Paired Data Values and Matched Pairs t Procedures
There are situations where data values are collected in pairs. In these situations, instead of the interest in the individual observations, the interest is now the differences in values of some variable for each pair.
Paired data values is when the data values have been observed in natural pairs. The following are three of the many ways for paired data values to occur:
1. Two different variables are measured for each individual and the difference between these two data values is examined. 2. Each individual is measured twice. The two measurements of the same characteristic are made under different conditions, such as different times. 3. Similar individuals are placed in pairs, where each member of the pair receives a different treatment. The same response variable is measured and compared for the two individuals in each pair.
Matched Pairs t Procedure The matched pairs t procedure will be used to investigate and estimate any differences between the responses to the two treatments. Instead of making one comparison for the variables of interest, one comparison for each of the n pairs will be made.
The parameter of interest is called the true mean of the difference of all pairs in the population, denoted as μd. The population mean difference μd is estimated by the sample mean difference x̄d, and the population standard deviation σd is estimated by the sample standard deviation of differences sd.
In using t procedures, the assumption is that the differences follow a normal distribution with a mean μd and a standard deviation σd. It is also assumed that the selected pairs are from a simple random sample of all possible pairs from the population.
Confidence Intervals and Matched Pairs Setting The method for constructing confidence intervals on the matched pairs setting is the same method for constructing confidence intervals in a one-sample case. However, in a matched pairs setting, differences are being examined rather than individual observations.
The following formula is the confidence interval:
(x̄d - t*(sd/√n), x̄d + t*(sd/√n)
Example Many drivers buy premium gasoline instead of regular in the belief that they will get better gas mileage. To test this belief, we obtain a sample of 8 cars. Each car is run for one tank on regular gas and one take on premium gas, with the order randomly determined. The mileage (in miles per gallon) is measured for each car for each type of gasoline. The data collected is the following:
A 95% confidence interval for the true mean difference in mileage for premium and regular gasoline. Assume that the differences d = P - R follow a normal distribution. The mean and standard deviation are calculated: Σxd = 4 + 0 + 2 + 1 + 2 - 1 + 5 + 3 = 16 x̄d = Σxd/n = 16/8 = 2.0 Σ(xd - x̄d)² = (4 - 2)² + (0 - 2)² + (2 - 2)² + (1 - 2)² + (2 - 2)² + (-1 - 2)² + (5 - 2)² + (3 - 2)² = 28 Σ(xd - x̄d)²/(n - 1) = 28/7 = 4.0 sd = √4 = 2.0 The confidence interval is constructed: Since C = 95, α/2 = 0.05/2 = 0.025. Therefore, t* = 2.365. (x̄d - t*(sd/√n), x̄d + t*(sd/√n) (2 - 2.365(2/√8), 2 + 2.365(2/√8) (0.33,3.67) If repeated samples of eight cars were taken and a confidence interval was calculated for each sample in a similar manner, then 95% of such intervals would contain the true mean difference in fuel economy for premium and regular gasoline.
When it comes to which sample's values are subtracted from which sample's values, it does not matter which order they are subtracted. The signs of the differences simply switch, but the confidence interval will provide the same information.
Hypothesis Testing and Matched Pairs Setting Using the P-Value Method The difference between the hypothesis testing using one sample and hypothesis testing using matched pairs is the hypotheses themselves.
The null hypothesis and alternative hypothesis always use the true mean difference of zero. Additionally, the alternative hypothesis depends on the order of the difference and the question.
For example, if the question wants to compare two products or objects and see which is better or greater in a characteristic, the hypothesis test will be right-sided, because the differences should be greater than zero to show that one product/object is better/greater than the other product/object, where d = assumed better product - regular product. The hypothesis test is left-sided if d = regular product - assumed better product.
If the question wants to know whether a product or object reduces a characteristic, the hypothesis test will be left-sided, because the differences should be less than zero to show that the product/object reduces the characteristic, where d = regular treatment - assumed reducing treatment. The hypothesis test is right-sided if d = assumed reducing treatment - regular treatment.
If the question wants to compare whether two products or objects differ in some way, the hypothesis test will be two-sided, because the differences should not equal zero to show that the product/object produced a difference in characteristic.
Test Statistic Since the null hypothesis uses a true mean of zero, the test statistic t is changed:
t = (x̄d - μd0)/(sd/√n) = (x̄d)/(sd/√n)
Interpretation of the P-Value For true mean difference, the interpretation of the mean is changed to the following:
"If there was no difference in mean (units) between the (two samples), the probability of observing a sample mean difference at least as (high/low/extreme) as (sample mean difference + units) would be between (p-value)."
