I have 2 columns of of the concentration of fat in fish which is measured by 2 different labs. I need to know if the difference of the results from these 2 labs are statistically significant. The data are collected from measurement of fat in 34 identical fish samples.

The problem is that these 34 samples consist of 3 types of fish, and each type accumulates different amount of fat averagely. Is there any sensitive method to take the influence pf fish types into consideration?

One idea came in my mind is to subtract one of the groups from the other one and run a t-test, one sample, with a hypothetical mean value=0. Would that make sense?

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    $\begingroup$ Are you interested mainly in the paired comparisons lab1-lab2? Or are you mainly interested in the group comparisons (fish1-fish2)? $\endgroup$ – user2974951 Oct 19 '18 at 10:10
  • $\begingroup$ Sorry if it was not clear. I'm interested in the comparison between the labs $\endgroup$ – Hadi Oct 19 '18 at 11:04
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    $\begingroup$ As long as each row of your data compares the same fish for both labs then you can use a paired test, either paired t-test if your data is approximately normal, or Wilcoxon signed-rank test for the non-parametric version. $\endgroup$ – user2974951 Oct 19 '18 at 11:10
  • $\begingroup$ Well, that's a good suggestion but I'm not sure if it would be the most sensitive way because although the means of each group and variances are quite close to each other, but the huge variances (about 14.2) compared to means (about 8.3) shows that it might ignore rather considerable variances between the pairs. I thought of making a new column of differences of lab2-lab1 and then run a one-sample t-test with hypothetical mean=0. So, we can avoid the big deviations between the amount of fat between fish types. How about that? $\endgroup$ – Hadi Oct 19 '18 at 12:06

Get the difference between labs, and use it as response variable.

Fit a linear model (also called ANOVA) with fish type as covariate, (dummy variable is needed). If fish type is not significant, you can come back to paired t test, or fit a model without covariate, and the p-value for intercept is the same as p from paired t test.

If fish type has effect on difference, it means the differences are not same across the fish type, and you need to deal them separately.

  • $\begingroup$ Thank you @a_statistician It was a great idea. Since I need to use a constant value for the dummy covariate, the slope of the concentration vs. covariates (=1 here for example) calculates DIV/0. I am using the concept described here to calculate the p-value. Do you have any solution for this problem? $\endgroup$ – Hadi Oct 22 '18 at 12:45
  • $\begingroup$ I am not sure that software is OK. what is DIV/0? $\endgroup$ – user158565 Oct 22 '18 at 14:31
  • $\begingroup$ DIV/0 is the error when a value is divided by zero. When it tries to calculate the slope, (y=Fat content, x=covariate=1 for each type of fish), DIV/0 happens because to calculate the slope (y2-y1)/(x2-x1) while x2=x1 ,which is the constant value of dummy covariate in our case here. This error happened only for one parameter (Fat) and other parameters generated rather meaningful p-values (for both Covariate and between groups). From those results, can I conclude that the influence of the type of fish is not significant when the p-value of the covariate for each test is p-value>alpha=0.05? $\endgroup$ – Hadi Oct 23 '18 at 12:01

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