Answer to Question #215513 in Statistics and Probability for Gayathri Devendran

Question

Out of 8000 graduates in a town; 800 are females and out of 1600 graduates employees 120 are females use X2 to determine if any destination made in appoinment in the basic sex. Ch square value table 3.84

Answer 1

$$\begin{matrix} \text{Number of} & \text{Number of} & \text{Total number of} \\ \text{Female\ \ \ \ \ \ } & \text{male\ \ \ \ \ \ \ \ \ \ } & \text{employees\ \ \ \ \ \ \ \ \ } \\ \text{employees} & \text{employees} & \text{} \end{matrix}$$

$$\begin{matrix} 800 \ \ \ \ \ \ \ \ \ \ \ & 7200\ \ \ \ \ \ \ \ \ \ \ & 8000 \ \ \ \ \ \ \ \ \ \ \ \\ 120 \ \ \ \ \ \ \ \ \ \ \ & 1480\ \ \ \ \ \ \ \ \ \ \ & 1600 \ \ \ \ \ \ \ \ \ \ \ \\ 920 \ \ \ \ \ \ \ \ \ \ \ & 8680\ \ \ \ \ \ \ \ \ \ \ & 9600 \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}$$

We know that,

Observed value is represented by O which is given

and expected value is represented by E

So,

$\\\ \\\ \\\\ E_1=\dfrac{920\times8000}{9600}=766.7$

$E_2= \dfrac{8680\times8000}{9600}=7233.3\\$

$E_3= \dfrac{920\times1600}{9600}=153.3$

$\\$$\\\ E_4=\dfrac{8680\times1600}{9600}=1446.7$

now,

$$\begin{matrix} \text{Number of} & \text{Number of} & \text{Total number of} \\ \text{Female\ \ \ \ \ \ } & \text{male\ \ \ \ \ \ \ \ \ \ } & \text{employees\ \ \ \ \ \ \ \ \ } \\ \text{employees} & \text{employees} & \text{} \end{matrix}$$ $$\begin{matrix} 766.7 \ \ \ \ \ \ \ \ \ \ & 7233.3 \ \ \ \ \ \ \ \ \ \ \ & 8000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 153.3 \ \ \ \ \ \ \ \ \ \ & 1446.7\ \ \ \ \ \ \ \ \ \ \ & 1600 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 920 \ \ \ \ \ \ \ \ \ \ \ \ & 8680\ \ \ \ \ \ \ \ \ \ \ \ \ \ & 9600 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}$$

Finally the table will be :

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c :c} O & E & O-E & (O-E)^2 & \dfrac{(O-E)^2}{E} \\ \hline 800 & 766.7 & 33.3 & 1108.9 & 1.45 \\ 120 & 153.3 & -33.3 & 1108.9 & 7.23 \\ 7200 & 7233.3 & 33.3 & 1108.9 & 0.15 \\ 1480 & 1446.7 & 33.3 & 1108.9 & 0.77 \\ \hdashline & & & Total & 9.60 \end{array}$$

The degree of freedom $=(R-1)(C-1)=(2-1)(2-1)=1$

The critical value of $\chi^2$ at $\alpha=0.05$ for $1$ d.f.  3.84 (given in question)

Since the calculated value of $\chi^2$ that is $\chi_{cal}^2=$  $9.60$ is greater than the critical value of $\chi^2$ at $\alpha=0.05$ for $1$ d.f. that is  $(3.841),$ then the null hypothesis is rejected and the alternative hypothesis is accepted.

Hence we can say that there is a distinction is made in appointment on the basis of sex.