Question

A computer engineer identifies four ways that a certain job can be done. To determine how long it

takes operators to do the job when each of these methods is used, the engineer asks four operators

to do the job using the method A, another four operators to do the job using method B, and so on.

Each operator’s time (in seconds) is shown below:

A B C D

19 18 21 22

17 16 20 23

22 15 19 21

20 14 19 20

Construct the relevant analysis of variance table and test the hypothesis that the average time of all

operators are equal at 1% level of significance. (Given that F 0.01(3, 12) = 5.95)

Answer 1

Let

$H_o$ : The average time of all operators are equal at 1% level of significance.

$H_a$ :The average time of all operators are not equal at 1% level of significance.

The variance table is given by-

The combines sample size $n=n_1+n_2+n_3+n_4$

$=4+4+4+4\\=16$

The number of independent groups $K=4$

The mean of the combined sample-

$\bar{x}=\dfrac{x_1+x_2+x_3+x_4}{4}$

$=\dfrac{19.5+15.75+19.75+21.5}{4}$

$=\dfrac{76.5}{4}=19.125$

The Mean square for treatment(MST)-

$=\dfrac{n_1(x_1-\bar{x})^2+n_2(x_2-\bar{x_2})^2+n_3(x_3-\bar{x_3})^2+n_4(x_4-\bar{x_4})^2}{k-1}$

$=\dfrac{4(19.5-19.125)^2+4(15.75-19.125)^2+4(19.75-19.125^2+4(21.5-19.125)^2}{4-1}$

$=\dfrac{4(0.1404+11.388+0.3908+0.104)}{3}$

$=18.078$

The Mean square for Error(MSE)-

$=\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2+(n_3-1)s_3^2+(n_4-1)s_4^2}{n-K}$

$=\dfrac{(4-1)(0.0351)+(4-1)(2.847)+(4-1)(0.0976)+(4-1)(0.410)}{16-4}$

$=\dfrac{3(0.0351)+3(2.847)+3(0.0976)+3(0.410)}{12}$

$=0.8744$

Then, $F=\dfrac{MST}{MSE}$

$=\dfrac{18.078}{0.8744}=20.066$

For Annova test the degree of freedom $df_1=n_1-1=4-1=3$ and $df_2=n-K=16-4=12$

The Tabulated value of F at degree of freedome 3 and 12 at 0.01 level of significance is 5.95

Conclusion: As The calculated value of F is greater than the tabulated F at 1% level of significance.

So H_o is rejected. This means that the average time of all operators are not equal at 1% level of significance.