The different instructional techniques have no effect on student improvement as evidenced by pre- and posttest scores.

Set alpha = 0.05. Remember, alpha is the probability of Type I error: the error associated with rejecting a true null hypothesis. In our class, we would make such an error if the instruction techniques really don't matter, but we reject the null hypothesis based on our data.
1. Since we have no idea of the underlying value of the variance, which set of densities shall we use?

2. How many degrees of freedom do you think we will ultimately have?

3. One issue is to calculate the "pooled" variance and "pooled" standard deviation. Since we expect that the variances of the improvement scores by both groups could be a bit different, we'd want an overall value that reflects the variance of both groups. Such a formula is given by

sp = sqrt(((n1-1)*(s1^2)+(n2-1)(s2^2))/(n1+n2-2))

where n1 is the number of students in the treatment group, n2 is the number of students in the control group, s1 is the standard deviation of the scores in the treatment group, and s2 is the standard deviation of the scores in the control group. Calculate the individual standard deviations for the treatment group and the control group. That is, calculate the sample standard deviation for the treatment group and the sample standard deviation for the control group.

4. Now, use the formula above to calculate the pooled standard deviation.