Example Using the previous example, a hypothesis test is conducted using the p-value method. Let the level of significance α = 0.05. H₀: μd = 0 vs. Hₐ: μ > 0 H₀: The average fuel economy is the same for regular and premium gasoline. Hₐ: The average fuel economy is higher for premium gasoline than for regular gasoline. Reject the null hypothesis if the p-value ≤ α = 0.05. Calculating the test statistic: t = (x̄d - μd0)/(sd/√n) = (x̄d)/(sd/√n) = (2.0)/(2.0/√8) ≈ 2.83 p-value = P(T(7) ≥ t) = P(T(7) ≥ 2.83) ≈ [0.01,0.02] If there was no difference in average fuel economy between the premium and regular gasoline, the probability of observing a sample mean difference at least as high as 2.0 mpg would be between 0.01 and 0.02. Since p-value = 0.05 < 0.01 < 0.02, the null hypothesis is rejected. With a 5% level of significance, there is sufficient evidence to conclude that the average fuel economy is higher/better for premium gasoline than for regular gasoline. If the differences were d = R - P, the value of the test statistic would be -2.83, with a p-value of P(T(7) ≤ -2.83). This yields the same result.
Hypothesis Testing and Matched Pairs Setting Using the Critical Value Method Matched pairs can also use the critical value method in hypothesis testing.
Example A pharmaceutical company is testing a new blood pressure medication. The systolic blood pressures of 20 patients suffering from hypertension are measured before and after taking the medication and the differences d = After - Before are calculated. The sample mean difference is calculated to be x̄d = -10.9, and the sample standard deviation of differences to be sd = 3.7. Researchers would like to conduct a hypothesis test to determine whether the medication is successful in reducing systolic blood pressure. It is assumed that the differences follow a normal distribution. The hypothesis test uses the critical value method. Let the level of significance α = 0.10. H₀: μd = 0 vs. Hₐ: μd < 0 H₀: The medication has no effect on the mean blood pressure of patients with hypertension. Hₐ: The medication reduces the mean blood pressure of patients with hypertension. Reject the null hypothesis if t ≤ -t*. Since α = 0.10, -t* = -1.328. t = (x̄d - μd0)/(sd/√n) = (x̄d)/(sd/√n) = (-10.9)/(3.7/√20) ≈ = -13.17 Since t ≤ -t*, the null hypothesis is rejected. At a 10% level of significance, we have sufficient evidence that the medication reduces the mean blood pressure of patients with hypertension.
Example If the p-value method was used in the above example, the p-value would calculate to p-value = P(T(19) ≤ t) = P(T(19) ≤ -13.17) = 0.0005 ≤ α = 0.10.
The following examples use the confidence interval method, the p-value method, and the critical value method:
Example Many psychology studies have examined the difference between the first and second children in a family. Suppose a researcher would like to compare their academic performance. A high school's GPA of a sample of 30 pairs of siblings is examined and the difference in GPAs d = first child - second child is calculated. The average difference x̄d is calculated to be 0.12 and the standard deviation is calculated to be sd = 0.31. It is assumed the average GPA differences follow a normal distribution. To create a 99% confidence interval of the true mean difference in GPAs: (x̄d - t*(sd/√n), x̄d + t*(sd/√n)) Since C = 99, α = 0.01. Then t* = 2.756. (0.12 - (2.756)(0.31/√30), 0.12 + (2.756)(0.31/√30)) (-0.04,0.28) Since μd0 = 0 is inside the confidence interval, the null hypothesis is failed to be rejected.
Example The p-value method is now used to conduct the previous example. Let the level of significance α = 0.01. H₀: μd = 0 vs. Hₐ: μd ≠ 0 H₀: There is no difference in average GPAs between the first and second child. Hₐ: There is a difference in average GPAs between the first and second child. Reject the null hypothesis if p-value ≤ 0.01. t = (x̄d - μd0)/(sd/√n) = (x̄d)/(sd/√n) = (0.12)/(0.31/√30) ≈ 2.12 p-value = 2(T(29) ≥ t) = 2(T(29) ≥ 2.12) ≈ 2[0.02,0.025] = [0.04,0.05] Since p-value > t, the null hypothesis is failed to be rejected. At a 1% level of significance, there is insuffcient evidence to conclude that there is a difference in the average GPAs between the first and second child.
Example The above example will now use the critical value method to conduct a hypothesis test. Let the level of significance α = 0.01. H₀: μd = 0 vs. Hₐ: μd ≠ 0 Reject the null hypothesis if t ≤ -t* or t ≥ t*. Since α = 0.01, t* = 2.756 and -t* = -2.756. t = (x̄d - μd0)/(sd/√n) = (x̄d)/(sd/√n) = (0.12)/(0.31/√30) ≈ 2.12 Since t < t* and t > -t*, the null hypothesis is failed to be rejected.
