September 16, 2026
Clinical Question:
Does a treatment reduce average systolic blood pressure?
Tools: R, ggplot2, and Plotly
The dataset is simulated for educational purposes.
Clinical researchers often want to know whether an observed difference provides evidence of a real difference.
Examples:
Hypothesis testing helps us use sample data to evaluate a claim about a population.
The population is the entire group we want to study.
Example: All patients who could receive a particular treatment.
The sample is the smaller group included in the study.
Example: 30 patients in our simulated dataset.
We use information from the sample to learn about the population.
Our dataset contains information about:
Control: 15 patients
Treatment: 15 patients
Systolic blood pressure (SBP)
Measured in mmHg.
Suppose we want to know:
Does the treatment reduce average systolic blood pressure below 130 mmHg?
We will focus on the treatment group.
Our reference value is:
\[ \mu_0 = 130 \text{ mmHg} \]
We will use a one-sample t-test.
\[ H_0: \mu = 130 \]
The population average systolic blood pressure is 130 mmHg.
\[ H_a: \mu < 130 \]
The population average systolic blood pressure is less than 130 mmHg.
Because we are looking for a reduction, this is a one-sided test.
Before performing the test, we choose a significance level.
\[ \alpha = 0.05 \]
The significance level is our cutoff for making a statistical decision.
\[ p < \alpha \Rightarrow \text{Reject } H_0 \]
\[ p \geq \alpha \Rightarrow \text{Fail to reject } H_0 \]
The p-value tells us how unusual our sample result would be if the null hypothesis were true.
A p-value does not tell us the probability that the null hypothesis is true.
We can perform the hypothesis test using t.test().
treatment = clinical$systolic_bp[ clinical$group == "Treatment" ] result = t.test( treatment, mu = 130, alternative = "less" ) result
One Sample t-test
data: treatment
t = -6.8961, df = 14, p-value = 3.685e-06
alternative hypothesis: true mean is less than 130
95 percent confidence interval:
-Inf 126.4756
sample estimates:
mean of x
125.2667
R gives us: - t-statistic - degrees of freedom - p-value - confidence interval - sample mean
For a one-sample t-test:
\[ t = \frac{\bar{x}-\mu_0} {s/\sqrt{n}} \]
Where:
For our treatment group:
\[ \bar{x} = 125.27 \text{ mmHg} \]
\[ s = 2.66 \]
\[ n = 15 \]
Therefore:
\[ \boxed{t = -6.896} \]
125.27 mmHg
-6.896
3.685e-06
0.05
We compare the p-value with our significance level.
\[ p = 3.685e-06 \]
\[ \alpha = 0.05 \]
Because:
\[ p < 0.05 \]
we reject \(H_0\).
There is enough statistical evidence to conclude that the treatment group’s average systolic blood pressure is below 130 mmHg.
The t-statistic is away from the center of the distribution.
This provides evidence against the null hypothesis.