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	B	C	D	E
4.0	6.5	3.9	6.1	5.0
4.0	6.9	3.8	6.3	5.1
3.8	7.2	4.6	6.5	5.2
4.0		4.5		4.9

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

Answer 1

Null hypotheses H0 : The average amount of soft drink is equal for all five machines.

Alternative hypotheses H1 : At least one average amount of soft drink is different for five machines.

Level of significance = 0.01

Degree of freedom of group = Number of groups - 1 = 5 - 1 = 4

Degree of freedom of error = Number of observations - Number of level = (4 + 3 + 4 + 3 + 4) - 5 = 13

Let T_i be the total amount for machine i, n_i be number of observations of machine i.

Let G be the total amount of all observations and N be total number of observations.

$\sum$X² is sum of squares of all observations.

T1 = 15.8, T2 = 20.6 , T3 = 16.8, T4 = 18.9, T5 = 20.2

G = 15.8 + 20.6 + 16.8 + 18.9 + 20.2 = 92.3

$\sum$X² = 62.44 + 141.7 + 71.06 + 119.15 + 102.06 = 496.41

SST = $\sum$X² - G²/N = 496.41 - 92.3²/18 = 23.11611

SSTR = $\sum$T²/n - G²/N = (15.8² /4 + 20.6² /3 + 16.8² /4 + 18.9² /3 + 20.2² /4) - 92.3²/18 = 22.20944

SSE = 23.11611 - 22.20944 = 0.90667

MSTR = SSTR / DF for group = 22.20944 / 4 = 5.55236

MSE = SSE / DF for error = 0.90667 / 13 = 0.06974385

Test statistic, F = MSTR / MSE = 5.55236 / 0.06974385 = 79.61