# What paired test should I use and how do I interpret it?

The question: The data file contains observations of the wavelengths (in nanometers) of the main color in two kinds of shoe polish. (Sample size = 100). Are the wavelengths significantly different?

Now, I figured out I should use a two sided paired sample test. That would mean either Wilcox or a t-test. How do I figure out whether I can use the T-test, or should stick to a nonparametric test like the Wilcoxon signed ranks test? Should I test for a normal distribution? On a side note: how do I interpret the results of such tests? If the p-value is high, does that mean they are significantly different or the other way round?

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From your description, it's not clear to me why you want to use a paired samples test, it sounds like you should be using an independent samples test. – gung Nov 20 '12 at 21:53
Assuming your data are actually paired (are they?), the two tests you list are not the only possibilities. Low p-values suggest a difference. – Glen_b Nov 20 '12 at 23:16
Does your hypothesis relate directly to a mean or just some general sense of a location difference? Would you assume the shapes are the same apart from location, or not? – Glen_b Nov 20 '12 at 23:49

I do not see why this should be a paired t-test. There is no connection between the 100 pairs. Paired would be if you have 100 shoes and each shoe is tested with a different laser...ok, that's not a good example ;):

shoe | laser 1 | laser 2 | Delta
1      34 nm     38 nm      4 nm
2      33 nm     29 nm     -4 nm
...


But maybe I do not understand your description and it is a paired setting. But I am afraid that you have to clearify it in this case.

Can you specify what difference in nm (wavelength unit) is considered big enough? Using R you could use:

polish.1 = c(25,24,36,24,32)
polish.2 = c(22,36,27,28,30)
BigDifference = 2
t.test(
polish.1,
polish.2,
alternative = "two.sided",
paired = FALSE,
mu = BigDifference
)


    Welch Two Sample t-test

data:  polish.1 and polish.2
t = -0.7171, df = 7.951, p-value = 0.4938
alternative hypothesis: true difference in means is not equal to 2
95 percent confidence interval:
-8.125667  7.325667
sample estimates:
mean of x mean of y
28.2      28.6


Welch means that it is not assumed that the standard deviation of the two data sets is equal.

So the result in the example means that it is not likeley that the wavelength is different (p-value is about 50 %).

Of course, we can go deeper in the subject of effect size and so on. But I want to wait what you and the others reply/comment.

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