example

An athletic shoe company believes that they have developed a shoe that will help short-distance runners lower their times in races. They recruited 24 runners. Each runner was given a new pair of the athletic shoes. The runners were encouraged to use these shoes and their favorite pair of running shoes equally in practice for two weeks. After two weeks, the runners ran two 100-meter dashes with five hours between races. For each runner, a coin was flipped. If the coin landed heads up, the runner wore his or her favorite running shoes in the first race; otherwise, he or she wore his or her newly developed shoes. In the second race, each runner wore the pair of shoes that was not used in the first race. The times for the runners are given in Table 18.1.

  1. Explain why this study has a matched-pairs design. Include a clear statement describing what constitutes a pair.
  2. Find the difference in observations from each pair.
  3. Estimate the mean and standard deviation of the differences in time to run a 100-meter dash when wearing the favorite running shoes compared to the new running shoes.
  4. Find the standard error of the estimated mean of the differences in time to run a 100-meter dash when wearing the favorite running shoes compared to the new running shoes.
  5. Is the assumption reasonable that the differences are normally distributed?
  6. To which population may inference be drawn from this study?

Table 18.1 Race times with different shoes

Table 18.2 Race time differences

Solution

  1. The two treatments are the favorite running shoes and the newly developed running shoes. Each treatment is applied to a runner. Thus, the favorite running shoes and the newly developed running shoes are paired by runner. A pair consists of the running times for the two treatments from a single runner. The order in which the shoes were used was randomized for each runner, a critical step in conducting the study.
  2. The differences in the two treatments are computed for each runner (see Table 18.2).
  3. The estimated mean difference in the running times using the favorite shoes versus using the newly developed shoes is

    (–0.06 + 0.64 +………+ 0.34)

    = 0.2050

    The estimated variance of these differences is , and the estimated standard deviation is 0.3068.

  4. The standard error of the estimated differences in the two treatments is .
  5. Because the sample size is small, it is difficult to determine whether or not the observed differences are normally distributed. Although formal tests exist for determining normality, we will not study them here. Instead, we will rely on examining graphs to determine whether there are indications that the data may not be normal. Figures 18.1, 18.2, and 18.3 show a histogram, a dotplot, and a boxplot, respectively. The histogram looks fairly symmetric and unimodal. With only 24 observations, the shape of a population is often not fully captured in a histogram of the data. The dotplot appears to be centered at about 0.20. The values range from –0.39 to 0.84 with a higher concentration of dots in the center. From the boxplot, the data appear to be fairly symmetric without any outliers. In summary, we do not see any indication of skewness, outliers, or other features that would cause us to think that the assumption of normality is unreasonable.

Figure 18.1

Figure 18.2

Figure 18.2

  1. Because the runners were recruited and not randomly selected from some population, the population to which inference may be drawn is the runners in this study.
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