Lab Four: Initial statistical tests: t-tests, non-parametric tests, simple ANOVA

Transcription

1 Lab Four: Initial statistical tests: t-tests, non-parametric tests, simple ANOVA Lab Objectives After today s lab you should be able to: 1. Use PROC FORMAT to divide a variable into categories for use in statistical tests requiring categorical variables, such as the t-test. 2. Use PROC TTEST in SAS to perform a two-sample t-test, a one-sample t-test, and a paired t-test. 3. Interpret results from PROC TTEST, including equality of variance F-test, pooled variance p-value, and unpooled variance p-value. 4. Understand output from PROC TTEST well enough to fill in TTEST table via hand calculations. 5. Use PROC NPAR1WAY to perform non-parametric equivalent to the two-sample ttest (Wilcoxon sum-rank test). 6. Interpret results from PROC NPAR1WAY. 7. Use PROC ANOVA to run a simple ANOVA in SAS and interpret the output. 8. If time, use PROC GLM to generate least squares means and differences with confidence limits, and to make pair-wise comparisons adjusted for multiple comparisons. Recommended reading in Walker: Chapters 4-6,

2 LAB EXERCISE STEPS: Follow along with the computer in front 1. Double-click on to open the SAS editor file data creation code which should be saved in your stats210 folder from last week; run the libname statement: libname stats210 'C:\Documents and Settings\mitl-pc.LANE-LIB\My Documents\Stats210 ; 2. Using the Explorer Browser on the left hand side of your screen, double check that a stats210 library has been properly created, and that it contains the SAS dataset runners that we created Monday. 3. In a new editor, type the following code (should be able to cut and paste proc format step from data creation code file): proc format; value menses 0-9="oligo/amenorrheic" 10-20="eumenorrheic"; Formatting can be used to define categorical variables from numeric variables for the purposes of statistical tests. proc ttest data=stats210.runners; class menses1; var sumedi1; Note error that appears in the log when you run this code: ERROR: The CLASS variable has more than two levels. proc ttest data=stats210.runners; class menses1; format menses1 menses5. ; var sumedi1 menarch wards1; By invoking formatting here, SAS treats menses1 as a binary variable, as specified by the format. Alternatively, you could permanently create a new binary variable using an if-then statement in a datastep, and use this new binary variable in proc ttest: data stats210.runners; set stats210.runners; if 0<menses1<=9 then isirreg=1; else if menses1>9 then isirreg=0; proc ttest data=stats210.runners; class isirreg; var sumedi menarch wards1; 2

3 4. Examine the output: Upper and lower 95% confidence limits are given for each group mean and for the difference in means between the 2 groups. Notice that the confidence interval for the difference in means is: , which does not cross 0 (the null value). N=17 women in the oligo/amenorrheic group; 30 women in the eumenorrheic group. Mean edi scores are 19.5 for the oligo/amenorrheic group, and 7.3 for the eumenorrheic group., for a difference of The TTEST Procedure Statistics This is the pooled standard deviation,calculated as: (16) (29) s p = = s p = = This is the standard error of the difference between the group means, calculated using the pooled variance: s. e.( diff ) = = Lower CL Upper CL Lower CL Upper CL Variable menses1 N Mean Mean Mean Std Dev Std Dev Std Dev Std Err sumedi1 oligo sumedi1 eumen sumedi1 Diff (1-2) menarch oligo menarch eumen menarch Diff (1-2) wards1 oligo wards1 eumen wards1 Diff (1-2) Note degrees of freedom =N-2 with pooled variance; pooled variance always has higher degrees of freedom than unpooled, making it preferable to use when the variances between the two groups are close enough to each other (see t diff = = s. e.( diff ) = T-Tests 3.40 The sample standard deviation of wards triangle bone density is half as much in the oligo/amenorrheic runners as than in the eumenorrheic runners. It may not be valid to use the pooled variance t-test. A t 45 value of 3.40 corresponds to a p-value of Shows significantly higher disordered eating in the oligo/amenorrheic group. Variable Method Variances DF t Value Pr > t sumedi1 Pooled Equal sumedi1 Satterthwaite Unequal menarch Pooled Equal menarch Satterthwaite Unequal wards1 Pooled Equal wards1 Satterthwaite Unequal Equality of Variances Variable Method Num DF Den DF F Value Pr > F sumedi1 Folded F menarch Folded F wards1 Folded F An F-test tests the null hypothesis that two variances are equal (here: the variance of wards bone density in oligo/amenorrheic runners = the variance of wards bone density in eumenorrheic runners). A significant p-value for the F-test indicates variances are not equal and unpooled variance t-test above must be used. 3

