Statistics Standard Deviation

[Pekis]

Hello,

At the end of page 53 of the HP 15C Collector's Edition owner's handbook, there is this note:

When your data constitutes not just a sample of a population but all of the population,
the standard deviation of the data is the true population standard deviation (denoted σ).
The formula for the true population standard deviation differs by a factor of sqrt((n − 1) / n)
from the formula used for the s function. The difference between the values is small
for large n, and for most applications can be ignored. But if you want to calculate the
exact value of the population standard deviation for an entire population, you can easily
do so: simply add, using Sigma+, the mean (x dash) of the data to the data before pressing
s. The result will be the population standard deviation. (If you subsequently
correct any of your accumulated data values, remember to delete the first mean value
and add the corrected one.)

The proof of this trick is easy math (I checked it ) but I never had heard of it. And you ?

Thanks

[ttw]

I had never heard of it before. It's easy, adding the mean doesn't change the mean. It does change N so re-creates the population mean. I just scale by N/(N+1) as don't remember having only the data available for the proposed computation.

[Matt Agajanian]

Hi.

Although fully described, I remember reading this technique of adding x̄, ȳ to calculate σx and σy in HP-67's manual section on statistical functions. Thanks for the 15C CE description.

(07-28-2023 02:07 PM)Pekis Wrote:  Hello,

At the end of page 53 of the HP 15C Collector's Edition owner's handbook, there is this note:

When your data constitutes not just a sample of a population but all of the population,
the standard deviation of the data is the true population standard deviation (denoted σ).
The formula for the true population standard deviation differs by a factor of sqrt((n − 1) / n)
from the formula used for the s function. The difference between the values is small
for large n, and for most applications can be ignored. But if you want to calculate the
exact value of the population standard deviation for an entire population, you can easily
do so: simply add, using Sigma+, the mean (x dash) of the data to the data before pressing
s. The result will be the population standard deviation. (If you subsequently
correct any of your accumulated data values, remember to delete the first mean value
and add the corrected one.)

The proof of this trick is easy math (I checked it ) but I never had heard of it. And you ?

Thanks

[Pekis]

The proof:
Knowing that:

\(\sigma^{2}x=\frac{M}{n^{2}}\)

\(s^{2}x=\frac{M}{n(n-1)}\)

where \(M=nb-a^{2}\)
\(a=\sum_{}^{}x\)
\(b=\sum_{}^{}x^{2}\)

If we add \(\bar{x}\) to the dataset, then:

\(n^{'}=n+1\)

\(a'=a+\bar{x}=a+\frac{a}{n}=a\frac{n+1}{n}\)

\(b'=b+\bar{x}^{2}=b+\frac{a^{2}}{n^{2}}\)

\(s^{2^{'}}x=\frac{M'}{n'(n'-1)}\)

As for M':

\(M'=n'b'-a'^{2}=(n+1)(b+\frac{a^{2}}{n^{2}})-a^{2}\frac{(n+1)^{2}}{n^{2}}\)

\(=\frac{n+1}{n^{2}}(bn^{2}+a^{2}-a^{2}(n+1))=\frac{n+1}{n^{2}}(bn^{2}-a^{2}n)=n\frac{n+1}{n^{2}}(bn-a^{2})=\frac{n+1}{n}M\)

Then

\(s^{2^{'}}x=\frac{\frac{n+1}{n}M}{n'(n'-1)}=\frac{\frac{n+1}{n}M}{(n+1)n}=\frac{M}{n^{2}}=\sigma^{2}x\) QED