In a clinical trial, let's say I want to test $$H_0: \mu_1 \leq \mu_2$$ $$H_1: \mu_1 > \mu_2$$
$\mu_1$ belongs to the placebo group and $\mu_2$ belongs to the trt group. I used an independent two-sample t-test to compare the two means. I wonder how to express the confidence interval here. And should the point estimate by $\bar{X}_1-\bar{X}_2$? I'm not familiar with statistics and clinical trials so I need some help.
If you use statistical software to test the hypothesis $H_0: \mu_1 \le \mu_2$ against $H_a: \mu_1 > m_2$ you will typically get a 95% confidence interval for $\mu_1 - \mu_2$ as part of the output.
Suppose you have data similar to the fictitious data, simulated using R statistical software below:
set.seed(2022)
x1 = rnorm(1000, 50, 7)
x2 = rnorm(1000, 55, 8)
summary(x1)
Min. 1st Qu. Median Mean 3rd Qu. Max.
25.67 45.28 49.65 50.01 54.79 76.36
length(x1); sd(x1)
[1] 1000 # sample size
[1] 6.987677 # sample standard deviation
summary(x2)
Min. 1st Qu. Median Mean 3rd Qu. Max.
25.86 49.61 54.99 54.97 60.66 79.58
length(x2); sd(x2)
[1] 1000
[1] 8.190649
boxplot(x1, x2, horizontal=T, col="skyblue2")
Then a Welch two-sample t test, which does not assume that treatment and control populations have the same variance, goes as shown below. Because the P-value is near $0,$ we reject $H_0$ in favor of the one-sided alternative $H_0.$ The 95% one-sided 95% confidence interval is $(-\infty, -4.40)$ so that $\mu_1 - \mu_2,$ estimated by $\bar X_1 - \bar X_2 = 50.00961 - 54.97116 = -4.96155$ is likely less than the upper bound $-4.40128 \approx -4.40.$
Welch Two Sample t-test
data: x1 and x2
t = -14.573, df = 1949.6, p-value < 2.2e-16
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
-Inf -4.40128
sample estimates:
mean of x mean of y
50.00961 54.97116
If you want a 2-sided 95% CI, then you can get it as part of the output for a 2-tailed (or 2-sided) test, specifically $(-5.63, -4.29),$ which is centered at $\bar X_1 - \bar X_2 = 50.00961 - 54.97116 = -4.96155 \approx -4.96.$
t.test(x1,x2)$conf.int
[1] -5.629265 -4.293849
attr(,"conf.level")
[1] 0.95
Output from Minitab statistical software for summarized data above, is shown below, where 0.000 means $< 0.0005:$
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 1000 50.01 6.99 0.22
2 1000 54.97 8.19 0.26
Difference = μ (1) - μ (2)
Estimate for difference: -4.960
95% upper bound for difference: -4.400
T-Test of difference = 0 (vs <):
T-Value = -14.57 P-Value = 0.000 DF = 1949
Again here, the two-sided 95% CI $(-5.628, -4.292)$ accompanies the two-sided test.
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 1000 50.01 6.99 0.22
2 1000 54.97 8.19 0.26
Difference = μ (1) - μ (2)
Estimate for difference: -4.960
95% CI for difference: (-5.628, -4.292)
T-Test of difference = 0 (vs ≠): T-Value = -14.57 P-Value = 0.000 DF = 1949
Both programs use the following formula for the 2-sided 95% confidence interval of $\mu_1-\mu_2:$
$$\bar X_1 - \bar X_2 \pm t^*\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}},$$ where $n_i, \bar X_i, S^2_i$ are the sample size, mean, and variance, respectively, of the $i$th sample, and $t^*$ cuts probability $0.025$ from the upper tail of Student's t distribution with the appropriate degrees of freedom. In my example $t^*=1.96.$