In: Statistics and Probability
compare the concentration (in ppm) of a particular item in lake water
Construct a nova table and determine if the concentrations differ between the lakes
lake1 | lake2 | lake3 |
12.7 | 8.3 | 20.3 |
9.2 | 17.2 | 16.6 |
10.9 | 19.1 | 22.7 |
8.9 | 10.3 | 25.2 |
6.4 | 19.9 |
The following table is obtained:
Lake 1 | Lake 2 | Lake 3 | |
12.7 | 8.3 | 20.3 | |
9.2 | 17.2 | 16.6 | |
10.9 | 19.1 | 22.7 | |
8.9 | 10.3 | 25.2 | |
6.4 | 19.9 | ||
Sum = | 48.1 | 54.9 | 104.7 |
Average = | 9.62 | 13.725 | 20.94 |
484.91 | 835.63 | 2233.99 | |
St. Dev. = | 2.355 | 5.232 | 3.224 |
SS = | 22.188 | 82.1275 | 41.572 |
n = | 5 | 4 | 5 |
Number of treatment, k = 3
Total size, N = 14
Grand mean =
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:
Source of Variation | SS | df | MS | F |
Between Groups | 327.2646 | 2 |
327.2646/2 = 163.6323 |
12.338 |
Within Groups | 145.8875 | 11 |
145.8875/11 = 13.2625 |
|
Total | 473.1521 | 13 |
Null and Alternative hypothesis:
H1: At least one mean is different.
Test statistic:
F = 12.338
p-value = 0.0015
As p =0.0015 < we, reject the null hypothesis.
There is enough evidence to claim that at least one mean is different.