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So I am asked to calculate this, but I can't find anything about it on the internet.

This is the question asked:

Find Sd (standard deviation of the differences)

Listed below are ages of actresses and actors at the time that they won Oscars for categories of Best Actress and Best Actor. This is paired sample data.

Actress: 22, 37, 28, 63, 32

Actor: 44, 41, 62, 52, 41

I need to use a formula, but I can't find the correct one. How do I calculate the standard deviation of the differences? What is the formula, can someone help me?

No sample standard deviation is given, so I am clueless right now.

What I've done already is: I calculated the differences of the two samples, which are:

-22, -4, -34, +11 and -9...

After that, I've subtracted the mean, which was -11.6... so eventually, you'll get:

-10.4, 7.6, -22,4, +22,6, +2,6...

When you square these numbers and add them up, you get the variance and you can take the square root of the variance. This should give me the sample standard deviation, but that gives me 11.4 instead of the 17.2 the answer model is suggesting. What am I doing wrong?

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  • $\begingroup$ Why on earth is this being downvoted? $\endgroup$
    – Siyah
    Commented Dec 15, 2016 at 22:01
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    $\begingroup$ You have presumably been down-voted because the answer is so readily available from a quick search of this site: stats.stackexchange.com/questions/29170/… $\endgroup$
    – Michael Lew
    Commented Dec 15, 2016 at 22:17
  • $\begingroup$ Are they paired so that actress aged 22 was the same year as actor 44? $\endgroup$
    – mdewey
    Commented Dec 15, 2016 at 22:17
  • 2
    $\begingroup$ If you want to discuss downvoting, you could raise a question on Meta. All I can do is comment generally. Public criteria are doesn't show any research effort; unclear or not useful. People should be using those criteria, but no one can tell if anyone is downvoting for other reasons unless they say so. I can sympathise with the downvotes, as, for example, I can't see how this could be useful to anyone else. The flavour is just: I got what appears to be the wrong result. Without more information, the reply is just that you must have made a mistake. Now, don' shoot the messenger. $\endgroup$
    – Nick Cox
    Commented Dec 15, 2016 at 22:34
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    $\begingroup$ In your edit above you are sometimes using stop as decimal separator and sometimes using comma. I can't tell whether that is trivial or a contributor to your problem. We can't check the rest of your calculation until you give it. $\endgroup$
    – Nick Cox
    Commented Dec 15, 2016 at 22:41

2 Answers 2

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Here is a step-by-step calculation in Stata's Mata. Slightly non-standard point is that :- is elementwise subtraction. Note that the calculation is using $n - 1$ to divide the sum of squared deviations for a sample of size $n$. 

: Actress = (22, 37, 28, 63, 32)

: Actor = (44, 41, 62, 52, 41)

: diff = Actress :- Actor

: diff
         1     2     3     4     5
    +-------------------------------+
  1 |  -22    -4   -34    11    -9  |
    +-------------------------------+


: diff :- mean(diff')
           1       2       3       4       5
    +-----------------------------------------+
  1 |  -10.4     7.6   -22.4    22.6     2.6  |
    +-----------------------------------------+

: (diff :- mean(diff')):^2
            1        2        3        4        5
    +----------------------------------------------+
  1 |  108.16    57.76   501.76   510.76     6.76  |
    +----------------------------------------------+

: sum((diff :- mean(diff')):^2)
  1185.2

: sum((diff :- mean(diff')):^2) / 4
  296.3

: sqrt(sum((diff :- mean(diff')):^2) / 4)
  17.2133669
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  • $\begingroup$ Exactly what I was looking for, thanks. I did make a calculation error. Glad to see where I made the mistake! $\endgroup$
    – Siyah
    Commented Dec 15, 2016 at 22:44
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What is the standard deviation of (-22, -4, -34, +11, -9)? Looks like 17.2 to me. Perhaps you have made a simple calculation error.

Why subtract the mean? That won't change the standard deviation at all.

(I have to say that I find the pairing in the example to be a bit puzzling. Yes, the data may well be paired by year of award, but the hypothesis that requires that pairing is not the obvious "Does the Academy choose younger women than men?", but a more obscure thing like "Does the Academy tend to choose women who are younger than the men that they choose in that particular year?". One shouldn't pair data just because one can find an obscure justification. The pairing should be sensible and should reduce the over all variance.)

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  • $\begingroup$ I agree with the last para. In fact my first reaction to the question was that the SD of the differences was meaningless unless it was based on all pairwise differences. Then the point about the same year being understood was made. The problem does seem contrived (and very poorly explained, on the evidence of the OP). But this is marginal to the thread. $\endgroup$
    – Nick Cox
    Commented Dec 15, 2016 at 22:44

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