One Sample T-Test With Howell Data, IQ of Students in Vermont

Transcription

1 One Sample T-Test With Howell Data, IQ of Students in Vermont data howell; infile 'C:\Users\Vati\Documents\StatData\howell.dat'; input addsc sex repeat iq engl engg gpa socprob dropout; IQ_diff = iq - 100; run; *Want to test null hypothesis that mean IQ is 100; proc means mean stddev n skewness kurtosis t prt CLM; var iq IQ_diff; run; One Sample T-Test With Howell Data, IQ of Students in Vermont The MEANS Procedure Variable Mean Std Dev N Skewness Kurtosis t Value Pr > t Lower 95% Upper 95% iq IQ_diff Data CI1; t = 0.19 ; n = 88 ; df = n-1; d = t/sqrt(n); ncp_lower = TNONCT(t,df,.975); ncp_upper = TNONCT(t,df,.025); d_lower = ncp_lower/sqrt(n); d_upper = ncp_upper/sqrt(n); output; run; proc print; var d d_lower d_upper; run; Obs d d_lower d_upper Vermont students mean IQ (M = , SD = 12.98) did not differ significantly from 100, t(87) = 0.19, p =.85, d =.02. A 95% CI for the difference in means runs from in in IQ units and from -.19 to.23 in standard deviation units

2 Experiment 2 of Karl's Dissertation Correlated t-tests, Visits to Mus Tunnel vs Rat Tunnel, Three Nursing Groups data Mus; infile 'C:\Users\Vati\Documents\StatData\tunnel2.dat'; input nurs $ 1-2 L1 3-5 L2 6-8 t t v_mus 15- v_rat 17-18; *t_mus=sqrt(1.575 * t1 +.5); *t_rat=sqrt(1.575 * t2 +.5); *L_mus=LOG10(1.575 * L1 + 1); *L_rat=LOG10(1.575 * L2 + 1); v_diff=v_mus - v_rat; *t_diff=t_mus - t_rat; *L_diff=L_mus - L_rat; proc sort; by nurs; proc means mean stddev n skewness kurtosis t prt CLM; var V_mus V_rat V_diff; by nurs; run; Experiment 2 of Karl's Dissertation Correlated t-tests, Visits to Mus Tunnel vs Rat Tunnel, Three Nursing Groups The MEANS Procedure nurs=mm Variable Mean Std Dev N Skewness Kurtosis t Value Pr > t Lower 95% Upper 95% v_mus v_rat v_diff The MM mice made significantly more visits to the Mus-scented tunnel (M = 22.44, s = 12.82) than to the Rattus-scented tunnel (M = 7.56, s = 5.89), t(15) = 5.64, p <.001, d = I did not report the unstandardized confidence intervals here, as I felt they would not add much to understanding the results. nurs=mr Variable Mean Std Dev N Skewness Kurtosis t Value Pr > t Lower 95% Upper 95% v_mus v_rat v_diff

3 The MR mice made significantly more visits to the Mus-scented tunnel (M = 22.31, s = 10.21) than to the Rattus-scented tunnel (M = 7.56, s = 5.73), t(15) = 7.80, p <.001, d = nurs=rr Variable Mean Std Dev N Skewness Kurtosis t Value Pr > t Lower 95% Upper 95% v_mus v_rat v_diff For the RR mice, the number of visits to the Mus-scented tunnel (M = 23.44, s = 8.51) did not differ significantly from the number of visits to the Rattus-scented tunnel (M = 24.75, s = 8.10), t(15) = 0.62, p =.54, d =.. Notice that house mice reared with house mice or with deer mice avoided the rat-scented tunnels, but those reared with rats did not. Avoiding the scent of rat makes a lot of sense when you consider that rats eat mice. proc corr nosimple; var V_Mus; with V_rat; by nurs; run; We need these correlations to run Algina s code for computing d with confidence intervals. The CORR Procedure nurs=mm 1 With Variables: v_rat 1 Variables: v_mus Pearson Correlation Coefficients, N = Prob > r under H0: Rho=0 v_mus v_rat

4 nurs=mr 1 With Variables: v_rat 1 Variables: v_mus Pearson Correlation Coefficients, N = Prob > r under H0: Rho=0 v_mus v_rat nurs=rr 1 With Variables: v_rat 1 Variables: v_mus Pearson Correlation Coefficients, N = Prob > r under H0: Rho=0 v_mus v_rat Data CI2; m1= ; m2= ; s1= ; s2= ; r= ; n= ; prob=.95 ; v1=s1**2;

