# Solved (Free): A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded

#### ByDr. Raju Chaudhari

Mar 9, 2021

A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level.

Patient A B C D E F
Before 161 162 165 162 166 171
After 158 159 166 160 167 169

#### Solution

Let $x$ denote blood pressure after medication and $y$ denote blood pressure before medication. The sample size $n = 6$. Let $d=x-y$. So $\overline{d}= -1.3333$ and $s_d = 1.8619$.

x y $d$ $d-\overline{d}$ $(d-\overline{d})^2$
158 161 -3 -1.6667 2.7778
159 162 -3 -1.6667 2.7778
166 165 1 2.3333 5.4444
160 162 -2 -0.6667 0.4444
167 166 1 2.3333 5.4444
169 171 -2 -0.6667 0.4444

#### Step 1 Hypothesis

The hypothesis testing problem is
$H_0 : \mu_d = 0$ against $H_1 : \mu_d < 0$ ($\textit{left-tailed}$)

#### Step 2 Test Statistic

The test statistic is
 \begin{aligned} t=\frac{\overline{d} -\mu_d}{s_d/\sqrt{n}} \end{aligned}

#### Step 3 Level of Significance

The significance level is $\alpha = 0.01$.

#### Step 4 Critical Value(s)

As the alternative hypothesis is $\textit{left-tailed}$, the critical value of $t$ for $5$ degrees of freedom and $\alpha = 0.01$ level of significance $\text{is}$ $\text{-3.365}$.

The rejection region (i.e. critical region) is $\text{t < -3.365}$.

#### Step 5 Computation

The test statistic for testing above hypothesis testing problem under the null hypothesis is

 \begin{aligned} t&=\frac{\overline{d} -\mu_d}{s_d/\sqrt{n}}\\ &= \frac{-1.3333-0}{1.8619/\sqrt{6}}\\ &= -1.7541 \end{aligned}

#### Step 6 Decision

The test statistic is $t =-1.7541$ which falls $\textit{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

OR

$p$-value approach:

The test is $\textit{left-tailed}$ test, so p-value is the area to the $\textit{left}$ of the test statistic ($t=-1.7541$). That is p-value = $P(t\leq -1.7541 ) = 0.0699$.

The p-value is $0.0699$ which is $\textit{greater than}$ the significance level of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.

Thus there is no sufficient evidence to support that the medication is effective.