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Posted 11-20-2015 09:02 AM
(2648 views)

Dear Community,

I recently recently performed a Ttest on my data following the Proc below:

Proc Ttest;

Class Trt;

Var PFG Male Female;

Run;

I am able to get only one F value, yet the reviewer of my manuscript would like me to report me the F-values with their two DF each. Am not sure how to go about this procedure.

I will really appreciate your help if you can guide me on the procedure to follow.

Best regards,

Peter

6 REPLIES 6

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Dear Rick,

Many thanks for the feedback.

The reviewer of the manuscript states that i should report all F-values with their two df each. No other information is given. My assumption is they want to check the credibility of the statisitics provided.

Being a basic user of SAS, i am not sure how to program my SAS Procedure to obtain this information.

I apprecite your feedback. Am able to get the information you stated i.e. Satterthwaite (unequal varince) and equal varince test results. Also the folded F test results. I have looked at the link you gave

Is there a standard Procedure in SAS to do this automatically in SAS?

Best regards,

Peter.

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The Equality of Variance is printed automatically by PROC TTEST. For example, run the following statements:

```
proc ttest data=sashelp.class;
class sex;
var height weight;
run;
```

Look in the output for the Equality of Variance tables. There is one for the HEIGHT variable and another one for the WEIGHT variable.

Equality of Variances | ||||
---|---|---|---|---|

Method | Num DF | Den DF | F Value | Pr > F |

Folded F | 8 | 9 | 1.03 | 0.9527 |

The first one says that DF1=8 and DF2=9. The p-value is large, which indicates that the data does not reject the null hypothesis that the variance of height is equal for men and women. DF1=8 because there are N1=9 females; DF2=9 because there are N2=10

males.

The second table is similar, except DF1=9, which means that the males are being consider as the numerator when testing the ratio of variances.

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Many thanks for that informative email. I really appreciate.

I am getting the same kind on information when i run my analysis.

Going by your example, it means that F(DF1, DF2) = F(8, 9) =1.03 right?

What does it mean when the DF1 and DF2 are the same? and when the P-value is small enough to enable rejection of the null hypothesis?

Best regards,

Peter

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Dear Slchen,

Thanks too for the feedback. What then does it mean when i have similar DF for both Numerator and denominator?

Here are the details of my results

pooled DF=14

Satterthwaite (unequal) DF = 11.96

Folded F: Num DF = 7, Den DF=7, F value = 2.41.

So in this case, is my F(DF1, DF2) = F(7,7)= 2.41? If yes, what does this mean?

Best regards,

Peter.

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