Question

The following data represents the sale (Rs. 1,000) per month of 3 brands of a
toilet soap allocated among 3 cities:
Cities
Brands A B C
I 42 48 30
II 42 54 57
III 29 42 29
At 5% level of significance, test whether the mean sales of 3 brands are equal.

Answer 1

Answer on Question #64886 – Math – Statistics and Probability

Question

The following data represents the sale (Rs. 1,000) per month of 3 brands of a toilet soap allocated among 3 cities:

At 5% level of significance, test whether the mean sales of 3 brands are equal.

Solution

$H_0$: The mean sales of 3 brands are equal.

$H_a$: Not all mean sales of 3 brands are equal.

Using the one-way analysis of variance (ANOVA)

$$df_T = 3 - 1 = 2,$$ $$df_{total} = 9 - 1 = 8,$$ $$df_E = 8 - 2 = 6,$$ $$\begin{aligned} SS_{total} &= 42^2 + 48^2 + 30^2 + 42^2 + 54^2 + 57^2 + 29^2 + 42^2 + 29^2 \\ &\quad - \frac{(42 + 48 + 30 + 42 + 54 + 57 + 29 + 42 + 29)^2}{9} = \\ &= 884.222, \end{aligned}$$ $$SS_T = \frac{(42 + 48 + 30)^2}{3} + \frac{(42 + 54 + 57)^2}{3} + \frac{(29 + 42 + 29)^2}{3} - \frac{(42 + 48 + 30 + 42 + 54 + 57 + 29 + 42 + 29)^2}{9} = 477.556,$$ $$SS_E = SS_{total} - SS_T = 884.222 - 477.556 = 406.666,$$ $$MS_T = \frac{SS_T}{3 - 1} = \frac{477.556}{2} = 238.778,$$ $$MS_E = \frac{SS_E}{9 - 3} = \frac{406.666}{6} = 67.778.$$

The test statistic is

$$F = \frac{MS_T}{MS_E} = \frac{238.778}{67.778} = 3.522.$$

Using the tables of the F distribution the critical value at 5% significance level is

$$F_{\alpha}(3 - 1,9 - 3) = F_{cr}(2,6) = 5.143.$$

The test statistic is less than critical value. Thus, we don't reject the null hypothesis at 5% level of significance.

There is not sufficient evidence to conclude that mean sales of 3 brands are not equal.

**Answer:** There is not sufficient evidence to conclude that mean sales of 3 brands are not equal.

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