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I have the following exercise that shows $n=6$ numbers:

$$ 1.40, 1.55, 1.35, 1.50, 1.29, 1.64 $$

Is data normally distributed at the 5% significance level?

Surely $\overline{x} = 1.455$, $s=0.1315674732$ and $n_1 = \cdots = n_6 = 1$ but how can i continue?

Doria
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    There are many tests. Search for some of your key words, such as [normal goodness of fit](https://stats.stackexchange.com/search?q=normal*+goodness+of+fit). See, in particular, the comment thread at https://stats.stackexchange.com/questions/353230. – whuber May 18 '19 at 15:09

1 Answers1

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It is difficult to judge normality with only $n = 6$ observations. Many small samples, regardless of origin, will "pass" popular normality tests unless there happens to be marked skewness or far outliers.

A Shapiro-Wilk test does not reject normality for your data. In R:

x = c(1.40,1.55,1.35,1.50,1.29,1.64)
shapiro.test(x)

        Shapiro-Wilk normality test
data:  x
W = 0.97174, p-value = 0.9039

 boxplot(x, horizontal=T, col="skyblue2")

enter image description here

This failure to reject normality is hardly a proof of normality. Below we see that samples of size six from uniform and exponential populations often "pass" tests of normality (with P-values above 5%).

set.seed(2019)
pv.u = replicate(10^5,shapiro.test(runif(6))$p.value)
mean(pv.u > .05)
## 0.93769
pv.e = replicate(10^5,shapiro.test(rexp(6))$p.value)
mean(pv.e > .05)
[1] 0.78049
BruceET
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