Statistical test for difference in means using only group mean and sample size I have aggregated survey data for various groups, but do not have access to the raw responses. Is there an appropriate test for statistical significance between group means with only access to the means and the number of responses for each group? I found that a Z-test for proportions is similar, but only deals with binary categories. The survey data I have involves interval data 1-10.
I have included two groups within the data. The columns include all possible information I have access to, the "total people in group" column may not be useful. Survey respondents are also segmented into groups by job function, geography, and ethnicity; however, I believe this simple example will help me with the remaining groups.

 A: There is not a reliable test for difference in means between your two groups without stringent assumptions about variances or data varability.
Consider: if the variance in average response is, say, about 36 in both groups, then a t test would find no evidence of difference in means at the $\alpha=0.05$ level. However, if the variance in both groups is, say, about .036, then you would find evidence of a difference in means at the $\alpha = 0.05$ level.
Why is this? Because hypothesis tests (confidence intervals also) rely on estimates of the standard error, which depends on the estimated variance to calculate the test statistic (there's a simpler form if the variances in both groups can be assumed equal, but this works whether or not they are equal):
$$ t = \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\frac{s^{2}_1}{n_1} + \frac{s^2_{2}}{n_2}}}$$
The $s^2$s in the above equation is what you are missing, and why you cannot perform a test for group differences in means (or group inferiority/superiority of means, or group equivalence of means).
