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

Marks: 6M

Year: Dec 2015

$\beta$ ∋

Step 1 :

Null hypothesis : $(H_0) \ The\ die\ is\ unbiased$

Alternative Hypothesis ($H_A)$: The die is not unbiased.

Step 2 : Test Statistics

On the hypothesis that the die is unbiased we should expect the frequency of each number to be 132/6 =22 $$\mathrm{\therefore } {X_{cal}}^2 = \sum{\frac{{(O-E)}^2}{E}} = 198/22 = 8.91$$

Step 3:

L.O.S ($\propto )\ $= 0.05

Degree of freedom = n-1 = 6-1 =5

Step 4 :

$\mathrm{\therefore }$ Critical value ${\left(X_{\propto }\right)}^2=11.0705$

Step 5 : Decision

Since $X^2_{cal}\lt\ $ ${\left(X_{\propto }\right)}^2$

${\boldsymbol{H}}_{\boldsymbol{O}}$ is accepted.

$\boldsymbol{\mathrm{\therefore }}$ The die is unbiased.