Statistics emergency! Please help!?

Question

How do you perform a G-test on the following data?

Category | Observed
A___________12
B___________6
C___________9

The formula I have for G-test is

G = 2 x [sum of]O ln (O/E)

where O is the observed and E is the expected (9)

And to get G for the data apparently you divide by a correction factor:

Correction factor: 1 + [(number of categories)² - 1]/6(number of observed frequencies)(degrees of freedom)]


But at the end I got G to be -0.206. Surely it can't be negative? The critical value for 5% significance level at 2 degrees of freedom is 5.99.

Answer 1

I get

G = 2 * (12 * ln(12/9) + 6 * ln(6/9) + 9 * ln(9/9))
G = 2 * (3.452185 - 2.432791 + 0)
G = 2.038788

The reference I was looking at says nothing of a correction factor. Using what you've written the correction factor looks to be 1 + (3^2 - 1) / 6 = 2.33333

G' = G / correction factor = 0.8737663


the critical value come from the chi-square and either way, using G or G' you have a test statistic less than the critical value so we will conclude the null hypothesis is plausible. In other words, it is plausible the expected value for each of the three categories is 9.

Answer 2

I don't know that much about it, but this is what I found. Check the source.

How the test works

The test statistic is calculated by taking an observed number (O), dividing it by the expected number (E), then taking the natural log of this ratio. The natural log of 1 is 0; if the observed number is larger than the expected, ln(O/E) is positive, while if O is less than E, ln(O/E) is negative.

Answer 3

Is this a two-tailed test?

If so then it can be negative, I think. I did the exam for this last year.