4 5. Other types of ttests. Test the null hypothesis that the wards triangle bone density in runners is different than.90 (the average in the general population) (=one sample ttest). Then test the null hypothesis that the change in bone mineral content during the 2-year trial was different than 0 (paired ttest). Note the use of titles for the results. proc ttest H0=.90 data=stats210.runners; var wards1; title 'one sample ttest'; proc ttest data=stats210.runners; paired bmc3*bmc1; title 'paired ttest'; 6. Examine and discuss output. one sample ttest Sample mean with 95% confidence limits Statistics Sample standard deviation with 95% confidence limits = Lower CL Upper CL Lower CL Upper CL Variable N Mean Mean Mean Std Dev Std Dev Std Dev Std Err wards T 46 = = T-Tests Variable DF t Value Pr > t wards No evidence that wards triangle bone density is different in runners than the average in the general population of young women. paired ttest Statistics Sample standard deviation of the difference = Lower CL Upper CL Lower CL Upper CL Difference N Mean Mean Mean Std Dev Std Dev Std Dev Std Err bmc3 - bmc T 46 = = T-Tests Difference DF t Value Pr > t bmc3 - bmc Borderline evidence of real change in wards triangle bone density over 2- year study period. 4

5 DUE FROM LAB THREE AND FOUR: using a calculator, or using SAS as a calculator, fill in the following and hand in just this page of the lab to me by end of class Thursday July 15 th : Fill in the Following Chart: The TTEST Procedure Statistics Lower CL Upper CL Lower CL Upper CL Variable isirreg N Mean Mean Mean Std Dev Std Dev Std Dev Std Err SPINE SPINE SPINE Diff (1-2) T-Tests Variable Method Variances DF t Value Pr > t SPINE Pooled Equal SPINE Satterthwaite Unequal

6 7. Use interactive data analysis to examine the distribution of the variable sumedi1. (solutions analysis interactive data analysis distribution(sumedi1)). Notice does not appear very Gaussian. 8. Run Kolmogorov-Smirnov and other tests of normality using the normal option in proc univariate: proc univariate normal data=stats210.runners; var sumedi1; Tests for Normality Test --Statistic p Value Shapiro-Wilk W Pr < W < Kolmogorov-Smirnov D Pr > D < Cramer-von Mises W-Sq Pr > W-Sq < Anderson-Darling A-Sq Pr > A-Sq < P-value for the null hypothesis that the variable is normally distributed. P<.05 is evidence against the assumption of normality. Suggests that you should try non-parametric test. 9. Try Wilcoxon rank-sum test for the null hypothesis that sumedi1 is the same in both groups. data stats210.runners; set stats210.runners; if mencat=1 or mencat=2 then isirreg=1; else isirreg=0; Since we are using this particular split of groups a lot, we might as well create a permanent new variable, rather than having to keep invoking the binary format. Note use of the if-then and if-then-else statements. proc npar1way data=stats210.runners wilcoxon; class isirreg; var sumedi1; title ' ; See next page for explanation of this code Note: if you had a very small sample size, you would want an exact test rather than a normal approximation: add the statement: exact wilcoxon; before the run statement. But if you try it here, you will tie up SAS because it involves lengthy computations. 6

