# A problem about confidence interval

How can I solve this?

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.

My attempt: I think I need to use the identity for the pooled estimator $$S_{p}^{2}$$ given by $$S_{p}^{2}=\frac{(n_1-1)^{2}+(n_2-1)S_{2}^{2}}{n_1 + n_2-2}$$ where $$S_{i}^{2}$$ is the sample variance from the $$i$$-th sample, $$i=1,2,\ldots,n$$.

First step: I am just beginning to deal with these kinds of problems and I don't know how to approach the problem. So, my question is how can I solve the problem using the definition of $$S_{p}^{2}$$ that I have learned?

• Where does $S_p^1$ come from? I guess a typo? – TrungDung Nov 19 '20 at 10:07
• Yes, That was a typo. I corrected. – user302981 Nov 19 '20 at 10:08
• That is the correct first step. The next search term you are looking for is "confidence interval between two means" which should give you what you need to do next. You might also like to add the self-study tag and read its wiki. – mdewey Nov 19 '20 at 16:41
• @mdewey I corrected the tag, thank you. Can you explain more about my approach for this problem? – user302981 Nov 19 '20 at 16:48
• There should be a derivation of this in your notes or textbook. If not, I like JBStatistics videos on YouTube. – Dave Nov 19 '20 at 17:19

Denote the lowered points of the blood pressure by the two drugs as $$X_1$$ and $$X_2$$ respectively, where both $$X_1$$ and $$X_2$$ are normal distribution with sample means $$\bar{X}_1=11$$ and $$\bar{X}_2=12$$, and STD $$\sigma_1=6$$ and $$\sigma_2=8$$ respectively. Obviously, the difference between $$\bar X_1$$ and $$\bar X_2$$ is also normal distribution with variance equal to $$\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}$$. Thus, the confidence interval is $$[a\space b]$$, where $$a=\bar{X}_1-\bar{X}_2 - z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$$,
$$b=\bar{X}_1-\bar{X}_2 + z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$$,
where $$z_{\alpha/2}=1.96$$ for confidence level of 95%.
• Thank you for you explication! Now, I understand, but I have a question, what's the value for $n_{1}$ and $n_{2}$? – user302981 Nov 21 '20 at 10:21