2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ If you define $d=\mu_1-\mu_2$ then the null hypothesis becomes $d \le 0$ and the alternative $d>0$, with $\bar X_1-\bar X_2$ being the natural estimator of $d$. You are trying to find a confidence interval for $d$. $\endgroup$
    – Henry
    May 14 at 11:21
  • $\begingroup$ Notice that you ask about a one-sided 2-sample t test, but about a two_sided confidence interval. See my Answer. Also, you do not say whether you are suing the pooled 2-sample t test or the Welch 2-sample t test. Unless you have solid advance evidence that the two populations have the same variance, you should use the Welch test. $\endgroup$
    – BruceET
    May 14 at 14:56

1 Answer 1

1
$\begingroup$

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")

enter image description here

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 $t^*=1.96.$

$\endgroup$
4
  • $\begingroup$ Thank you for your detailed answer! It is a one-sided test so I was pretty confused. $\endgroup$ May 15 at 2:34
  • 1
    $\begingroup$ Is one-sided test CI is different from the formula you provided? $\endgroup$ May 15 at 2:46
  • $\begingroup$ Yes, a one-sided CI is usually used along with a one-sided test. The formula for the one-sided interval is different: No $\pm,$ just $+$ to get an upper bound, in this problem (and $t^*$ cuts 5% of the area from the upper tail of the standard normal density curve. $\endgroup$
    – BruceET
    May 15 at 3:58
  • $\begingroup$ I guess my case is a one-sided test then. There is not much information about a one-sided CI in a clinical trial so I was very confused. Thank you. $\endgroup$ May 15 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.