Before/After Treatment Results Captopril is a drug designed to lower systolic blood pressure. When subjects were treated with this drug, their systolic blood pressure readings (in mm Hg) were measured before and after the drug was taken. Results are given in the accompanying table (based on data from “Essential Hypertension: Effect of an Oral Inhibitor of Angiotensin-Converting Enzyme,” by MacGregor et al., British Medical Journal, Vol. 2). Using a 0.01 significance level, is there sufficient evidence to support the claim that captopril is effective in lowering systolic blood pressure?

Subject	A	B	C	D	E	F	G	H	I	J	K	L
Before	200	174	198	170	179	182	193	209	185	155	169	210
After	191	170	177	167	159	151	176	183	159	145	146	177

The systolic blood pressure measurements of a sample of 12 subjects are recorded as:“before the drug Captopril is taken” and “after the drug Captopril is taken”.

It is claimed that the drug Captopril is effective in lowering systolic blood pressure levels.

Corresponding to the given claim, the following hypotheses are set up:

Null Hypothesis:

\({H_0}:{\mu _d} = 0\)

Alternative Hypothesis:

\({H_1}:{\mu _d} > 0\)

The test is right-tailed.

Where \({\mu _d}\)be the mean difference between the systolic blood pressure levels before and after taking the drug.

The following table shows the differences in the systolic blood pressure levels for each matched pair:

The mean value of the differences is computed below:

\(\begin{aligned} \bar d &= \frac{{\sum\limits_{i = 1}^n {{d_i}} }}{n}\\ &= \frac{{9 + 4 + ...... + 33}}{{12}}\\ &= 18.58\end{aligned}\)

The standard deviation of the differences is computed below:

\(\begin{aligned} {s_d} &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \bar d)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( {9 - 18.58} \right)}^2} + {{\left( {4 - 18.58} \right)}^2} + ....... + {{\left( {33 - 18.58} \right)}^2}}}{{12 - 1}}} \\ &= 10.10\end{aligned}\)

The mean value of the differences for the population of matched pairs \(\left( {{\mu _d}} \right)\) is considered to be equal to 0.

The value of the test statistic is computed as shown:

\(\begin{aligned} t &= \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ &= \frac{{18.58 - 0}}{{\frac{{10.10}}{{\sqrt {12} }}}}\\ &= 6.371\end{aligned}\)

The degrees of freedom are computed below:

\(\begin{aligned} df &= n - 1\\ &= 12 - 1\\ &= 11\end{aligned}\)

The critical value of t at\(\alpha = 0.01\)and degrees of freedom equal to 11 for a right-tailed test is equal to 2.7181.

The corresponding p-value is equal to 0.00003.