Question

the average number of articles produced by two machines per day are 200 and 250 with standard deviation of 20 and 25 respectively on the basis of 25 days production. Can you regard both the machines equally efficient at 1% level of significance?

Answer 1

We have

$$\overline{x_s}=200, \overline{y_s}=250, \overline{\sigma_{x_s}}=20, \overline{\sigma_{y_s}}=25,\\ n=m=25, \alpha=0.01.\\ H_0: \overline{x_p}=\overline{y_p}\\ H_1: \overline{x_p}\neq\overline{y_p}\\$$

Assume that variances of the populations are equal.

We will use the following criterion:

$T=\frac{\overline{X}-\overline{Y}}{\sqrt{(n-1)S_x^2+(m-1)S_y^2}}\sqrt{\frac{nm(n+m-2)}{n+m}}$

$\text{where } \overline{X} \text{ and } \overline{Y} \text{ are random values of sample means},\\ S_x^2 \text{ and } S_y^2 \text{ are corrected sample variances respectively}.\\ \text{This random value has Student's t-distribution with } k=n+m-2 \text{ degrees of freedom}.$

$k=n+m-2=25+25-2=48\\ t_{cr}=t_{cr}(\alpha;k)=t_{cr}(0.01;48)=2.68 \text{ (two-sided critical value)}.\\ t_{ob}=\frac{\overline{x_s}-\overline{y_s}}{\sqrt{(n-1)\sigma_x^2+(m-1)\sigma_y^2}}\sqrt{\frac{nm(n+m-2)}{n+m}}=\\ \frac{200-250}{\sqrt{24(20)^2+24(25^2)^2}}\sqrt{\frac{(25^2)48}{50}}=-7.80\text{ (observed value)}\\ t_{ob}<-t_{cr} \text{ (observed value is in the critical region)}$

The null hypothesis is rejected in favor of the alternative hypothesis.

The machines are not equally efficient at 1% level of significance.

Now we will prove that variances of the populations are equal.

$H_0: \sigma_{p_x}^2=\sigma_{p_y}^2\\ H_1: \sigma_{p_x}^2<\sigma_{p_y}^2$

We will use the following criterion:

$F=\frac{S_b^2}{S_l^2}\\ \text{ where } S_b^2 \text{ is bigger corrected variance} \text{ and } S_l^2 \text{ is fewer corrected variance}.$

$\text{This random value has F-distribution with } k_1=m-1, k_2=n-1\text{ degrees of freedom}.\\ F_{cr}=F_{cr}(\alpha;k_1;k_2)=F_{cr}(0.01;24;24)=2.6591.\\ F_{ob}=\frac{25^2}{20^2}=1.5625.\\ F_{ob}<F_{cr} \text{ and } H_0 \text{ is true}.$