Answer to Question #155314 in Statistics and Probability for Rohith Kumar

Question

Out of 8000 graduates in a town 800 are females, out of 1600

graduate employees 120 are females. Use 2  at 5% level to

determine if any distinction is made in appointment on the basis

of sex:

Answer 1

$$\begin{matrix} \text{Number of} & \text{Number of} & \text{Total number of} \\ \text{Female\ \ \ \ \ \ } & \text{male\ \ \ \ \ \ \ \ \ \ } & \text{employees\ \ \ \ \ \ \ \ \ } \\ \text{employees} & \text{employees} & \text{} \end{matrix}$$ $$\begin{matrix} 800 \ \ \ \ \ \ \ \ \ \ \ & 7200\ \ \ \ \ \ \ \ \ \ \ & 8000 \ \ \ \ \ \ \ \ \ \ \ \\ 120 \ \ \ \ \ \ \ \ \ \ \ & 1480\ \ \ \ \ \ \ \ \ \ \ & 1600 \ \ \ \ \ \ \ \ \ \ \ \\ 920 \ \ \ \ \ \ \ \ \ \ \ & 8680\ \ \ \ \ \ \ \ \ \ \ & 9600 \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}$$

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

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

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

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

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

Observed value (O)

Expected value(E)

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

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. $=3.841$

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

Therefore there is a distinction is made in appointment on the basis of sex.