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Testing for a difference in standard deviation


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Hello Everyone

I've encountered the following problem - I have this test that was administrated to a sample of participants, in a 2X2 design (both within and between participants). Each test generated a mean and standard deviation for each participant. Now, I want to check if there is a statistical difference in the standard deviation of the test (the value that was generated for each participant), both between and within groups. Can I use Anova in this case? Or should I use another test?

Thanks a lot!

Edited by Curious guy
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Hi curious guy. ANOVA tests for MEAN differences, not differences in variances/ standard deviations. You want to use an F-test. Use the VARIANCES, and divide the larger by the smaller for your F-ratio.

Wait! I just noticed you have a sd for EACH participant? Each participant has a set of values associated with it? That would be different. In that case, you could treat the sd as your X and use ANOVA. You'll be testing for the mean differences in the set of sds.

Edited by Ellies
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Hi Ellies and kgbfan. Thanks for your replies!

 

 

Ellies - Yes, that's exactly my question. Is it correct to use anova to compare means of SD's? It sounds logical to me, but I'm worried I might be overlooking something. 

 

kgbfan - I have over 20 SD's to compare. Are you sure Levene's test is a possible solution?

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Hi Ellies and kgbfan. Thanks for your replies!

 

 

Ellies - Yes, that's exactly my question. Is it correct to use anova to compare means of SD's? It sounds logical to me, but I'm worried I might be overlooking something. 

 

kgbfan - I have over 20 SD's to compare. Are you sure Levene's test is a possible solution?

 

 

I't's a bit unclear to me what data you have.   Are you only given the SD's or do you also have the data points that were used to calculate the SD's?  Also does each patient have one SD or multiple?

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I't's a bit unclear to me what data you have.   Are you only given the SD's or do you also have the data points that were used to calculate the SD's?  Also does each patient have one SD or multiple?

I only have one SD for each participant. Not the actual data points which were used to calculate the SD. 

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I only have one SD for each participant. Not the actual data points which were used to calculate the SD. 

Oh, I see, then I guess ANOVA could be appropriate.  Two assumptions for ANOVA are that the data are 1) normally distributed and 2) that the variances are comparable between both groups, although ANOVA still holds up pretty well for large sample sizes when the former is violated .  If you cannot assume homogeneity of variances, I'd opt for Welch's  Test, instead.

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Yes, but that's the variances of the sets of standard deviations that need to be homogeneous. Not the variances of the sds used as data values... :)

So you're likely to be just fine with ANOVA.

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Yes, but that's the variances of the sets of standard deviations that need to be homogeneous. Not the variances of the sds used as data values... :)

So you're likely to be just fine with ANOVA.

 

Well isn't he comparing sets of standard deviations?  Otherwise how can you compare individual data values using ANOVA?

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Yes, but the sds he is testing are the data values for the ANOVA. For the ANOVA to be applicable, the sets of sds need to have homogeneous variances. The test then will indicate if there is a significant difference in the means for the sets of sds.

He has four groups, in a 2x2 design. Each of these groups has a mean for the set of sds in the group. There is also a mean for each row and column, etc. Each group also has a variance; the variance of the group of sds. It is the variances of the groups that need to be homogeneous.

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Yes, but the sds he is testing are the data values for the ANOVA. For the ANOVA to be applicable, the sets of sds need to have homogeneous variances. The test then will indicate if there is a significant difference in the means for the sets of sds.

He has four groups, in a 2x2 design. Each of these groups has a mean for the set of sds in the group. There is also a mean for each row and column, etc. Each group also has a variance; the variance of the group of sds. It is the variances of the groups that need to be homogeneous.

 

OK yes, that is what I meant. :)

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