1
$\begingroup$

Can I say that the values are close enough to be normally distributed? The histogram does not look normally distributed at all, but the Q-Q plot is not so far away. My sample size is 30. The shapiro-Wilk test has a p-value = 0,044. The Kolmogorov-Smirnov test has a p-value= 0,2. Zskewness= 1,97 Zkurtosis = 0,47

histogram

Q-Q plot

edit:

27.8 
26.4 
25.6 
28.8 
26.1 
30.2 
31.6 
31.4 
27.9 
29.3 
31.1 
27.3 
26.3 
26.8 
35.1 
27.8 
26.5 
27.2 
31.4 
31.9 
28.2 
27.2 
29.4 
29.5 
30.2 
26.0 
30.7 
35.0 
29.3 
28.2
Improve this question
$\endgroup$
13
  • 1
    $\begingroup$ "Enough normal" to do what? $\endgroup$
    – Michael M
    Jul 23, 2016 at 11:11
  • 1
    $\begingroup$ Which parametric test? It matters! $\endgroup$
    – Glen_b
    Jul 23, 2016 at 12:21
  • 2
    $\begingroup$ 30 values in histogram, 24 on probability plot??? $\endgroup$
    – Nick Cox
    Jul 23, 2016 at 12:32
  • 1
    $\begingroup$ @Nick Very observant! I believe I see six overplotted points (near 27.3, 27.9, 28.1, 29.1, 30.5, and 31.9, interpolating roughly). If you look closely, (a) these point symbols are slightly darker than the others and (b) they have a little too much vertical space relative to at least one neighbor. This suggests the probability plot indeed is trying to display 30 points but (mistakenly, IMHO) plots ties at common values of "expected normal". (I therefore am very suspicious of the KS test results: if the SPSS implementation treats ties in this manner, it's not sufficiently powerful.) $\endgroup$
    – whuber ♦
    Jul 23, 2016 at 13:07
  • 2
    $\begingroup$ @Michael M I don't know about "generally" and I do recognise a common argument. Nevertheless I'd give contrary advice. What could be more prudent than to check before testing how far assumptions are satisfied in a set of data, and then decide what to do? The most important issue is to to make a good analysis of the immediate set of data, not worry about how often the strategy might go wrong long-term. Also, on your advice one would never use any but the weakest possible tests! "Spying on the data" is a loaded term. But certainly acknowledge which decisions were made in which order and why. $\endgroup$
    – Nick Cox
    Jul 23, 2016 at 14:07

2 Answers 2

Reset to default
1
$\begingroup$

Short answer: Yes, probably.

Longer answer: It depends on what you're trying to achieve with the test what your data looks like, how it was gathered, how much you know about its theoretical distribution, and so on and so on. At the very least, compare parametric tests with and without heteroskedasticity and outlier corrections to see if they significantly affects test results. Finally, remember that interpretation of parametric tests should consider the whole picture, not just p values.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thank you for your answer! Since the sample size is 30, can I just use the central limit theorem instead? I see that the cut off to use this is usually 30 or just above 30. $\endgroup$
    – user123867
    Jul 23, 2016 at 12:52
  • 1
    $\begingroup$ 30 is not a magic number! $\endgroup$
    – Nick Cox
    Jul 23, 2016 at 14:07
  • $\begingroup$ No, but is it big enough to use the central limit theorem? $\endgroup$
    – user123867
    Jul 23, 2016 at 14:29
  • 1
    $\begingroup$ Same question, same answer. It's not like freezing or melting. But the bigger deal is what you want to do with this and the rest of your data. Marginal normality is very rarely an assumption, but conditional normality sometimes is. To get a better answer, show us all your data (it seems that there isn't much). $\endgroup$
    – Nick Cox
    Jul 23, 2016 at 15:53
  • $\begingroup$ Thanks, but (1) presumably the commas are decimal points so that this is your main variable (2) an image doesn't allow easy copy and paste and I won't type in 30 numbers; (3) much the key is the other variables defining the rest of the t test and ANOVA you intend. $\endgroup$
    – Nick Cox
    Jul 23, 2016 at 17:16
1
$\begingroup$

You've helpfully posted 30 values in the first instance. They aren't named so I've called them $y$ and I'm presuming that they are the response or outcome you want to explain. We can look at its so-called marginal distribution, but the bigger deal for the $t$ tests and ANOVA you have in mind would be the conditional distributions of $y$ for each distinct value of any predictors or covariates.

A normal probability plot (normal quantile-quantile plot) should look more like this than the plot first posted (but now removed).

enter image description here

There are different conventions (1) for which variable goes on which axis and (2) on whether the expected values are presented using the observed mean and standard deviation (SD), or estimates thereof, or in standard normal terms (with mean 0 and SD 1).

Setting those small points aside, the main detail is to confirm your reports to the effect that the distribution is slightly skew. As the values cover a fairly small relative range (about 35/25) no (standard) transformation will make much difference to that skewness, as over a short interval such transformations are close to linear in the observed values. It is shown above that a logarithmic transformation reduces the skewness slightly; in practice I wouldn't bother unless there were independent scientific or substantive grounds for working on logarithmic scale.

It's an assertion based on experience that the degree of normality is too slight to worry about much and I would go ahead with $t$ tests and ANOVA. But I would rather have data for your other variables to be able to check that conclusions here are robust to minor non-normality.

EDIT: The fuller data are now revealed as two groups of 30 on the same outcome. Given yet more anonymity, I am calling the group identifier x with values of 0 and 1, although nothing depends on any such names or values.

A superficial analysis suggests that the groups are so close in their means (29.007 versus 29.843) that any kind of test is virtually redundant.

But the extra 30 values have a mild outlier that inclines me to using logarithmic scale to dampen its effects a little, and quantile plots hint at structure that is more than (not even!) a shift of distributions:

enter image description here

enter image description here

Collectively, group 1 is lower at lower values and high at higher values. The small difference in means arises because these effects almost cancel. (The pairing here, it must be emphasised, is quantile to quantile, not part of the observation design.)

All logarithms here are natural.

Mild protest: You posted 30 extra values and then deleted them! Why do that? No one can use them in answers if you do that. Here are all the data as I have them: 

x   y
0   27.8
0   26.4
0   25.6
0   28.8
0   26.1
0   30.2
0   31.6
0   31.4
0   27.9
0   29.3
0   31.1
0   27.3
0   26.3
0   26.8
0   35.1
0   27.8
0   26.5
0   27.2
0   31.4
0   31.9
0   28.2
0   27.2
0   29.4
0   29.5
0   30.2
0   26
0   30.7
0   35
0   29.3
0   28.2
1   25.8
1   27.5
1   25.2
1   27.2
1   28
1   27.3
1   29.8
1   28.7
1   28.7
1   33.4
1   32.8
1   33.8
1   30.8
1   32.2
1   33.7
1   25.5
1   25.4
1   28.8
1   26.3
1   26.8
1   32.1
1   27.9
1   30.1
1   31.6
1   29.5
1   33.6
1   31.8
1   29.1
1   32.4
1   39.5
Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ So are these paired responses? You should tell us more, and not leave the data and design cryptic! $\endgroup$
    – Nick Cox
    Jul 24, 2016 at 8:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.