5 v2=s2**2; s12=s1*s2*r; se=sqrt((v1+v2-2*s12)/n); pvar=(v1+v2)/2; nchat=(m1-m2)/se; es=(m1-m2)/(sqrt(pvar)); df=n-1; ncu=tnonct(nchat,df,(1-prob)/2); ncl=tnonct(nchat,df,1-(1-prob)/2); ul=se*ncu/(sqrt(pvar)); ll=se*ncl/(sqrt(pvar)); output; proc print; title1 'll is the lower limit and ul is the upper limit'; title2 'of a confidence interval for the effect size'; title3 'MM group only' ; var es ll ul ; run; ll is the lower limit and ul is the upper limit of a confidence interval for the effect size MM group only Obs es ll ul House mice who had been reared with house mice visited the tunnels scented with house mouse significantly more often (M = 22.44, SD = 12.81) than tunnels scented with rat (M = 7.56, SD = 5.89), t(15) = 5.64, p <.001, d = 1.49, 95% CI [.74, 2.22]. Independent Samples T-Tests on Mouse-Rat Tunnel Difference Scores Foster Mom is a Mouse or is a Rat data Mus2; set Mus; if nurs NE 'RR' then Mom = 'Mouse'; else if nurs = 'RR' then Mom = 'Rat'; proc ttest; class Mom; var v_diff; run; *proc mixed; *class Mom; *model v_diff = Mom / ddfm = satterth; *repeated / group=mom; *lsmeans Mom / pdiff CL; *run;

6 title2 'Foster Mom is a Mouse, Pups Exposed to Rat Scent (MR) or Not (MM)'; run; Independent Samples T-Tests on Mouse-Rat Tunnel Difference Scores Foster Mom is a Mouse or is a Rat The TTEST Procedure Variable: v_diff The dependent variable here is a measure of preference for visiting the Mus-scented tunnel rather than the Rattus-scented tunnel. Mom N Mean Std Dev Std Err Minimum Maximum Mouse Rat Diff (1-2) Mom Method Mean 95% CL Mean Std Dev 95% CL Std Dev Mouse Rat Diff (1-2) Pooled Diff (1-2) Satterthwaite Method Variances DF t Value Pr > t Pooled Equal Satterthwaite Unequal Equality of Variances Method Num DF Den DF F Value Pr > F Folded F

7 data CI3; t= 5.96 ; df = 46 ; n1 = 32 ; n2 = ; ***********************************************************************************; d = t/sqrt(n1*n2/(n1+n2)); ncp_lower = TNONCT(t,df,.975); ncp_upper = TNONCT(t,df,.025); d_lower = ncp_lower*sqrt((n1+n2)/(n1*n2)); d_upper = ncp_upper*sqrt((n1+n2)/(n1*n2)); output; run; proc print; var d d_lower d_upper; run; Obs d d_lower d_upper Mice that had been fostered onto a mouse mother had a significantly stronger preference for visiting mouse-scented rather than rat-scented tunnels (M = 14.81, SD = 9.03) than did mice who had been fostered onto a rat mother (M = -1.31, SD = 8.40), t(32.1) = 6.11, p <.001, d = 1.82, 95% CI [1.11, 2.52]. The graphics provided by SAS in the htm output can be very helpful with respect to evaluating the normality assumption. For each group we get a histogram overlaid with 1.) a smoothed curve representing the distribution of the observed scores, and 2.) a normal curve with the same mean and standard deviation of the observed scores.

8 data Mus3; set Mus; if nurs NE 'RR'; proc ttest; class nurs; var v_diff; run; Independent Samples T-Tests on Mouse-Rat Tunnel Difference Scores Foster Mom is a Mouse, Pups Exposed to Rat Scent (MR) or Not (MM) Variable: v_diff nurs N Mean Std Dev Std Err Minimum Maximum MM MR Diff (1-2) nurs Method Mean 95% CL Mean Std Dev 95% CL Std Dev MM MR Diff (1-2) Pooled Diff (1-2) Satterthwaite Method Variances DF t Value Pr > t Pooled Equal Satterthwaite Unequal Method Equality of Variances Num DF Den DF F Value Pr > F Folded F The sample sizes are equal and the ratio of the larger variance to the smaller variance less than 4, so I employ the pooled variances t test.

