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The Matched-Pairs Design for Comparing Two Treatment Means Study Guide (page 3)

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Updated on Oct 5, 2011

Confidence Intervals on the Difference in Two Treatment Means

If the goal is to provide an interval of reasonable values for the mean difference in two treatment means based on a matched-pairs design, we want to set a confidence interval on that mean difference. The conditions for inference are that the differences (Di's) are a random sample from a population of differences. Second, the Di's are either normally distributed or the sample size is large enough (at least 30) to assume that the average of the Di's is approximately normally distributed by the Central Limit Theorem.

The methods for statistical inference using the Di's are computationally identical to those for one population; the interpretation is all that differs. Recall is an unbiased estimate of the difference in the treatment means, μ1–μ2D. The standard error of this estimate is where . Using the form point estimate ± multiplier × standard error, a 100(1 –α)% confidence interval on μD = μ1 – μ2 has the form where t* with (n – 1) degrees of freedom is the proper tabulated value to give 100(1 – α)% confidence.

Example

Look again at the study comparing running shoes earlier in this lesson. Find a 90% confidence interval on the mean difference in the race times for a 100-meter dash using the athlete's favorite running shoes and the new running shoes.

Solution

The 24 runners were recruited not randomly selected from all runners, so our inference will be restricted to these 24 runners (review the table in Lesson 2). We may believe that these runners are representative of all runners and thus attempt to broaden our scope of inference, but we need to be very careful in doing so. If we think these runners may differ from a broader population of runners, the 24 runners must be taken as the population of interest. The random assignment of treatment order is necessary for the first condition of inference to be satisfied. Note: Unless treatments are randomly assigned within a pair, we do not have a random sample of differences, which is the first condition. Although we have not formally tested whether or not the population of differences has a normal distribution, the graphs constructed in the previous example suggest that it is not an unreasonable assumption, so we will assume that the differences in running times are at least approximately normally distributed.

Based on n=24 runners, we found = 0.2050 and sD = 0.3068. For a 90% confidence interval, α = 0.10, so we want to put 5% of the probability in each tail. Looking in the t-table in Lesson 12, the t-value at the intersection of the row corresponding to 23 degrees of freedom and the column showing 0.05 in the upper tail, we have t* = 1.714. The confidence interval on μF – μN is 0.2050 ± 1.714 or 0.2050 ± 0.1096. Therefore, we estimate that, on average, the new running shoes allow runners to complete the race in 0.21 fewer seconds compared to their favorite running shoes, and we are 90% confident that this estimate is within 0.11 seconds of the true mean difference in the times to run the 100-meter dash using the new running shoes and the runners' favorite shoes.

Hypothesis Tests Concerning the Difference in Two Treatment Means

Tests of hypotheses concerning the difference in two treatment means are based on the same philosophy as the hypothesis tests discussed earlier in this book. Five steps are followed to conduct a hypothesis test.

Step 1: Specifying the Hypotheses

For a matched-pairs design, the null hypothesis, Ho, is that the difference in the treatment means is do; that is, Ho: μ1– μ2 =do. Although it is common for do=0 (implying that the means or proportions are equal), this is not necessary; do can be any value. The alternative is that the difference is less than, greater than, or equal to do.

Step 2: Verify Necessary Conditions for a Test and, if Satisfied, Construct the Test Statistic

The conditions for testing hypotheses about the difference in two treatment means are the same as conditions for testing confidence intervals. First, the differences (Di's) must be a random sample from some population of differences. Second, we must satisfy the condition of normality. That is, the Di's are normally distributed or the sample size is sufficient that the sample mean difference is approximately normal by the Central Limit Theorem.

The test statistic has the now familiar form 

For the paired design, this becomes

.

We know the distribution of this test statistic is at least approximately a t-distribution with (n – 1) degrees of freedom if the null hypothesis is true.

Step 3: Find the p-Value Associated with the Test Statistic

If the null hypothesis about the difference in two treatment means is true, the test statistic has either an exact or an approximate t-distribution. The distribution of the test statistic when the null hypothesis is true is called the null distribution. If the null hypothesis is not true, the test statistic is not distributed according to the null distribution and is more likely to assume a value that is "unusual" for a random observation from that distribution. The p-value is the probability of determining the probability of observing a value as extreme as or more extreme than the test statistic from a random selection of the standard normal distribution.

How do we measure how unusual a test statistic is? It depends on the alternative hypothesis. These are summarized in Table 18.4.

Table 18.4

Step 4: Decide Whether or Not to Reject the Null Hypothesis

Before beginning the study, the significance level of the test is set. The significance level is the largest acceptable probability of a type I error. If the p-value is less than the significance level, the null hypothesis is rejected; otherwise, the null is not rejected.

Step 5: State Conclusions in the Context of the Study

Statistical tests of hypotheses are conducted to determine whether or not sufficient evidence exists to reject the null hypothesis in favor of the alternative hypothesis.

Example:

For the running shoe study presented in the previous lesson, the company wants to be able to claim that the new shoe reduces the mean of the 100-meter race times for runners. Is there statistical evidence to support this claim?

Solution:

Step 1: Specifying the Hypotheses

Let the subscripts F and N represent the runners' favorite shoes and the newly developed shoes, respectively. The company wants to know whether the mean of the race times is lower for the new shoes. (Learning that the mean is greater would certainly not be a strong promotional point.) Thus, the hypotheses of interest are Ho: μF – μN = 0 versus Ho: μF – μN > 0. Notice we could have written these as Ho: μF = μN and Ho: μF > μN .The two sets of hypotheses are equivalent.

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