this problem is from the wackerly, but i have a particular question, in this problem we have to estimate the variance with $S^2_p$ or can we use the information standard deviations that they gave us.
Here is the problem:
Two new drugs were given to patients with hypertension. The first drug lowered the blood pressure of 16 patients an average of 11 points, with a standard deviation of 6 points. The second drug lowered the blood pressure of 20 other patients an average of 12 points, with a standard deviation of 8 points. Determine a 95% confidence interval for the difference in the mean reductions in blood pressure, assuming that the measurements are normally distributed with equal variances.
Assuming the given sample standard deviations $S_1,S_2$ have respective denominators $16-1$ and $20-1$, the pooled sample variance is \begin{align} S_p^2 & = \frac{(16-1)S_1^2 + (20-1)S_2^2}{(16-1)+(20-1)} = \frac{(15\times6^2) + (19\times 8^2)}{(16-1)+(20-1)} \\[12pt] & = \frac{540 + 1216}{34} = \frac{1756}{34} \approx 7.186589^2. \end{align}
Recall that $((16-1)+(20-1)) \dfrac{S_p^2}{\sigma^2} \sim \chi^2_{(16-1)+(20-1)}.$
You need to use $S_p$ with a t-distribution with $(16-1)+(20-1) = 34$ degrees of freedom.