9 Preference for the Mus-scented tunnel for the MM mice (M = 14.88, s = 10.55, n = ) did not differ significantly from that for the MR mice (M = 14.75, s = 7.57, n = ), t(30) = 0.04, p =.97, d =.01, 95% CI [-.68,.71] Independent t on WTLOSS Data data wtloss; input program $ ; cards; A 25 A 21 A 18 A 20 A 22 A 30 B 15 B 17 B 9 B 12 B 11 B 19 B 14 B 18 B B 10 B 5 B 13 proc ttest; class program; var loss; run; program N Mean Std Dev Std Err Minimum Maximum A B Diff (1-2) program Method Mean 95% CL Mean Std Dev 95% CL Std Dev A B Diff (1-2) Pooled Diff (1-2) Satterthwaite Method Variances DF t Value Pr > t Pooled Equal Satterthwaite Unequal Equality of Variances Method Num DF Den DF F Value Pr > F Folded F

10 data CI4; t= 4.54 ; df = ; n1 = 6 ; n2 = 12 ; ***********************************************************************************; d = t/sqrt(n1*n2/(n1+n2)); ncp_lower = TNONCT(t,df,.975); ncp_upper = TNONCT(t,df,.025); d_lower = ncp_lower*sqrt((n1+n2)/(n1*n2)); d_upper = ncp_upper*sqrt((n1+n2)/(n1*n2)); output; run; proc print; var d d_lower d_upper; run; Obs d d_lower d_upper Participants who completed the Marine Corps weight loss program (M = 22.67, s = 4.27, n = 6) lost significantly more weight than those who completed the Boy Scouts weight loss program (M = 13.25, s = 4.09, n = 12), t(9.7) = 4.47, p =.001, d = 2.27, 95% CI [1.00, 3.50]. When considering these results, one must question how the selective attrition might have affected the means. Perhaps weight loss was greater in Program A simply because Program A was more effective in chasing off those who were not losing weight. options pageno=min nodate formdlim='-'; proc format; value bk 0='Kerry' 1='Bush'; run; title 'Income, IQ, and Presidential Voting'; run; data Voting; infile 'C:\D\StatData\Bush-Kerry2004.txt'; input IQ state $ income vote candidate $; format vote bk. ; proc corr; var vote iq income; run; proc ttest; class vote; var iq income; run; Data CI; t= 2.61 ; df = 49 ; n1 = 20 ; n2 =31 ; d = t/sqrt(n1*n2/(n1+n2));

11 ncp_lower = TNONCT(t,df,.975); ncp_upper = TNONCT(t,df,.025); d_lower = ncp_lower*sqrt((n1+n2)/(n1*n2)); d_upper = ncp_upper*sqrt((n1+n2)/(n1*n2)); output; run; proc print; var d d_lower d_upper; run; Obs d d_lower d_upper Income, IQ, and Presidential Voting options pageno=min nodate formdlim='-'; proc format; value bk 0='Kerry' 1='Bush'; run; title 'Income, IQ, and Presidential Voting'; run; data Voting; infile 'F:\StatData\Bush-Kerry2004.txt'; input IQ state $ income vote candidate $; format vote bk. ; proc corr; var vote iq income; run; proc ttest; class vote; var iq income; run; The CORR Procedure 3 Variables: vote IQ income Simple Statistics Variable N Mean Std Dev Sum Minimum Maximum vote IQ income For the vote variable, 0 coded the state voting for Kerry, 1 coded the state voting for Bush.

12 Pearson Correlation Coefficients, N = 51 Prob > r under H0: Rho=0 vote IQ income vote IQ income There are two significant correlations in the matrix above. Voting for Bush is negatively correlated with average IQ and average income in the state. We can test the null hypothesis that the correlation between voting preference and IQ is zero this way: t r n 2 1 r The obtained value of t will be identical to that for the pooled t test. The TTEST Procedure Variable: IQ vote N Mean Std Dev Std Err Minimum Maximum Kerry Bush Diff (1-2)

13 vote Method Mean 95% CL Mean Std Dev 95% CL Std Dev Kerry Bush Diff (1-2) Pooled Diff (1-2) Satterthwaite Method Variances DF t Value Pr > t Pooled Equal Satterthwaite Unequal Method Equality of Variances Num DF Den DF F Value Pr > F Folded F The estimated mean IQ of residents of states which voted for Kerry (M = 101.0, s = 1.99, n = 20) was significantly higher than that of states which voted for Bush (M = 99.3, s = 2.42, n = 32), t(46.1) = 2.71, p =.009, d =.75, 95% CI [., 1.33]. One could use the point biserial r as the effect size estimator here, or its square (which can be interpreted as a proportion of variance in IQ explained by the model). The value of that r here is Variable: income vote N Mean Std Dev Std Err Minimum Maximum Kerry Bush Diff (1-2)

14 vote Method Mean 95% CL Mean Std Dev 95% CL Std Dev Kerry Bush Diff (1-2) Pooled Diff (1-2) Satterthwaite Method Variances DF t Value Pr > t Pooled Equal Satterthwaite Unequal Method Equality of Variances Num DF Den DF F Value Pr > F Folded F Average income in states voting for Kerry (M = $34,672, s = $5,002), n = 20) was significantly greater than that in states voting for Bush (M = $28,333, s = $2,879, n = 31), t(27.2) = 5.14, p <.001, d = 1.65, 95% CI [.99, 2.29].