Solved: A clinical trial is planned to compare an experimental medication designed to lower blood pressure to a placebo. Before starting the trial, a pilot study

A clinical trial is planned to compare an experimental medication designed to lower blood pressure to a placebo. Before starting the trial, a pilot study is conducted involving seven participants. The objective of the study is to assess how systolic blood pressure changes over time untreated. Systolic blood pressures are meassured at baseline and again 4 weeks later. Is there a statistically significant difference in blood pressure over time? Run the test at a 5% level of significance.

Solution

Let $x$ denote SBP at baseline and $y$ denote SBP 4 week later.

We use paired t-test.

The sample size $n = 7$. Let $d=x-y$. So $\overline{d}= -0.8571$ and $s_d = 3.4365$.

Step 1 Hypothesis

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

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 $\alpha = 0.05$.

Step 4 Critical Value(s)

As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $t$ for $6$ degrees of freedom and $\alpha = 0.05$ level of significance $\text{are}$ $\text{-2.447 and 2.447}$.

The rejection region (i.e. critical region) is $\text{t < -2.447 or t > 2.447}$.

Step 5 Computation

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

$$ \begin{aligned} t&=\frac{\overline{d} -\mu_d}{s_d/\sqrt{n}}\\ &= \frac{-0.8571-0}{3.4365/\sqrt{7}}\\ &= -0.6599 \end{aligned} $$

Step 6 Decision

Traditional approach:

The test statistic is $t =-0.6599$ which falls $\textit{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

OR

$p$-value approach:

The test is $\textit{two-tailed}$ test, so p-value is the area to the $\textit{extreme}$ of the test statistic ($t=-0.6599$). That is p-value = $2*P(t\geq 0.6599 ) = 0.5338$.

The p-value is $0.5338$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

That is our data does not support the claim that there is a statistically significant difference in systolic blood pressure over time.

Further Reading

• Statistics
• Descriptive Statistics
• Probability Theory
• Probability Distribution
• Hypothesis Testing
• Confidence interval
• Sample size determination
• Non-parametric Tests
• Correlation Regression
• Statistics Calculators