Question

1.A tea company appoints four salesman A.B.C.D and observes their sales in three seasons Summer, Winter and Monsoon. The figures(in lakhs) are given in the following table:

Seasons. A B C D

Summer. 45. 40. 38. 37

Winter. 43. 41. 45. 38

Monsoon. 39. 39. 41. 41

Carry out an analysis of variance.

Answer 1

Seasons A B C D Totals

Summer. 45 40 38 37 160

Winter. 43 41 45 38 167

Monsoon 39 39 41 41 160

Totals 127 120 124 116 487

There are $4$ salesmen and $3$ seasons. Therefore, $n=3*4=12.$

The correction factor, $c.f={487^2\over 12}=19764.083333$

Total sum of squares, $SST=\sum\sum y_{ij}^2-c.f=19841-19764.083333=76.91667$

Between salesmen sum of squares, $R_1={(127^2+120^2+124^2+116^2)\over3}-c.f=19787-19764.08333=22.916667$

Between seasons sum of squares,$R_2={(160^2+167^2+160^2)\over4}-c.f=19771-19764.08333=8.166667$

Error sum of squares, $ESS=SST-R_1-R_2=76.91667-22.916667-8.166667=45.833333$

The mean sum of squares for seasons is ${8.166667\over 2}=4.0835$

The mean sum of squares for salesmen is ${22.916667\over 3}=7.639$

The mean error sum of squares is ${45.83333\over 6}=7.639$

We have the following ANOVA table,

The hypotheses tested are,

$a)$ For the seasons

$H_0:$ The seasons are not different

$Against$

$H_1:$ At least one of the seasons is different.

The test statistic is $F_1=1.87$

$F_1$ is compared with the $F$ distribution table value at $\alpha=0.05$ with numerator degrees of freedom equal to 6 and denominator degrees of freedom is 2. The table value is,,

$F_{0.05}(6,2)=19.33$.

The null hypothesis is rejected if $F_1\gt F_{0.05}(2,6)$.

Since $F_1=1.87\lt F_{0.05}(2,6)=5.99$, we fail to reject the null hypothesis and conclude that sales in the 3 seasons are not different.

$b)$

For the seasons

$H_0:$ The salesmen are not different

$Against$

$H_1:$ At least one of the salesmen is different.

The test statistic is $F_2=1$

$F_2$ is compared with the $F$ distribution table value at $\alpha=0.05$ with numerator degrees of freedom equal to 3 and denominator degrees of freedom equal to 6. The table value is,

$F_{0.05}(3,6)=4.76$.

The null hypothesis is rejected if $F_2\gt F_{0.05}(3,6)$.

Since $F_2=1\lt F_{0.05}(3,6)=4.76$, we fail to reject the null hypothesis and conclude that there is no difference in sales between the 4 salesmen.