ANOVA and post-hoc analysis

[ARI]

Hello there!!!!

I'm struggling once again with statistics. This time I wonder if someone can advise me.

I run a taste panel panel with (14 subjects) to whom I gave 6 food different products with different hardness and asked them to score them from 1 to 10 according to their perception regarding to hardness can I run ANOVA to determine if the samples can be differentiated according their hardness? how should introduce the data in SPSSS and which kind of post-hoc test I should run after ANOVA?

Hope you can help me please thanks in advance

[lumhearts]

I'll make a suggestion. Run a regular one-way repeated measures ANOVA. You will need to enter 6 different columns into SPSS, one for each food. Then put in a factor level of 6 into the SPSS dialog box. When you do this type of analysis on 6 different measures is actually called profile analysis. If the test is significant, run paired t tests on all pairs and use a Bonferroni correction if you want to.

Also, you should check your assumptions.

[ARI]

Many thanks for your suggestion I followed a bit of a different strategy before checking your reply and I am just wondering if that makes sense. Do you mind to tell me?

First of all I wrote down all the scores in a colum, for scores regarding to the same product i put the same number in another colum as a factor. The I compare means using ANOVA and as I obtained that the assumption of variances is violated in my case (through levene's test) I then use the Welch F ratio then I use Game's Howell post-hoc test for multiple comparisons of means Do you think that approach is correct?

If the answer is yes I am now struggling with other issue

the table of comparison that I obtained gave as a result the following results and I don't know how to present them or how to form subset groups do you think you can help me with this?

the result was as follows

MEAN PRODUCT 1= 2
MEAN PRODUCT 2=1=4
MEAN PRODUCT 3=4=5
MEAN PRODUCT4=2=3
MEAN PRODUCT5=3=6
MEAN PRODUCT 6=5

the equal symbol I am using to mean that the mean between the product are not significant different

Moreover I am a bit concern because it seems to me that products can not be differentiated according to hardness is that correct?. before knowing that the assumption of homogenety was violated I run a Student Newman Keuls test and results were better appearing just 3 subsets. but now it seems I have too many

Can you advise me about this possibly I am messing around but I hope you can understan my question?

Many thanks in advance once again

[lumhearts]

Hi Ari,
I think what you've done is very good. The Welch and G-H are right on. However, if your sample sizes are equal across the groups, you should not care about equal variances.

The main thing though is that when you review the assumptions of ANOVA, you will see that the groups need to be independent. And yours are not because each group actually contains the same people. So (I think) the better course would be profile analysis. Also, and maybe more important, profile analysis should give you more statistical power than the ANOVA you ran. Try it and see.

[ARI]

Many thanks for the quick answer, I however do not know how to perform the profile analysis would you mind to give me a clue. Also If I use ANOVA in the circumstances I did and I report it. Am I gonna be wrong?

Thanks a lot I am really struggling about how to analyse my data

[ARI]

Sorry I just realised that actually your first suggestion is what a Profile analysis is (ANOVA with repeated measurements)

Thanks again I will do it hopefully it works

[lumhearts]

Yeah - to put it nicely - the regular ANOVA would not be right. Try the repeated measures and see what you get.

[ARI]

I have done it and looks much better however I don't understand what do you mean about " run paired t tests on all pairs and use a Bonferroni correction " Is that the same as select from the tag "options" confidence interval adjustment bonferroni? and one more thing hoping I am not annoying you can I use this procedure to compare the means of any other variable like in this case "hardness" when I am using the same people?

Seriously many thanks

[lumhearts]

Good. Post-hoc tests for repeated measures are actually just paired t-tests. So just go to the -> compare means -> paired, and put in all the different 2-group combinations for the six foods. But because you are running 15 t-tests now instead of the original ANOVA, you need to correct so as to not surpass alpha of .05. So, use .05/15 = .0033 as your new alpha. This is called Bonferroni adjustment. It is sometimes a matter of opinion (I personally don't agree with) but most researchers/editors want to see it. The confidence interval that you mentioned only applies when you also have a grouping variable, which I don't think you have.

Just ask if you have other questions.

[ARI]

I am really sorry but I am confused now I don't understand how to compare the means :(

[lumhearts]

Was the profile analysis significant?

[lumhearts]

If it is significant. Go to Analyze -> Compare Means -> Paired Samples t Test. Then place food1 in the Variable1 column and food2 in the Variable2 column. Then for the next Pair, put food1 and food3 in. Then food1 and food4...all the way to food5 and food6. This should be 15 pairs total.

[ARI]

Many many thanks for all your time to answer my questions I run the test according to my little knowledge I think was significan and run the pair test but I still confused and overall I don't know how to report my result. I am attaching a document with the outcome I got do you mind to have a look and guide me a bit.

thousend thanks!!!!

[Junes]

Hi ARI,

Hope no-one minds if I step in to help.

Disclaimer: still learning myself, might get things wrong

When you do a lot of multiple comparisons, you increase the risk of "accidentally" getting some significant results when in fact there are no differences in the population (Type I error).

There are ways to correct for this. Bonferroni is by far the simplest, but unfortunately it's also quite conservative, most would say too much so (which means you'll get fewer significant results).

What you do is this: you divide your significance threshold alpha (usually .05) by the number of comparisons. In this case 15. This gives you a new threshold value of .05 / 15 = 0.0033 . For all comparisons, you reject your null if the p is below this number. Thus, in your case, it would mean all differences are significant except pairs 1, 10, 11, 12, and 15.