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Science Forum Index » Statistics - Math Forum » Normal Probability Plot... Z-Score Computation
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| HopefulProdigy |
 Posted: Wed Jan 24, 2007 2:19 pm |
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Guest
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Ok, so the deal is, my AP Stat class went back for a day to review normality assessments (We've just completed P and T hypothesis tests).
One of the assessment possibilities is to make a "Normal Probability Plot" of the data. I completely understand how it works, and how the x value is what the Z-score for that point would have been had the distribution been normal.
But no one knew how that score was calculated. So online I found:
***
The normal probability value z for the jth value (rank) in a variable with N observations is computed as:
z = invNorm [(3*j-1)/(3*N+1)]
***
(http://www.statsoft.com/textbook/glosn.html)
(invNorm being the inverse normal cumulative distribution function, converting probability into z)
This works amazingly accurately when compared to the z-scores of the normal probability plot.
The only thing is that I don't understand how [(3*j-1)/(3*N+1)] calculates the normal probability.
I see that it does in fact fairly accurately find it, but I don't understand why.
So the bottom line is, can anyone explain to me,
"How, and why, does [(3*j-1)/(3*N+1)] fairly accurately predict the normal probability of a data point in a data set?" |
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| HopefulProdigy |
| Posted: Wed Jan 24, 2007 3:36 pm |
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WOW!
That works PERFECTLY!!! (with c=.5)
Thank you so much, I seriously don't know how to thank you.
Wow. |
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| Jack Tomsky |
| Posted: Wed Jan 24, 2007 7:13 pm |
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Quote: WOW!
That works PERFECTLY!!! (with c=.5)
Thank you so much, I seriously don't know how to
thank you.
Wow.
Try c=1/3.
Jack |
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| Ray Koopman |
| Posted: Wed Jan 24, 2007 8:59 pm |
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On Jan 24, 4:19 pm, HopefulProdigy <i...@ipoo.org> wrote:
Quote: Ok, so the deal is, my AP Stat class went back for a day to review normality assessments (We've just completed P and T hypothesis tests).
One of the assessment possibilities is to make a "Normal Probability Plot" of the data. I completely understand how it works, and how the x value is what the Z-score for that point would have been had the distribution been normal.
But no one knew how that score was calculated. So online I found:
***
The normal probability value z for the jth value (rank) in a variable with N observations is computed as:
z = invNorm [(3*j-1)/(3*N+1)]
***
(http://www.statsoft.com/textbook/glosn.html)
(invNorm being the inverse normal cumulative distribution function, converting probability into z)
This works amazingly accurately when compared to the z-scores of the normal probability plot.
The only thing is that I don't understand how [(3*j-1)/(3*N+1)] calculates the normal probability.
I see that it does in fact fairly accurately find it, but I don't understand why.
So the bottom line is, can anyone explain to me,
"How, and why, does [(3*j-1)/(3*N+1)] fairly accurately predict the normal probability of a data point in a data set?"
The general problem is to translate the observed ordinal position
j=1,...,N into a proportion p in such a way that j and N+1-j get
complementary proportions, and neither p=0 nor p=1 is produced.
The general formula in common use is p = (j-c)/(N+1-2c),
where c is a small constant whose value can be argued about.
Try a few different values of c. How much does it seem to matter? |
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| Old Mac User |
| Posted: Thu Jan 25, 2007 11:15 am |
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It's a delightful approximation that began with a very simple version
and has been improved over the years. Some tinkering here... adding a
little twist there... and behold. OMU
On Jan 24, 7:19 pm, HopefulProdigy <i...@ipoo.org> wrote:
Quote: Ok, so the deal is, my AP Stat class went back for a day to review normality assessments (We've just completed P and T hypothesis tests).
One of the assessment possibilities is to make a "Normal Probability Plot" of the data. I completely understand how it works, and how the x value is what the Z-score for that point would have been had the distribution been normal.
But no one knew how that score was calculated. So online I found:
***
The normal probability value z for the jth value (rank) in a variable with N observations is computed as:
z = invNorm [(3*j-1)/(3*N+1)]
***
(http://www.statsoft.com/textbook/glosn.html)
(invNorm being the inverse normal cumulative distribution function, converting probability into z)
This works amazingly accurately when compared to the z-scores of the normal probability plot.
The only thing is that I don't understand how [(3*j-1)/(3*N+1)] calculates the normal probability.
I see that it does in fact fairly accurately find it, but I don't understand why.
So the bottom line is, can anyone explain to me,
"How, and why, does [(3*j-1)/(3*N+1)] fairly accurately predict the normal probability of a data point in a data set?" |
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