7 Explanation of code: Requests non-parametric statistics (= npar ) that are analogues of 1-way ANOVA Same as proc ttest proc npar1way data=stats210.runners wilcoxon; class isirreg; var sumedi1; title ' ; By using the wilcoxon option, you request only the wilcoxon statistics out of a slew of statistics that SAS will give you otherwise. If you don t clear the title or give this analysis a new title, SAS will use the last title assigned, which was paired ttest. OUTPUT: The NPAR1WAY Procedure Wilcoxon Scores (Rank Sums) for Variable sumedi1 Classified by Variable isirreg is Sum of Expected Std Dev Mean Irreg N Scores Under H0 Under H0 Score ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ Average scores were used for ties. Wilcoxon sum-rank test assumes that the two groups have the same dispersion for sumedi1 (analogous to ttest homogeneity of variances assumption) Sum of ranks from 1-47 of the group with 30 and the group with 16. Normal approximation uses the estimate of expected sum of scores and standard deviation of sum of scores from above: Z = = Another approximation is to rank the data from 1-47 and then run a simple twosample ttest to compare the mean ranks of the groups. Wilcoxon Two-Sample Test Statistic Normal Approximation Z One-Sided Pr > Z Two-Sided Pr > Z t Approximation One-Sided Pr > Z Two-Sided Pr > Z Expected sum of ranks for a group of 30 and a group of 16 if there is no difference between the two groups in sumedi1 (null hypothesis). Note: for more explanation about the calculation of expected sum of scores and standard deviation of sum of scores, read chapter 13 of your book. Z includes a continuity correction of 0.5. null hypothesis. The Compare with results of previous t-test for sumedi1; not very different!: sumedi1 Pooled Equal

8 10. Use PROC ANOVA to compare mean neck (neck of hip) bone density at baseline between 3 groups of menstrual regularity (amenorrheic, oligomenorrheic, eumenorrheic, contained in the categorical variable mencat that should already exist in your dataset). proc anova data=stats210.runners; class mencat; model neck1=mencat; ANOVA TABLE: Dependent Variable: NECK1 Divide each SS by its df to get mean SS Sum of Source DF Squares Mean Square F Value Pr > F Could look up F=1.22 in an F- Table for 2, 44 degrees of freedom Model Error Corrected Total Between group sum of squares (variation due to group) F-statistic= / =1.22 =between group variability divided by within group variability Within group sum of squares (variation due to random error) Total sum of squares=between group SS + within group SS Null hypothesis: mean1=mean2=mean3 Alternative hypothesis: at least two means differ Indicates not enough evidence to show that any groups differ. 8

9 11. Try ANOVA for the outcome variable sumedi1 (though not perfectly normally distributed outcome ). And examine output at least two of the groups differ. proc anova data=stats210.runners; class mencat; model sumedi1=mencat; 12. To figure out which groups are different after adjusting the p-value post-hoc for having done 3 pairwise comparisons (using a scheffe adjustment): Proc glm= General linear model more powerful than ANOVA...does ANOVA plus we are actually making a linear regression model : model sumedi1=mencat with sumedi1 as the outcome and mencat as the categorical predictor. proc glm data= stats210.runners; class mencat; model sumedi1=mencat; lsmeans mencat/pdiff adjust=scheffe cl; automatically adjust my p-values for all pairwise comparisons using a scheffe adjustment Generates the same ANOVA table as above, plus the following: The GLM Procedure Least Squares Means Adjustment for Multiple Comparisons: Scheffe lsmeans LSMEAN mencat SPINE LSMEAN Number mean sumedi1 score for each group pdiff Least Squares Means for effect mencat Pr > t for H0: LSMean(i)=LSMean(j) Dependent Variable: sumedi1 After adjusting for multiple comparisons, only groups 2 and 3 (oligomenorrheic and eumenorrheic) are significantly different at p<.05 level. i/j

10 cl sumedi1 mencat LSMEAN 95% Confidence Limits Difference Simultaneous 95% Between Confidence Limits for i j Means LSMean(i)-LSMean(j) % confidence intervals for the mean sumedi1 scores. Note that the confidence interval for group 1 is wide, because it is a very small group; use interactive data analysis distribution tables frequency counts to find that n=4. 95% Confidence intervals for the difference in means between group i and group j. Note that 2 vs. 3 does not cross Note the difference in p-values of differences if we hadn t adjusted for multiple comparisons: proc glm data= stats210.runners; class mencat; model sumedi1=mencat; lsmeans mencat/pdiff cl; 10