Answer to Question #279173 in Statistics and Probability for Snehil

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.