Hi;

My G*Power setup is as follows and my method is A.R.E.

1) What does ARE stand for in G*Power?

2) my degree of freedom is 53.2960000. My question is that is my DF Ok? I didn't set it up specifically and the value is awkward.

3) how can i interpret this with my results? E.g. Given a b and c, the mean difference is k, so that I can reject/accept null hypothesis. What are my parameters?

Best regards.

My G*Power setup is as follows and my method is A.R.E.

1) What does ARE stand for in G*Power?

2) my degree of freedom is 53.2960000. My question is that is my DF Ok? I didn't set it up specifically and the value is awkward.

3) how can i interpret this with my results? E.g. Given a b and c, the mean difference is k, so that I can reject/accept null hypothesis. What are my parameters?

```
t tests - Means: Wilcoxon-Mann-Whitney test (two groups)
Options: A.R.E. method
Analysis: A priori: Compute required sample size
Input: Tail(s) = One
Parent distribution = min ARE
Effect size d = 0.9076923
α err prob = 0.05
Power (1-β err prob) = 0.95
Allocation ratio N2/N1 = 1
Output: Noncentrality parameter δ = 3.3748581
Critical t = 1.6739508
Df = 53.2960000
Sample size group 1 = 32
Sample size group 2 = 32
Total sample size = 64
Actual power = 0.9541689
```

Best regards.

Do more with

EXPERT OFFICE^{®} is a registered trademark of EXPERTS EXCHANGE^{®}

If [ a ] sample size N is required to achieve a specified power for the Wilcoxon signed-rank test and a samples size N0 is required in the t test to achieve the same power, then the ratio N0/N is called the efficiency of the Wilcoxon [ / Wilcoxon-Mann-Whitney ] signed-rank test relative to the one-sample t test. The limiting efficiency as sample size N tends to infinity is called the asymptotic relative efficiency (A.R.E. or Pitman efficiency) of the Wilcoxon signed rank test relative to the t test. ...

Note, that the A.R.E. of the Wilcoxon signed-rank test to the one-sample t test is identical to the A.R.E of the Wilcoxon rank-sum test to the two-sample t test ... . If ... [the distribution of differences between base case and test case has] a normal distribution, then the A.R.E. is 3/pi = 0.955. This shows, that the efficiency of the Wilcoxon test relative to the t test is rather high ... [when] the assumption of normality made in the t test is true. It can be shown that the minimal A.R.E. [ all reasonable cases of distributions ] ... is 0.864.

That is when using the Wilcoxon test you are no worse off than 0.86 times the value from the t-test, and in cases where the data is badly not normal you are usually much better off (ARE > 1).

Here the GPower is looking at the distribution of H which is the difference in distributions of F (your base situation) and G (your testing case), where G is F shifted, that is G(x) = F(x − D) and D is the shift amount. The ARE gives details as sample size increases to infinity.

2) Degrees of freedom typically describe how much

3) Interpretation

You set the α err probability *(reporting a difference found when it was just due to sampling variations) * of 5%.

You set the power probability *(Not missing out an actual difference due to sampling variations)* to be at least 95% (This is usually considered high if it involves real money in collection data - often 80% is regarded as a *good* value to use).

You set the size of the difference that you consider significant (significant in the sense that it is important in the real world, not talking about statistical significance here) to be 0.9077 times the actual standard devation. That is if there is a shift of this size (or greater) between standard case and your treatment case you want to be able to pick it up. This implies that if the shift is less than this amount you are willing to have less than the power percentage chance of finding it.

You set a one sided test which means that you only test for changes in one direction (this has more power than a 2 sided test), and if the results go the other way (treatment group times were higher than control group times) all you can say is that there was no significant decrease in timing.

You want the sample size for both the control group and the treatment group to be the same.

For those settings -

You will need a sample size of (at least) 32 in each of the 2 groups.

The actual power (for a sample size of 32+32=64) is 96%.

In the calculations, the program used noncentrality parameter of 3.3748581 with Df = 53.2960000 producing a non-central t value of 1.6739508.

Ian

## Premium Content

You need an Expert Office subscription to comment.Start Free Trial