Other Types of t-tests

Recapitulation Recapitulation 1. Still dealing with random samples. 2. However, they are partitioned into two subsamples. 3. Interest is in whether the means of some variable differ significantly between the two subsamples. 4. For large samples, the sampling distribution of mean differences is normally shaped. 5. For smaller samples, the sampling distribution takes the shape of one of the Student’s t distributions.

6. In either case, the significance tests involve a test of the null hypothesis (H 0 ) that in general (i.e., in the universe) the means do NOT differ. Symbolically, H 0 :  2 -  1 = The alternate hypothesis (H 1 ) can either be nondirectional H 1 :  2 -  1  0.0 or directional, either H 1 :  2 -  1 > 0.0 or H 1 :  2 -  1  0.0

Reject a null hypothesis (H 0 ) when either: 1. the value of the statistical test (  2, z, t, or F) exceeds the critical value at the chosen  -level; or, 2. the p-value for the statistical test is smaller than the chosen value of .

An Example of a T-Test PPD 404 TTEST PROCEDURE Variable: MANUFPCT PERCENT OF LABOR FORCE IN MANUFAC AGECITY2 N Mean Std Dev Std Error Newer Older Variances T DF Prob>|T| Unequal Equal For H0: Variances are equal, F' = 1.40 DF = (37,24) Prob>F' =

An Example of a T-Test PPD 404 TTEST PROCEDURE Variable: MANUFPCT PERCENT OF LABOR FORCE IN MANUFAC AGECITY2 N Mean Std Dev Std Error Newer Older Variances T DF Prob>|T| Unequal Equal For H0: Variances are equal, F' = DF = (37,24) Prob>F' =

The tests that we have performed thus all assumed that the “subuniverse” variances were equal. Remember, parameters (e.g., universe variances) are almost always unknown. Therefore, we let subsample variances be proxies for the unknown universe variances. If the two variances in the universe are greatly different—as indicated by statistically significant differences between the subsample variances—, then a different method must be used to estimate the standard error of the difference.

A hypothesis test exists to decide whether the two variances are significantly different. In this F' test, a ratio is computed for the larger to the smaller subsample variance. If the variances are roughly equal, then the value of F' should be roughly equal to 1.00 (i.e., a / a = 1). If the subsample variances are not equal, then the F' ratio will become greater than 1.0. At some point (i.e., beyond the critical value), the value of the F' statistic becomes so much greater than 1.0 that it is unlikely that the variances are equal. The test is:

The null hypothesis is that the two variances are equal (i.e., there is no difference between them): H 0 :  2 -  1 = 0.0 If this hypothesis must be rejected, this means that the variances are treated as UNEQUAL, and an adjustment must be made in estimating the standard error of the difference. Let's test the variances from the frostbelt and sunbelt cities to see if they may safely be presumed to be equal:

The sample statistics were: Frostbelt CitiesSunbelt Cities _ _ Y 2 = Y 1 = 2.84 s 2 2 = 9.98 s 1 2 = N 2 = 37 N 1 = 26

This is clearly greater than 1.0, but is it sufficiently greater to infer that the null hypothesis of equal variances must be rejected? We need a sampling distribution for F'. This is found in Appendix 3 (pp ). Notice that the tables in this appendix are three- dimensional tables, i.e., their dimensions are  -levels (pages), n 1 (columns), and n 2 (rows). How can we use these tables? First, decide on the chance of being wrong in rejecting the equal-variance null hypothesis. Let's stick with  = This means we are dealing with the sampling distributions for F' on p The columns identify critical values for the numerator degrees of freedom (the larger variance), the rows the denominator degrees of freedom. In this case, the larger variance is for frostbelt cities, and df = 25 (26 - 1) while the smaller is for sunbelt cities, df = 36 (37 - 1).

Thus we look for Column 25 (n 1 ) and Row 36 (n 2 ). There is no Column 25, but df = 24 (Column 24). There also is no Row 36, but we can interpolate. Our critical value is 60 percent of the way between 1.89 and 1.79, 1.83 (by interpolation). Since an F' of 5.77 is GREATER than the F' critical value (1.83), we reject the null hypothesis and conclude that the universe variances are probably UNEQUAL. This means that instead of pooling to estimate the standard error of the difference (as above), we should calculate the t-value as follows:

where s 2 2 and s 1 2 are the variances (standard deviations squared) for the two subsamples, and N 2 and N 1 are the respective subsample sizes. Our t-statistic is calculated as

This is slightly less then t = calculated without adjusting for unequal variances. Is it statistically significant? To decide, we first must also make an adjustment in the value of the number of degrees of freedom in order to be sure that we are using the correct sampling distribution for this application:

With unequal population variances, both the calculation of t and the determination of degrees of freedom (i.e., the appropriate sampling distribution) must be adjusted. This time, the critical value (in Appendix 2, p. 543) at alpha = 0.05 (one-tailed test) is 12 percent of the way from to Interpolating gives [0.12(1.697 – 1.684)], or Supplying the negative sign gives us the critical value for the left-hand tail, Since lies beyond – in the region of rejection, we REJECT the null hypothesis that frostbelt and sunbelt cities lost population at the same rate.

