Shapiro-Wilks test and the order statistic

In summary, the conversation discusses the interpretation of the order statistic in the context of the Shapiro-Wilk test. The participants confirm that if the data is sorted in ascending order, then the ith order statistic is equal to the ith element in the sequence. However, the Wiki page mentions that this may not always be the case. The conversation also mentions the convenience of finding the special case where the order statistic is equal to the element when computing the test.
  • #1
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TL;DR Summary
How to interpret the order statistic in the context of SW
Given the Shapiro-Wilk test value W:
1654873563116.png

where I'm interested in the numerator. If my data is sorted in ascending order, my understanding is that $x_(i) = x_i$. Is that correct?
 
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  • #2
Mayhem said:
Summary: How to interpret the order statistic in the context of SW

Given the Shapiro-Wilk test value W:
View attachment 302652
where I'm interested in the numerator. If my data is sorted in ascending order, my understanding is that $x_(i) = x_i$. Is that correct?
Sounds right to me. From the wiki page, https://en.wikipedia.org/wiki/Shapiro–Wilk_test:
##x_{(i)}## (with parentheses enclosing the subscript index i; not to be confused with ##x_i##) is the ith order statistic, i.e., the ith-smallest number in the sample
 
  • #5
Orodruin said:
The Wiki page seems to be saying explicitly that in general ##x_{(i)} \neq x_i##.
But if the elements of the sequence are already ordered from smallest to largest, then ##x_{(i)} = x_i##.
 
  • #6
Fair enough.
 
  • #7
Orodruin said:
The Wiki page seems to be saying explicitly that in general ##x_{(i)} \neq x_i##.
Yes, but for computation, finding the special case where ##x_{(i)} = x_i## makes life easier, which I did, and I got the right results when debugging random samples against known calculators.
 

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