Question

A certain vending company's soft-drink dispensing machines are supposed to serve 6 oz of beverage. Various machines were sampled, and the resulting amounts of dispensed drink (in ounces) were recorded, as shown in the following table. Does this sample evidence provide sufficient reason to reject the null hypothesis that all five machines dispense the same average amount of soft drink? Use α = .01?

Machines
  A  Â	  B  Â	  C  Â	  D  Â	  E  Â
3.7	6.8	4.0	6.1	6.2
4.2	6.8	4.4	6.5	4.6
4.1	7.3	4.5	6.1	5.5
4.2		4.6		5.1

(a) Find the test statistic. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer bounds exactly.)
< p <

Answer 1

The following table is obtained:

	Group 1	Group 2	Group 3	Group 4	Group 5
	3.7	6.8	4	6.1	6.2
	4.2	6.8	4.4	6.5	4.6
	4.1	7.3	4.5	6.1	5.5
	4.2		4.6		5.1
Sum =	16.2	20.9	17.5	18.7	21.4
Average =	4.05	6.967	4.375	6.233	5.35
\sum_i X_{ij}^2	65.78	145.77	76.77	116.67	115.86
St. Dev. =	0.238	0.289	0.263	0.231	0.676
SS =	0.17	0.16666666666666	0.2075	0.10666666666665	1.37
n =	4	3	4	3	4

The total sample size is N = 18 . Therefore, the total degrees of freedom are:

df_{total} = 18 - 1 = 17

Also, the between-groups degrees of freedom are df_{between} = 5 - 1 = 4 , and the within-groups degrees of freedom are:

df_{within} = df_{total} - df_{between} = 17 - 4 = 13

First, we need to compute the total sum of values and the grand mean. The following is obtained

Also, the sum of squared values is

Based on the above calculations, the total sum of squares is computed as follows

The within sum of squares is computed as shown in the calculation below:

The between sum of squares is computed directly as shown in the calculation below:

Now that sum of squares are computed, we can proceed with computing the mean sum of squares:

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃ = μ₄ = μ₅

Ha: Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df_1 = 4 and df_2 = 4 , therefore, the rejection region for this F-test is

(3) Test Statistics

(4) Decision about the null hypothesis

Since it is observed that F = 33.133 < F_c = 5.205 , it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0 , and since p = 0 < 0.01, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 5 population means are equal, at the α=0.01 significance level.