T Tests: Comparison of Means. Most t tests involve the comparison of two populations with respect to the means of randomly drawn samples from the respective populations.
(N1-1)S12 + (N2-1)S22
(N1 + N2) - 2
Inserting our sample data into the formula, we have
(39)(36) + (21)(16) / 40 + 22 -2 = 1404 + 336/60 = 29. Thus sp2 equals 29.
Pooled estimate of the standard deviation of the sampling distribution of differences in sample means is in the denominator-what we computed on previous slide
X1 – X2
The numerator of t equals the mean of group 1 (40) minus mean of group 2 (35) or 5. This value, 5, is divided by the square root of (29/40 + 29/22) and t equals 3. 498. Can we reject the null hypothesis? In other words, how likely is it that we would obtain a value of t as large as 3.498 if the experimental and control groups were from the same population with respect to the variable of interest? Looking up in the table we find that a t of 3.498 is significant (p < .005, one-tailed, DF = 60) and we can reject the null hypothesis-can say that the experimental and control groups differ significantly.
So in this case , t would be equal to 5/ the square root of 36/40 + 16/22, or 3.919. This statistic requires that you compute a different DF before consulting the t distribution table
like Blalock, use
N1-1 and N2 -2
in the denominator
for unequal variances.
X1 – X2
Can we reject the null hypothesis that there are no differences
between males and females in months of previous experience?
Where XD-bar is the mean difference between pairs of scores, N is the # of pairs of scores, the XD are the differences between each of the matched pairs of scores
√∑(XD –XD)2 / √(N-1)
Note: this computing formula gives an equivalent result to pp. 152-154 in Levin and Fox
Calculate t for this data: XD = 5; ∑(XD –XD)2 = 354, N= 10, DF=N-1
t = 5
= 5/1.983 = 2.521
Mean difference in positivity after hearing a commercial against pirating movie files
Note that the mean is higher (e.g. in this
case a more positive attitude) after the commercial
This correlation indicates that
about 49% (1-(.692)2)of the variation in post-test attitudes could be explained by pre-test attitudes. Presumably the rest of the variation is explained by treatment plus error
We have a significant value of t, but look at that confidence interval ;-( Also, compare the means; does this seem like a major change? And compare the standard deviations; in both cases they are all over the place in the raw scores
So we can reject the null hypothesis of no differences between pre and post and conclude that our treatment increased negative attitudes towards downloading