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Hypothetically, I am trying to make a graph representing the abundance of 4 species in a quadrat. I want to display this as a % of the total number each species represents. Here is an example dataset:

             Rep 1    Rep 2    Rep 3   Total 
Species 1     10        32      21       63
Species 2      2        19      10       31
Species 3     27        28      20       75
Species 4     21         8      19       48
                                         217

Below I have worked out the table with the values displayed as percentages. I have then made a 'mean' percentage, which is made from the rep 1-3 % for each species.

            Rep 1   Rep 1 As %  Rep 2   Rep 2 As %  Rep 3   Rep 3 As %     Mean of all %'s      
Species 1     10      16.67%     32       36.78%      21       30.00%         29.03%
Species 2      2       3.33%     19       21.84%      10       14.29%         14.29%
Species 3     27      45.00%     28       32.18%      20       28.57          34.56%
Species 4     21      35.00%      8        9.20%      19       27.14%         22.12%
Totals:       60     100.00%     87      100.00%      70       100.00%        100.00%

I'd like to display this as a graph to show the average % that this species was viewed in each rep. Therefore I need SD or SE to show how much the % values varied around the mean. If I just calculate SD using the same values as I did for the mean will this work? I'm just not certain as it is a % I'm not sure if it would be changing what the SD is technically showing?

EDIT: Major edit to try and clear it up.

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    $\begingroup$ What do the "Rep" columns represent? Of what do you want to compute the standard errors? How were the data collected? $\endgroup$
    – whuber ♦
    Sep 1, 2017 at 21:31
  • $\begingroup$ Adding to @whuber's questions: Why are you combining the reps (species is cleary not the same across reps)? After you've told us what standard errors you want to compute, tell us why you want to include them in the graph. $\endgroup$
    – Peter Flom
    Sep 1, 2017 at 21:50
  • $\begingroup$ This is a hypothetical example to make a much larger dataset simpler. Each species number represents one single species across all reps. So the SD for example, of row 'Species 1' would be the variation in number of times Species 1 was observed in a given space and time. Aaargh, didn't realise enter submitted the comment. The reps would be for example, a photoquadrat. $\endgroup$
    – user2443444
    Sep 1, 2017 at 21:53
  • $\begingroup$ I have edited the question heavily to try and better represent my question. $\endgroup$
    – user2443444
    Sep 1, 2017 at 22:21
  • $\begingroup$ How did you obtain 16.67% in the first row, second column? $\endgroup$
    – user603
    Sep 2, 2017 at 7:50

1 Answer 1

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I think I have answered my own question as I was just getting confused. If I am still wrong please let me know...

        Rep 1   Rep 1 As %  Rep 2   Rep 2 As %  Rep 3   Rep 3 As %     Mean of all %'s      
Species 1     10      16.67%     32       36.78%      21       30.00%         29.03%
Species 2      2       3.33%     19       21.84%      10       14.29%         14.29%
Species 3     27      45.00%     28       32.18%      20       28.57          34.56%
Species 4     21      35.00%      8        9.20%      19       27.14%         22.12%
Totals:       60     100.00%     87      100.00%      70       100.00%        100.00%

I have converted into a % for each rep by dividing each value by total number of observations for that rep. If I calculate SD by using the three new values (in the example of Species 1 - 0.1667, 0.3678, and 0.3000) then this SD will still be valid as all the data has been standardised and is still proportionally related.

I think I was getting confused as I was unsure if converting to a % changed what the data represented, but in reality I have just divided all the numbers in the same way to a decimal, keeping them proportional and meaning SD calculations are done the same as always.

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  • $\begingroup$ I think the way you calculated the final column is correct, but it is probably labeled incorrectly. It is not the mean of the percentages, but the overall proportion, or you could call it the weighted mean. $\endgroup$
    – Sal Mangiafico
    Sep 2, 2017 at 16:25
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    $\begingroup$ I think calculating the standard deviation of the percentages is okay, but I'm not sure there's much sense in reporting a standard deviation for three values. If your goal is to visually present the variation across the Reps, you could plot the results of each of the three (as a proportion), say with a circle, and then add a line or a cross to indicate the overall proportion. (Cont...) $\endgroup$
    – Sal Mangiafico
    Sep 2, 2017 at 16:28
  • $\begingroup$ (... cont) On the other hand, if the Reps aren't very meaningful, you might present just the overall proportion, maybe with no error bars. Another idea, if the Reps aren't very meaningful, is to present just the overall proportion but use a confidence interval for the proportion for the error bars. Be sure to use a method for confidence intervals that is appropriate for proportions. $\endgroup$
    – Sal Mangiafico
    Sep 2, 2017 at 16:28

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