Comparaison of two independent samples with unequal variance and high sample size difference I am analysing a radiologic parameter on CT scans of 145 patients. These come from two different hospitals (hospital 1 > 120 patients and hospital 2 > 25 patients, total distribution is normal). A question that arises is the potential heterogeneity of the parameter due to the use of different CT scanners, e.g. from the hospital the patients are coming from. My promoter suggested I perform a t test  between the two groups (hospital 1 vs hospital 2) but that seems incorrect to me, 120 vs 25 patients with unequal variances. I performed a F test to show the variances were unequal, but this does not answer the initial question... Any idea of the test I could perform to appraise the difference between hospital of origin question ?
Thanks !
 A: Consider data sampled from two different normal populations using R,
summerized below. Sample means $\bar X_1 = 100.76$ and $\bar X_2 = 124.12$
differ. The question is whether they differ 'significantly' in a statistical sense.
As in your Question the sample sizes and standard deviations also differ.
summary(x1);  length(x1);  sd(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  65.92   91.06   99.39  100.76  111.30  138.52 
[1] 120        # size of Sample 1
[1] 14.92361   # sample SD of Sample 1
summary(x2);  length(x2);  sd(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  98.46  110.52  119.43  124.12  135.86  163.56 
[1] 25
[1] 17.11231
x = c(x1,x2);  hosp = rep(1:2, c(120,25))

The narrower boxplot corresponds to the smaller sample. The 'notches' in the sides
of the boxes are approximate nonparametric confidence intervals calibrated not to
overlap if the respective populations have significantly different centers at the 5% level. So it seems possible that a formal statistical test will find a significant
difference in population means.
boxplot(x ~ hosp, varwidth=T, notch=T, col="skyblue2", xlab="Hospital")


Here are results for a Welch two-sample two-sided t test of $H_0: \mu_1=\mu_2$ against
$H_a: \mu_1 \ne \mu_2.$ The P-value is near $0$ so we can reject $H_0$ at any
reasonable significance level.
t.test(x ~ hosp)

        Welch Two Sample t-test

data:  x by hosp
t = -6.3419, df = 32.046, p-value = 4.024e-07
alternative hypothesis: 
  true difference in means is not equal to 0
95 percent confidence interval:
 -30.86399 -15.85818
sample estimates:
mean in group 1 mean in group 2 
       100.7572        124.1183 

This test does not assume that population variances are equal.
A traditional two-sample t test that does assume equal variances
would have DF $n_1 + n_2 - 2 = 143$ degrees of freedom. The Welch
test allows for unequal variances as suggested by by the different
sample standard deviations by reducing the DF to about 32. Thus
the test has somewhat lower power than a pooled t test, but for our
fake data power is not an issue.
Note: The R code below shows how the data above were samples (simulated)
from two normal populations. If you like, you could run this code in R to get the same
datasets used above. Of course, results using your actual CT scan data may be different.
set.seed(828)
x1 = rnorm(120, 100, 15)
x2 = rnorm( 25, 120, 20)

