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I have a series of data looking like this, and I wonder if this can be considered to close to normally distributed, even if it has a tail to the right? Can I use t-tests and define "large" being over 1 standard deviation and such, or would that be wrong?

Count    1 536
Mean     27,8
Median   26,0
Mode     28,0
Stdv     14,4
Skew      0,9
Max      92,0
Min       0,0

Here is a histogram of the data:

enter image description here

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    $\begingroup$ There isn't enough information in this numerical summary to conclude, with any certainty, the shape of the distribution. We need a histogram and/or QQ plot of the data to answer this. The fact that there is little skew (although a skew of 0.9 in a sample size of 536 may indicate some problems) and the mean/median/mode are about about equal is a good sign but doesn't guarantee anything. This could be a very long-tailed (but roughly symmetric) distribution for all we know. $\endgroup$
    – Macro
    Commented Oct 9, 2012 at 13:12
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    $\begingroup$ One indication of non-normality is that the mean and median, while close to each other, are far from the middle of the distribution which would be 46. One thing that could cause this is an outlier or two at the high end. 92 is $\frac{92-27.8}{14.4} = 4.46$ standard deviations from the mean, which is pretty extreme with n = 1500. $\endgroup$
    – Peter Flom
    Commented Oct 9, 2012 at 13:15
  • $\begingroup$ Im not permitted to post any images, so i post a plot at this adress: space.hgo.se/gis/statistics.jpg $\endgroup$
    – Gustaf
    Commented Oct 9, 2012 at 13:35
  • $\begingroup$ The trend line is an moving average of 5 $\endgroup$
    – Gustaf
    Commented Oct 9, 2012 at 13:39
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    $\begingroup$ I would say you do not want to use t-tests here, unless the right tail is of great interest to you. Variables such as size often benefit from a log transformation. $\endgroup$
    – Peter Flom
    Commented Oct 9, 2012 at 13:53

1 Answer 1

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I think that you are mixing together a couple of different concepts. You ask about normality, using t-tests, and deciding if something is large.

First, no, the data shown is not considered close to normal, but if you are interested in using t-tests then that is not the important question. The population and sample distributions are less important than the sampling distribution when determining if t-tests can be used. What determines the shape of the sampling distribution is a combination of the population distribution, how the sample is obtained, and the sample size. The beauty of the Central Limit Theorem (and it is beautiful enough that I have a cross-stitch of it on my wall) is that as the sample size gets large the sampling distribution becomes more like the normal distribution, so for a large sample size and a simple random sample (or samples similar to SRS) we can use the t-tests and the approximation will be very good. The important thing to be large here is the sample size, not the data or standard deviation. If you believe that the data shown above represent the population (which needs to come from your knowledge of the science behind the data) and with a sample size of 1536 the normal based inference (t-test and others) will be very good approximations.

If your goal is to determine which farms are "large", then the t-test proceedures do not answer that question. That will be more determined by the science behind the data and the question of interest and you will not be able to use the normal or t distributions for this data looking at decisions about individual farms.

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  • $\begingroup$ Tank you for your answer. I know I asked two questions that are not related. You gracefully answered the first, thanks. The second about large and small, is that I think dividing the data into stdv would be a good way in describing it. Since the data, apart from the tail to the right, look quit normally distributed, I though of defining all to the right of 1 stdv as large and all to the left as small. I will probably get roughly the right proportions that I want (68% in the middle and the rest on each side). Since I'm not any master in statistics, I'm afraid of violating any statistic rule $\endgroup$
    – Gustaf
    Commented Oct 9, 2012 at 19:10
  • $\begingroup$ Generally when splitting data we would look at the mean + sd, rather than just the sd. Consider an IQ test with mean 100 and sd 10, we might consider someone smart if their IQ is > 110 = 100 + 10, if we consider everyone > 10 as smart then we have a genius population. This may have been what you meant, but it was unclear. $\endgroup$
    – Greg Snow
    Commented Oct 9, 2012 at 19:25
  • $\begingroup$ Yes that's what I mean. the mean + 1 stdv. Is it OK to do so with my data? $\endgroup$
    – Gustaf
    Commented Oct 9, 2012 at 19:33

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