Solved – Shapiro-Wilk test and t-test

hypothesis testingnormal distribution

I have two populations with n=18 and I'm trying to find out if it makes sense to compare them with a t-test. I ran a Shapiro-Wilk test in SigmaPlot 12.5 for both populations seperately and these are the results:

population1:    W-Statistic = 0.900       P  = 0.057    Passed
population2:    W-Statistic = 0.912       P  = 0.094    Passed

However, if I'm trying to run a t-test, it says:

Normality Test (Shapiro-Wilk)   Failed  (P = 0.003)

Here it seems that there is only one P-value for both populations, which is a bit confusing to me. Does anyone have an idea how the P-value might have been calculated here and why it can be that low, even if it is much higher for both populations tested seperately?

This is the underlying data…

pop1:       pop2:
6.0696      6.4659
6.8833      6.2842
5.9243      5.9193
6.5391      7.526
7.2505      6.71
6.4299      7.3117
4.9903      13.5116
4.8506      9.1565
4.7737      7.7016
6.9384      8.5998
6.6842      9.2543
6.614       10.2234
6.3128      9.7079
6.3533      7.8677
6.2728      8.7079
7.4372      9.405
7.2657      8.5998
7.1165      8.3411

Best Answer

Imagine samples from two normally-distributed populations with vastly different means. Pooling the samples thus ignoring which group the samples were from would give you a bimodal and decidedly non-normal distribution.

Related Solutions

Solved – Can a two-sample t-test be used with data that doesn’t follow a normal distribution

One simple way to convince yourself that the CLT applies or does not apply is with some simulations.

Here is some R code:

testfun <- function(n1=19, n2=15) {
    x <- rexp(n1, 1/3)
    y <- rt(n1, 5) + 3
    t.test(x,y)$p.value
}

out <- replicate(10000, testfun(n1=19, n2=15))
hist(out)
abline(v=0.05, col='red')
mean( out <= 0.05 )

This code defines a function (testfun) that generates data from 2 different distributions (t with 5 df and exponential ) that have the same mean (3 in this case) and runs the built in t.test function and returns the p-value.

The replicate then runs this 10,000 times and we look at the results. The histogram should be close to uniform, but in this case we see an excess of values close to 0. The mean function calculates the type I error rate (since the null is true in the simulations), for my run this was a little of 7% when it should be 5%. Is that far enough to cause you concern? or are you happy with that as a "close enough" approximation?

Of course you should probably run this generating data from distributions that are more reasonable for your study, it may be that for something less skewed than the exponential that the differences would be small enough to not worry about.

Solved – Reporting Shapiro-Wilk and Levene’s test results

I think it's fine to say "data were tested for normality and homogeneity of variance" if that's what you did; it will be understood.

While it's important that people understand that you cannot confirm that you have either normality or equality of variance, making sure they know this is not the purpose of what you're doing - you're not teaching people statistics and you're not responsible for making sure they understand that hypothesis tests can't confirm that the distributions that the data were drawn from are normal or have the same variance.

should not have included it in the question. I have been asked to include a test (such as Shapiro-Wilk) instead of using QQ-plots, but my sample size is relatively small.

In the same situation if I could not convince the editor of the pointlessness of the test, I would probably be inclined to do one and report it at face value but I would not be able to resist adding a sentence pointing out how pointless it is to do so with a small sample (or indeed, with a large one, for the converse reason).

Best Answer

Related Solutions

Solved – Can a two-sample t-test be used with data that doesn’t follow a normal distribution

Solved – Reporting Shapiro-Wilk and Levene’s test results

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