# Comparing means, having a large variance

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

• 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. Commented Jun 21, 2019 at 21:37
• 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. Commented Jun 21, 2019 at 22:41

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.