I am using Origin to run a two sample t-test for conductance for two different ionic solutions. One ionic solution has a sample size of 69, an average of 20 µS and a standard deviation of 0.7 µS. The other ionic solution has a sample size of 21, an average of 22 µS and a standard deviation of 0.7 µS.

When I run the two sample t-test, it gives me a p-value of 9E-12. This p value seems too small; might it be because the data I'm using is less than 1? Is there another way to calculate p values in this case?

  • 4
    $\begingroup$ You have an enormous effect size, so such small p-value is quite appropriate. $\endgroup$ – Andrey Kolyadin Dec 1 '16 at 16:06
  • 3
    $\begingroup$ To underline the implications of the answers: Values being less than 1 is completely irrelevant. For example, try multiplying by 1000 to get data in nS. The t-test result will be unaffected, as numerator and denominator are changed by the same factor. Conversely, if t-tests depended on the units you are using, they would fail completely as general purpose tests (and surely all good textbooks and programs would have to warn you about that). $\endgroup$ – Nick Cox Dec 1 '16 at 17:52

Remember that t-tests are based on standard errors of the means, which are smaller than the standard deviations; in your case, with 69 and 21 observations in the groups, standard errors are a good deal smaller. Standard deviations represent the spread of the individual observations. Standard errors represent the confidence you have in the mean value calculated over the number of observations, and are the standard deviations divided by the square root of the number of observations.

So your standard error for the 20 µS solution is less than 0.1 µS, and that for the 22 µS solution is less than 0.2 µS. With such strong confidence in the mean values, there is essentially no way that the 2 mean values could be the same. I haven't repeated the calculation, but your very small p-value doesn't seem surprising, as @AndreyKolyadin pointed out in a comment.

| cite | improve this answer | |

What @EdM said. If it helps, I've repeated the calculation in Minitab

Two-Sample T-Test and CI 

Sample   N    Mean  StDev  SE Mean
1       69  20.000  0.700    0.084
2       21  22.000  0.700     0.15

Difference = μ (1) - μ (2)
Estimate for difference:  -2.000
95% CI for difference:  (-2.355, -1.645)
T-Test of difference = 0 (vs ≠): T-Value = -11.46  P-Value = 0.000  DF = 33
| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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