In: Math
We will perform two tailed t-test if the underlying distribution assumptions are assumed to be in place.
Null hypothesis: μ1 - μ2 = 0 or μ1 = μ2
Alternative hypothesis: μ1 - μ2 ≠ 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.SE = sqrt[(s1^2/n1) + (s2^2/n2)]
= 2.89
Calculate Degrees of freedom (DF):
DF=(s1^2/n1 + s2^2/n2)^2 / { [ (s1^2 / n1)^2 / (n1 - 1) ] + [ (s2^2 / n2)^2 / (n2 - 1) ] }
= 2.74 (rounded to 3)
t = [ (x1 - x2) - d ] / SE
where x and x2 are two sample means and d is the difference between two means we want to perform statistical significant test for. In our case because we want to compare two means, this is 0.
hence t = (11-7)/2.89 = 1.39
Using the t-distibution we find out the probablity of P(t < -1.39) = .1294 and P(t > 1.39) =.1294
which is greater than the statistical significance level we assumed (.10). hence fail to reject the null hypothesis. which means that the drug effect cannot be ascertained at the significance level (10%).