Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: May 2014

$n_1=10\ and\ n_2=8\ (<30\ $ sample is small) Step 1 : Null hypothesis : $$(H_0)\ :{\sigma }^2_1=\ {\sigma }^2_2$$ (i.e. variance are not significantly different)

Alternative Hypothesis :$$\ {\sigma }^2_1\neq \ {\sigma }^2_2$$ (i.e. variance are significantly different)

Step 2 :

Test statistics:

Let $x_1$ and $x_2$ denote the increase in weights of rats fed on diets A and B respectively. $$\overline{X_1\ }=\ \frac{\sum{x_1}}{n_1}=\ \frac{64}{10}=6.4\ $$ $$\overline{X_2\ }=\ \frac{\sum{x_2}}{n}=\ \frac{40}{8}=5\ $$

$$S^2_1=\ \frac{{\sum{(X_1-\overline{X_1\ })}}^2}{n_1}$$ $$\mathrm{\therefore }n_1\ .\ S^2_1=\ {\sum{\left(X_1-\overline{X_1\ }\right)}}^2=102.4$$ $$S^2_2=\ \frac{{\sum{(X_2-\overline{X_2\ })}}^2}{n_2}$$ $$\mathrm{\therefore }n_2\ .\ S^2_2=\ {\sum{\left(X_2-\overline{X_2\ }\right)}}^2=82$$ $${\sigma }^2_1 = \frac{n_1\ .\ S^2_1}{n_1-1} = \frac{102.4}{10-1} = 11.3778$$

And $${\sigma }^2_2 = \frac{n_2\ .\ S^2_2}{n_2-1} = \frac{82}{8-1} = 11.7143$$ $$\mathrm{\therefore } F_{cal}=\ \frac{{\sigma }^2_1}{{\sigma }^2_2}=\ \frac{11.7143}{11.3778}=1.0296$$

Step 3 :

L.O.S = 5% (Two failed test)

Degree of freedom , $V_1=\ n_1-1=10-1=9$

And $V_2=\ n_2-1=8-1=7$

Value of F at (9,7) = $F_{\left(9,7,0.025\right)}$ = 3.6767

And Value of F at (7,9) = $F_{\left(7,9,0.025\right)}$ = 3.2927 $$\mathrm{\therefore } \frac{1}{F_{\left(9,7,0.025\right)}} = \frac{1}{3.6767}=0.2728$$

Step 4 :

Decision

Since $\frac{1}{F_{\left(9,7,0.025\right)}}\lt\ \ F_{cal}\lt\ F_{\left(7,9,0.025\right)}$

$H_0\ $ is accepted

$\boldsymbol{\mathrm{\therefore }}$ Variance are not significantly different at 5% L.O.S