Test of difference of two variences based on samples variances

Assuming equal variance in populations is important if we want to use simple formula for the difference of the means based on pooled variance  instead of long formula of degree of freedom and diffrent variances.

http://demosophy.org/estimators-for-the-difference-of-the-means-of-two-populations/

to check the assumption of equality we consider:

H0:     \sigma_1^{2}= \sigma _2^{2}

H1:      \sigma_1^{2}\neq \sigma _2^{2}

according to Cochran’s theorem s2 follows a scaled chi-square distribution:

(n-1)\frac{s^2}{\sigma^2}\sim\chi^2_{n-1}.
this means
{s}^{2}=\frac{{{\chi }^{2}}_{n-1}*{\sigma}^{2} }{n-1}
therefore”
\frac{{s_1}^{2}}{{s_2}^{2}}=\frac{{}\frac{{{\chi }^{2}}_{m-1}*{\sigma_1}^{2} }{m-1}}{\frac{{{\chi }^{2}}_{n-1}*{\sigma_2}^{2} }{n-1}}
if null hypothesis is true
\frac{{s_1}^{2}}{{s_2}^{2}}=\frac{{}\frac{{{\chi }^{2}}_{m-1} }{m-1}}{\frac{{{\chi }^{2}}_{n-1} }{n-1}}
and  we know that  if chi_m^2 and chi_n^2 be independent variates distributed as chi-squared with m and n degrees of freedom.We Define a statistic F_(n,m) as the ratio of the dispersions of the two distributions F_(n,m)=(chi_n^2/n)/(chi_m^2/m).

this means that
\frac{{s_1}^{2}}{{s_2}^{2}}
has a F_{m-1,n-1} distribution.

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