"+90+ 92)=80"
"\\bar{x}_2=\\dfrac{11}{9}(64+ 66 +74+ 78 +82 +85 +87"
"+ 92+ 93+ 95+ 97)=83"
"s_1^2=\\dfrac{1}{9-1}((66-80)^2+(67-80)^2+(75-80)^2"
"+(76-80)^2 +(82-80)^2+(84-80)^2"
"+(88-80)^2 +(90-80)^2 +(92-80)^2"
"=\\dfrac{367}{4}=91.75"
"+(78-83)^2 +(82-83)^2+(85-83)^2"
"+(87-83)^2 +(92-83)^2 +(93-83)^2"
"=\\dfrac{649}{5}=129.8"
"F=\\dfrac{s_2^2}{s_1^2}=\\dfrac{129.8}{91.75}\\approx1.4147"
For "\\nu_1=8" and "\\nu_2=10, F_{0.05}=3.36"
The calculated value of "F" is less than the tabulated value. So the null hypothesis is accepted.
Hence it may be concluded that two populations have the same variance.
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