We are comparing two sets of data for microbial growth. In one column I got 36, 23, 35, 30, and 34 In the second I got 29, 20, 30, 26, and 29. I am trying to compare them to see if they are statistically different. The way it was done before looks like they are doing a t test, but the way they did it was taking the difference of the logs of the numbers and then doing a mean of the differences and standard deviation mean. They then used this formula mean/(std deviation/(square root 4))=t to come up with a t value. I'm not very good with statistics but this gives me a t value of 8ish, which is above the 2.78 on the t table. I used excel to do a t test of the data without the logs and it's giving me numbers below 2.78, but now with playing with the numbers I'm not able to get any that fail, no matter how far apart they are. I guess what I'm asking is if the formula they used before me is an accurate way to do this. The numbers look using the way they do it if I had all of one sample get 50's, and the next set get all 49's and one 48, it still fails because the t value is 5ish and is greater than 2.78. That doesn't make sense to me because all the values are pretty much the same.

I hope I have given enough information. If you need any more let me know.

  • 1
    $\begingroup$ Forget about what was done in the past and report the result you got with your test. $\endgroup$
    – user78229
    Commented Nov 17, 2017 at 20:22

1 Answer 1


It's usually best to keep things simple. If we can assume your samples are independent, you can use a standard t-test. In R:

> x1 <- c(36, 23, 35, 30, 34)
> x2 <- c(29, 20, 30, 26, 29)
> t.test(x1,x2)

  Welch Two Sample t-test

  data:  x1 and x2
  t = 1.6, df = 7.5, p-value = 0.2
  alternative hypothesis: true difference in means is not equal to 0
  95 percent confidence interval:
   -2.199 11.799
  sample estimates:
  mean of x mean of y 
       31.6      26.8 

With a p-value of 0.2, the two samples are not significantly different (assumming alpha probability = 0.05). Alternatively, if the samples are related in some way, use a paired-sample t-test.

  • $\begingroup$ Thanks for your comment. I ended up going with doing an ANOVA test to compare the two results. Just took a little while to relearn how to do it and then prove that excel was doing it correctly. $\endgroup$
    – Jon
    Commented Nov 22, 2017 at 16:24
  • $\begingroup$ ANOVA returns the same p-value as a t-test, so your results should be the same. And with only two samples, there's no advantage of ANOVA over t-test. But if you are more comfortable with ANOVA, that's fine. $\endgroup$ Commented Nov 23, 2017 at 17:07
  • $\begingroup$ Ah...the person before me must have just had a totally wrong formula then because I was getting different results. They actually didn't have a p value at all and just had the test failed if t > 2.78. Of course they were also doing the differences between the values of the two tests and then doing the t calculation based off of those 5 numbers, so I'm not exactly sure what they were thinking. $\endgroup$
    – Jon
    Commented Nov 24, 2017 at 17:32

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.