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
Traditional approach:
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.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
