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

Marks: 6M

Year: May 2015

$n_1=10\ and\ n_2=13\ ($ $\lt$30 so it is a small sample )

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 values of two samples respectively. $$\overline{X_1\ }=\ \frac{\sum{x_1}}{n_1}=\ 15$$ $$\overline{X_2\ }=\ \frac{\sum{x_2}}{n}=\ 14$$ $$S^2_1=\ \frac{{\sum{(X_1-\overline{X_1\ })}}^2}{n_1}$$ $$\mathrm{\therefore }n_1\ .\ S^2_1=\ {\sum{\left(X_1-\overline{X_1\ }\right)}}^2=90$$ $$S^2_2=\ \frac{{\sum{(X_2-\overline{X_2\ })}}^2}{n_2}$$ $$\mathrm{\therefore }n_2\ .\ S^2_2=\ {\sum{\left(X_2-\overline{X_2\ }\right)}}^2=10$$ $${\sigma }^2_1 = \frac{n_1\ .\ S^2_1}{n_1-1} = \frac{90}{10-1} = 10$$ And $${\sigma }^2_2 = \frac{n_2\ .\ S^2_2}{n_2-1} = \frac{108}{13-1} = 9$$ $$\mathrm{\therefore } F_{cal}=\ \frac{{\sigma }^2_1}{{\sigma }^2_2}=\ \frac{10}{9}=1.1111$$

Step 3 :

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

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

And $V_2=\ n_2-1=13-1=12$

Value of F at (9,12) = $F_{\left(9,12,0.025\right)}$ = 3.07

And Value of F at (12,9) = $F_{\left(12,9,0.025\right)}$ = 3.80

$\mathrm{\therefore } \frac{1}{F_{\left(9,12,0.025\right)}} = \frac{1}{3.07}=0.3257$

Step 4 :

Decision

Since $\frac{1}{F_{\left(12,9,0.025\right)}}\lt\ \ F_{cal}\lt\ F_{\left(9,12,0.025\right)}$

$H_0\ $ is accepted

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