Normality Condition for a 2 sample difference in population means I was recently working on a statistics problem from AP Statistics FRQ 2018:Here is a picture
I don't understand how we got the normality condition which is required for the 2 sample difference in population means test. Specifically,  the central limit theorem states for a large sample size, greater than 30, the distribution of sample means will be normally distributed. However, we only took 1 sample and how do we even know the distribution of this sample is normally distributed?
Much thanks for any help provided.
 A: Minitab will accept summarized data. Here is the output for a Welch one-sided t test. This test assumes data are normal, but not that variances are equal. You can use formulas
from a textbook to verify these computations.
I will leave the interpretation of the
results up to you.
Two-Sample T-Test and CI 

Sample    N   Mean  StDev  SE Mean
1       110  217.0   34.0      3.2
2       100  186.0   29.0      2.9


Difference = μ (1) - μ (2)
Estimate for difference:  31.00
95% lower bound for difference:  23.81
T-Test of difference = 0 (vs >): 
  T-Value = 7.13  P-Value = 0.000  DF = 207

Because the two sample standard deviations are about the same a pooled test (easier to compute) will give similar results. But because sample sizes differ, the pooled T statistic will not be exactly the same.
Here is Minitab output for the pooled test:
Two-Sample T-Test and CI 

Sample    N   Mean  StDev  SE Mean
1       110  217.0   34.0      3.2
2       100  186.0   29.0      2.9

Difference = μ (1) - μ (2)
Estimate for difference:  31.00
95% lower bound for difference:  23.76
T-Test of difference = 0 (vs >): 
  T-Value = 7.07  P-Value = 0.000  DF = 208
Both use Pooled StDev = 31.7186

Note: 'Both' above means the T test and the CI both use the pooled standard deviation shown.
