### Oct. 21, 2008

# Two Sample T-test Continued

MJ<-read.csv("Marihuana.csv",header=T) attach(MJ) model <- aov(DSST~Group) To get all plots on one graph par(mfrow=c(2,2)) plot(model) names(model) [1] "coefficients" "residuals" "effects" "rank" [5] "fitted.values" "assign" "qr" "df.residual" [9] "contrasts" "xlevels" "call" "terms" [13] "model" model$fitted.values 1 2 3 4 5 6 7 8 -5.111111 -5.111111 -5.111111 -5.111111 -5.111111 -5.111111 -5.111111 -5.111111 9 10 11 12 13 14 15 16 -5.111111 0.250000 0.250000 0.250000 0.250000 0.250000 0.250000 0.250000 17 0.250000 model$fitted.values 1 2 3 4 5 6 7 8 9 -5.111111 -5.111111 -5.111111 -5.111111 -5.111111 -5.111111 -5.111111 -5.111111 -5.111111 10 11 12 13 14 15 16 17 0.250000 0.250000 0.250000 0.250000 0.250000 0.250000 0.250000 0.250000 library(car) qq.plot(model$residuals) par(mfrow=c(1,1)) qq.plot(model$residuals) We QQ plot residuals to determine if they are normal and can be used to compare w/ t sample t-test## Assumptions of the two-sample t-test

-independence (not too big a deal in t-test) -Normality of errors -mean = zero -homogeneity of variance (homostedasticity) tapply(DSST,Group,var) CU NS 30.21429 39.11111 30 - 39 is far less than 4-1 so homostedasticity assumption is met. library(car) levene.test(DSST,Group) Levene's Test for Homogeneity of Variance Df F value Pr(>F) group 1 0.0112 0.917 15**NOTE: SPSS does Levene's test by default and it runs it incorrectly!**par(mfrow=c(2,2)) plot(model) ?t.test

**NOTE: Equality of variances is assumed to be false, so you must change to true**t.test(DSST~Group,var.equal=T)

Two Sample t-test

data: DSST by Group

t = 1.866, df = 15, p-value = 0.08172

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-0.7625956 11.4848178

sample estimates:

mean in group CU mean in group NS

0.250000 -5.111111

**We fail to reject the null hypothesis, because it seem slike there is no difference between cronic users and nieve subjects.**

It seems like you don't gain anything from being a chronic user.

Either there are no significant differences, or we have a POWER issue.

**NOTE: Degrees of freedom for a two sample t-test is (n-sub-1 - 1)+(n-sub-2 - 2)**

t.test(DSST~Group)

Welch Two Sample t-test

data: DSST by Group

t = 1.8811, df = 15, p-value = 0.07951

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-0.713513 11.435735

sample estimates:

mean in group CU mean in group NS

0.250000 -5.111111

## Cohen's d for effect size of two sample t-test

Because we assume homogeneity of variance, we can use the pooled standard deviations of the two samples See notes on two-sample t-test for the pooled standard deviation**When we used the pooled SD we are estimating d and instead of calling it d-hat, we call it g in the two-sampled t-test.**g is an estimation of the effect difference between the two samples g is a point estimate for d library(MBESS) ?smd smd(Group.1=subset(DSST,Group=="NS"),Group.2=subset(DSST,Group=="CU")) [1] -0.906721

**NOTE: If you have a balanced design, we will get d-hat. If we have an unbalanced design we get g**

**in R**ci.smd(smd=-.906721,n.1=8,n.2=9) $Lower.Conf.Limit.smd [1] -1.898482

$smd

[1] -0.906721

$Upper.Conf.Limit.smd

[1] 0.1119626

ci.smd(smd=-.906721,n.1=9,n.2=8)

$Lower.Conf.Limit.smd

[1] -1.898482

$smd

[1] -0.906721

$Upper.Conf.Limit.smd

[1] 0.1119626

ci.smd(smd=.906721,n.1=9,n.2=8)

$Lower.Conf.Limit.smd

[1] -0.1119626

$smd

[1] 0.906721

$Upper.Conf.Limit.smd

[1] 1.898482