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I have asked question concerning comparison between two groups. Got really nice answers and I want to perform downstream analysis using really good gui11aume's suggestion.
But not everything is clear to me, hence I need help to clarify significance test I should apply.
What I want: To compare two means of percentage and find out if both compared groups contain same percentage of analyte.
What I have: 

Group   Mean   Size
  A     0.1     5
  B     2.3    124

Question: How can I do it?

gui11aume suggested to compare two means using t-test just like in this example. However I found that for t-test one can also use full string (not only pre-calculated mean). So, if it's possible to calculate t-test statistics from what I have - how can I do it (in R)?
And what is the difference between such t-test and cbeleites answer using prop.test? They seem similar to me.

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2 Answers 2

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If these means are percentages, you don't need the full stream. You can do a t-test with just proportion and N. There are lots of online calculators for this, or you can use prop.test in R. With full stream data you can use t.test in R. Or, you can make a table of successes/failures and use chisq.test.

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    $\begingroup$ It's just an example, so I will probably use R. But still, what is the difference between t.test and prop.test? And just out of curiosity - if I provide only two means and N for t.test - doesn't it nead SD? $\endgroup$
    – pogibas
    Commented Jun 13, 2013 at 18:16
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    $\begingroup$ If the values are proportions, then the variances are derived directly from the means. Offhand, I am not sure of the difference (if there is one) between prop.test and t.test, except for the interface. You could, of course, examine the code. $\endgroup$
    – Peter Flom
    Commented Jun 13, 2013 at 19:39
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Your previous question related to counts. A proportions test would be relevant to that.

Now you're talking about "percentage of analyte". That doesn't sound like count data to me, but continuous compositional data. The difference is critical.

You should clarify which kind of data this is.

If those means are continuous proportions rather than counts divided by counts, then you shouldn't use proportions tests or chi-square or anything else designed for counts.

On the other hand, t-tests assume normality and constant variance; the first might not matter too much (though one sample is very small), and it's possible to adjust somewhat for the second. However, compositional data is usually modelled as beta (or Dirichlet, when there's more than one components of interest and more than two components in total).

Even worse, you seem to suggest you only have means and the total number of observations. Without standard deviations or the original data you can't compute a t-statistic.

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