Currently I am working on a project in which i do have two different sample-sets with a limited size and a large variance. Furthermore, I do have a bigger data-set that gives information about the overall variance of the test. I would like to determine whether the two means are the same, taking into account the overall variance. I would like to get some advises how to solve such a problem. Please find a summary of one set of the data below.

Dataset 1: Mean=3.68, Variance=5.86 N=31

Dataset 2: Mean=4.34, Variance=6.98 N=16

Overall variance: 16.9

Thank you in advance for the help

  • $\begingroup$ You do not say whether you expect or hope that the two datasets show significantly different means. They don't---not using just the data in the two datasets, nor taking the variance from the larger study into account. See my Answer. $\endgroup$
    – BruceET
    Commented Jun 21, 2019 at 21:37
  • $\begingroup$ I always like to inspect a plot of the data points prior to determining the meaning of the summary statistics. Make a graph that show each observed value clustered by dataset and you will easily see why the t-test of BruceET gave a high P-value. $\endgroup$ Commented Jun 21, 2019 at 22:41

1 Answer 1


According to the Minitab output below, a Welch two-sample t test, based on current data (ignoring 'overall variance' from other sources), shows no significant difference at the 5% level between sample means, P-value $0.411 > 0.05.$

Two-Sample T-Test and CI 

Sample   N  Mean  StDev  SE Mean
1       31  3.68   2.42     0.43
2       16  4.34   2.64     0.66

Difference = μ (1) - μ (2)
Estimate for difference:  -0.660
95% CI for difference:  (-2.279, 0.959)
T-Test of difference = 0 (vs ≠): 
    T-Value = -0.84  P-Value = 0.411  DF = 28

A pooled two-sample t test, which may be valid here because sample variances seem similar, also gives P-value about 0.4.

Introducing a larger estimate of population variances from an allegedly similar situation would only increase the P-value. Using the standard deviation $\sqrt{16.9} = 4.11$ for both groups (in what is essentially a two-sample z test) gives P-value 0.60, far from significant at any reasonable level.

Not knowing details of the 'overall' variance, I would be reluctant to use it. For example, it could be a mixture distribution of populations with different means and variances, and hence possibly irrelevant to your experiment.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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