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

Marks: 8M

Year: May 2015

Step 1 :

Null hypothesis : $$(H_0)\ Accidents\ are\ uniformly\ distributed\ over\ the\ weeks.$$

Step 2 : Test Statistics

Total accidents = 84

Equally distributed expected accidents per day = 84/7 = 12 $${X_{cal}}^2 = \sum{\frac{{(O-E)}^2}{E}} =2.306$$

Step 3:

L.O.S = 5%

Degree of freedom = n-1 = 7-1 =6

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

Step 4 : Decision

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

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

$\boldsymbol{\mathrm{\therefore }}$ Accidents are uniformly distributed over the week.