Paired Differences and Regression

Transcription

1 Paired Differences and Regression Students sometimes have difficulty distinguishing between paired data and independent samples when comparing two means. One can return to this topic after covering simple linear regression. Usually students do not have difficulty identifying paired data in a regression context, so it may help to point out that every paired difference problem is a potential regression problem. To begin with, for any paired data you can make a scatter plot, so for any problem or example you want to use for reviewing paired differences, you could make a scatter plot. Here we will use data that give the times for the men and women winners of the New York City marathon in the years 1978 to Year Men Women This appeared in an introductory statistics textbook as a paired differences problem, but we could just as well make a scatter plot of the data. 1

2 The null hypothesis we usually test is equivalent to on average, or δ = 0 y x = 0 y = x algebraically. Translating this into a regression model with an equivalent hypothesis test, we fit ŷ = a + x + ε where a is both an intercept and the difference between x and y, while ɛ is an error term, the slope is forced to be 1, and we test whether a = 0. If we reject that, then we think there really is a difference, and we estimate that with the mean of the sample differences, which here is

3 > summary(diff) Min. 1st Qu. Median Mean 3rd Qu. Max A graph may clarify the situation. We ll use a scale that has y = x at 45º. Here the red line is the null model y = x while the black line is the model suggested by the data. y = x The null hypothesis says that the data are close enough to y = x that we can consider any deviations from that line to be random noise. The graph certainly contradicts that notion, as does the usual ttest. > t.test(diff) 3

4 One Sample ttest data: diff t = , df = 20, pvalue < 2.2e16 alternative hypothesis: true mean is not equal to 0 95 percent confidence interval: sample estimates: mean of x Having rejected the null, we can reasonably say there is a difference, and the output above gives a confidence interval for that difference. A natural question at this point is whether we could fit a regular regression model to data we might also analyze as paired differences. The answer is, Yes, always. Let s do it. lm(formula = Women ~ Men) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) * Men ** Residual standard error: on 19 degrees of freedom Multiple RSquared: , Adjusted Rsquared: Fstatistic: on 1 and 19 DF, pvalue: Here the software has fit the model ŷ = x We know from theory that this line minimizes the sum of the squared residuals, so in that sense it has to be a better fit than the model y = x However, the latter model has the virtue of simplicity, and so is often chosen when it is close enough for statisticians. 4

5 Here we can see that the regression line (in green) does not differ very much from the simple difference model within the range of the data. In fact, the rightmost data point is a bit of an outlier and if we were to delete it the regression line might then have a slope greater than one. Try it. For review problems you can do any paired difference data in your textbook as regression data. In each case, ask yourself if the simple constantdifference model is close enough for you. Hidden Assumptions Comparison with regression reveals some additional assumptions that are made by our simple difference model. These are not on the syllabus, but they do review regression assumptions and a common type of assumption that shows up in other places in an introductory course. That type is the assumption that sneaks in when a Greek letter, such as σ, is mentioned. Often the implicit assumption is that the parameter mentioned is a constant rather than a function. For example, in regression we do not want the spread about the line to vary when x varies. That s equally true for paired differences. Also, whenever we estimate a mean, we assume there is but one mean to estimate. For our data, the 5

6 regression line suggests that the difference is not constant but instead decreases with increasing x. If that s really true, a regression model might be better. We check for a different form of this illness in inference for a single mean when we plot the data and look for signs of bimodality. The usual cause of bimodality is that we have mixed together apples and oranges two fruits with different means. In that case, trying to estimate the mean is usually a mistake. Another example Here is the first example from Chapter 25 of BVD. The data are the miles driven by a number of employees commuting on a fourdayweek and a fiveday week. The hope was that the shorter work week would reduce commuting mileage. MTB > let c4=c3c2 MTB > name c4 diff. MTB > print c1c4 Row Name 5Day_mileage 4Day_mileage diff 1 Jeff Betty Roger Tom Aimee Greg Larry G Tad Larry M Leslie Lee MTB > ttest c4 Test of mu = 0 vs not = 0 Variable N Mean StDev SE Mean 95% CI T P diff ( , ) It appears that the fourdayweek DID reduce mileage (p = 0.017) by 982 miles on average. But is this a good model? MTB > regr c3 1 c2 The regression equation is 4Day_mileage = Day_mileage 6

7 Predictor Coef SE Coef T P Constant Day_mileage S = RSq = 95.4% RSq(adj) = 94.9% Analysis of Variance Source DF SS MS F P Regression Residual Error Total Unusual Observations Obs 5Day_mileage 4Day_mileage Fit SE Fit Residual St Resid R R denotes an observation with a large standardized residual. Note that the regression model gets the opposite sign for the constant term. If you look at the data, some people drove longer on the fourdayweek, and those people were typically people whose total mileage was small on either plan. Could the slope be 1? The ttest is t = ( )/ = indicating the slope is NOT 1. For R 2 for the constant difference model, compare the sum of squared residuals for the difference to that for Y. The latter is given on the regression printout as while for the differences: MTB > let c5=c4average(c4) MTB > sum c5 Sum of C5 = 0 MTB > let c6=c5*c5 MTB > sum c6 Sum of C6 = and so R 2 = ( )/ = = 72.5% compared to 95.4% for the regression model. Finally, a plot sheds considerable doubt about the model that says the difference is constant: 7

8 MTB > plot c4 c2 * * 0+ * * diff. * * * * * * * Day_mi We can also make a plot as we did for the previous dataset showing the y = x line, the offset constant difference model, and the regression line. 8

9 So, which model should we use? Sufficient unto the day is the model thereof. We should not be worried that the value of R 2 associated with the constant difference model is lower. We already know that it would have to be lower than the least squares line unless it coincided with that line. More troubling is the result of the hypothesis test of β 1 = 1which showed that a slope of one is unlikely. However, we might still use this model if it answered the question of interest with sufficient accuracy. It does seem to answer the question of whether the four day week reduces driving mileage. However, the suggestion that it reduces it more or less the same for everyone may be an oversimplification. An environmentalist seeking to reduce automobile emissions might be very happy to have this group driving 982 fewer miles per person on average. (This study appeared in the Journal of Environmental Health.) On the other hand, if you look at the individual employees, about two thirds of the employees would benefit from the change and one third would not. The regression model essentially says that the saving depends on how much you drive. Presumably the office needs to have everyone on the same schedule. When a third of the workers realize they may not benefit from the change, it may create some problems. As is so 9

10 often the case in statistics, what is the right approach depends on the context and purpose. We generally prefer definitive answers to it depends answers, but sometimes it depends is the best answer. 10