The appropriate statistical test is the unpaired or independent or
two-group Student t-test. Also appropriate is the one-way
analysis of variance (ANOVA) for between-cases. The difference between the means of the two
groups are compared against the estimated standard error of the mean
difference. That is,
where
The calculated value of Student's t is compared to the t-distribution with degrees of
freedom equal to 2(n-1) when there are n observations in each group.
In a study of pain mechanisms, behavioral neuroscientists injected a group of rats with a drug that would increase pain sensitivity if a certain neural pathway was involved in pain. Pain sensitivity was measured by the time (in seconds) it took for the rat to flick its tail away from a heat lamp. Longer times until tail flick indicate less pain sensitivity. Testing was always stopped after ten seconds before there could be any injury to the rats. Data from the injected group were compared to a placebo group in which rats received inert injections. Below are the data for fourteen rats, seven in each group:
Placebo: 2.6 10 9.5 7.4 6.9 8.5 5.2
Drug: 2.2 3.8 7.1 2.7 6.2 5.3 3.1
The neuroscientists want to know whether the drug increased pain sensitivity. If the drug had absolutely no effect, then we would expect the mean tail flick time in the placebo group to be the same as the mean tail flick time in the drug group.
In a study of pain sensitivity, rats injected with a drug believed to increase pain sensitivity flicked their tails away from a heat source in 4.3 seconds, on average; rats injected with a placebo flicked their tails away in 7.2 seconds, on average. The mean difference of 2.8 seconds is statistically significant (t(12) = 2.33, p = .038). Thus, the drug indeed increased pain sensitivity.
Use one variable for the independent variable and another for the dependent variable. The alternative in Method 2 uses separate variables for values of the dependent variable for each group.
> flick <- c(2.6,10,9.5,7.4,6.9,8.5,5.2,2.2,3.8,7.1,2.7,6.2,5.3,3.1) > group <- factor(c(rep("placebo",7),rep("drug",7))) > t.test(flick ~ group, var.equal=T) Two Sample t-test data: flick by group t = -2.3315, df = 12, p-value = 0.03797 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -5.4442552 -0.1843162 sample estimates: mean in group drug mean in group placebo 4.342857 7.157143 #tapply is useful for obtaining group statistics. it 'applies' a function to a variable # to each subgroup defined by a factor. the syntax is: # tapply(variable, factor, function) > tapply(flick, group, mean) drug placebo 4.342857 7.157143 > tapply(flick, group, sd) drug placebo 1.875151 2.585122
> placebo <- c(2.6,10,9.5,7.4,6.9,8.5,5.2) > drug <- c(2.2,3.8,7.1,2.7,6.2,5.3,3.1) > t.test(placebo,drug,var.equal=TRUE) Two Sample t-test data: placebo and drug t = 2.3315, df = 12, p-value = 0.03797 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.1843162 5.4442552 sample estimates: mean of x mean of y 7.157143 4.342857 > boxplot(placebo,drug,ylab='Tail flick (seconds)',col=c('red','blue'), + names=c('Placebo','Drug'))
#ADVANCED: checking the normality assumption > res <- c(placebo-mean(placebo), drug - mean(drug)) > qqline(res) > qqline(res, col='blue',lwd=2)
MTB > TwoSample 95.0 'control' 'drug'; SUBC> Alternative 0. Two Sample T-Test and Confidence Interval Twosample T for control vs drug N Mean StDev SE Mean control 7 7.16 2.59 0.98 drug 7 4.34 1.88 0.71 95% C.I. for mu control - mu drug: ( 0.12, 5.50) T-Test mu control = mu drug (vs not =): T= 2.33 P=0.038 DF= 12
Note: In contrast to the other computer programs above, Excel reports variances instead of standard deviations.
© 2002, Gary McClelland