The SAS Procedure for t-tests The SAS Procedure for t-tests libname old 'a:\'; libname library 'a:\'; options nonumber nodate ps=66; proc ttest data=old.cities; class agecity2; var manufpct; title1 'An Example of a T-Test'; title2; title3 'PPD 404'; run;

An Example of a T-Test PPD 404 TTEST PROCEDURE Variable: MANUFPCT PERCENT OF LABOR FORCE IN MANUFAC AGECITY2 N Mean Std Dev Std Error Newer Older Variances T DF Prob>|T| Unequal Equal For H0: Variances are equal, F' = 1.40 DF = (37,24) Prob>F' =

Evaluating these t-test results is done in two steps: First, decide whether the (population) variances are probably equal or unequal (i.e., is the F' test statistically significant?). Then, decide whether the appropriate t-test (for unequal or equal variances) calls for rejection of the null hypothesis of no difference between means.

Reject the null hypothesis (H 0 ) when either: 1. the value of the statistical test (  2, z, t, F', or F) exceeds the critical value at the chosen  -level; or, 2. the p-value for the statistical test is smaller than the chosen value of .

Comparisons thus far have been performed on subsamples whose observations were independent. That is, DIFFERENT observations made up each subsample such as frostbelt and sunbelt cities. Another application of the t-test involves the SAME sample measured at two different points in time. The classic example is BEFORE and AFTER observations on subjects in an experiment. Here the two sets of observations are RELATED rather than independent. These are called paired samples. Once again, a different type of t-test must be used.

Suppose that we have a random sample of five students from this class and that we want to test the hypothesis that the class as a whole (the universe) performed differently on the second examination compared with the first. Since we do not know exactly HOW the two scores differed (i.e., which was better than the other), our alternate hypothesis commits us to a two-tailed test. The null hypothesis is that the class performed the same on both midterms. Our data consist of Midterm One and Midterm Two scores for each student. The independent-sample t-test is inappropriate.

========================================= _ _ _ Student Exam 1 Exam 2 D D (D – D) (D - D)  = 47  =

The appropriate t-test is To estimate the standard error of the paired differences, we must first calculate the standard deviation for this sample. It is, as before when we discussed univariate statistics, the square root of the sum of the deviations squared divided by the number of degrees of freedom (here, D is used instead of Y to denote DIFFERENCES between scores rather than the scores themselves).

In the present example, N = 5, the mean difference is 9.4 (i.e., 47 / 5 = 9.4), and the sum of the deviations squared (the differences between the midterm score differences and the mean difference) is Thus, the sample standard deviation is obtained as follows:

As before, we use the value of the sample standard deviation (i.e., 6.656) to estimate the value of the standard error (the standard deviation of the sampling distribution of mean paired score differences). It is— again—the sample standard deviation divided by the square root of sample size, or

Now we can calculate the value of the t-statistic and make the usual hypothesis test by comparing t with the critical values of t. Let's assume that alpha is We have 4 degrees of freedom. Thus, in Appendix 2 (p. 543) we find the critical values for a two-tailed test to be  The value of t is: Under the null hypothesis,  = 0.0.

Since is GREATER than the critical value, , we REJECT the null hypothesis at the 0.05 level and infer that there is a statistically significant difference in scores between the first and second midterm. Our sample data suggest that in general scores went down on the second exam compared to the first.

The SAS Procedure for Paired Sample t-tests The SAS Procedure for Paired Sample t-tests libname new 'a:\'; libname library 'a:\'; options nonumber nodate ps=66; data new.scores; input student exam1 exam2; diffrnce = exam1-exam2; cards; ;

The SAS Procedure for Paired Sample t-tests (continued) The SAS Procedure for Paired Sample t-tests (continued) proc print data=new.scores; var _all_; title1 'A Paired-Sample T-Test'; title2 'Values in SCORES.SD2'; title3 'PPD 404'; proc means data=new.scores n mean stderr t prt; var diffrnce; title1 'A Paired-Sample T-Test'; title2 'Output'; title3 'PPD 404'; run;

A Paired-Sample T-Test Values in SCORES.SD2 PPD 404 OBS STUDENT EXAM1 EXAM2 DIFFRNCE

A Paired Sample T-Test Output PPD 404 Analysis Variable: DIFFRNCE NMeanStd ErrorT Prob>|T|