Before we go to test the means first we have to test their variability using F-test.
We test,
H0​:σ12​=σ22​vsH1​:σ12â€‹î€ =σ22​
The test statistic is,
Fc​=sa2​sb2​​=72.5139​=1.91724138
The table value is,
Fα,nb​−1,na​−1​=F0.05,4,4​=6.388233 and we reject the null hypothesis if Fc​>Fα,nb​−1,na​−1​
Since Fc​=1.91724138<F0.05,4,4​=6.388233, we accept the null hypothesis that the population variances for medicine type A and medicine type B are equal.
We now proceed to test whether the difference in means is significant.
tc​ is compared with the table value at α=0.05 with na​+nb​−2=5+5−2=8 degrees of freedom.
The table value is,
t20.05​,8​=t0.025,8​=2.306004
The null hypothesis is rejected if ∣tc​∣>t0.025,8​.
Now,
∣tc​∣=0.1538<t0.025,8​=2.306004, and we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that the two medicines differ significantly with regard to their effect and increasing weight at 5% significance level.