Recent content by Omerikooo

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    Interaction in linear regression

    I don't understand why not ? You can use BMI as a continuous variable and introduce an interaction variable with age group. For example Y = Intercept + Beta1*BMI +Beta2*BMI*(Age Category = Adolescents) + Beta3*BMI*(Age Category = Adult) This model would adjust BMI coefficient according to age...
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    Interaction in linear regression

    Okay. If you are okay with it no problem.
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    Interaction in linear regression

    Hello, I can help you about the interpretation but why do you include both obese and overweight at the same time. Couldn't you use only one variable including reference, overweight and obese. This would make better sense. For a female obese: Y = 4.4004 + 0.041*ethnicity - 0.071*age -...
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    Comparing two groups with categorical data- Need help in choosing the correct stat test to assess significance

    Okay, I'm not sure what that really means. Odds ratio can be calculated from contingency tables as you put in the second photo. But I'm also sure that it can be calculated via logistic regression analysis. There should be a relationship between them but I don't know really. I will try to do...
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    Comparing two groups with categorical data- Need help in choosing the correct stat test to assess significance

    To give odds you should do logistic regression. With chi or fisher you can only report significant frequency difference between groups with regards to your findings. Chi and fisher doesn't give Odds.
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    Comparing two groups with categorical data- Need help in choosing the correct stat test to assess significance

    Yes I mean a chi-square test for every finding. To run a Chi-square test your data is adequate. Expected values are intrinsically calculated in chi-square formula based on your contingency table, so no, you don't need to provide extra expected characteristics data, chi-square test takes care of...
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    Comparing two groups with categorical data- Need help in choosing the correct stat test to assess significance

    Hello, I think you should do repeated Chi-square tests (fisher if needed) for each outcome. For example fining A present/absent vs controls/cases. This will simply show if the frequency of your findings differ between cases and controls. BUT! I wouldn't stop there. Seeing the predictive value...
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    binomial problem

    Hey, Question asks the possibility of tossing the dice two times before tossing blue, in other words it asks. 1(all possible tosses) - the possibility of tossing the dice blue before the third toss. In this case there are 2 possibilities. You can toss the dice blue first and stop, the second...
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    Failure to calculate Kappa for a constant rater.

    This article by Hripcsak and Heitjanb tackles many issues I came up to when doing work about reliability analyses is good help. It also includes tetrachoric correlation discussion as well. https://core.ac.uk/download/pdf/82193386.pdf
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    Failure to calculate Kappa for a constant rater.

    I think tetrachoric correlation may be a solution for me.
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    College work Help

    please add the questions
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    Failure to calculate Kappa for a constant rater.

    Wow! Good source indeed. I failed to find any correlation that takes care of binary-binary cases but still thanks!
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    Help deciding which spss functions to use

    1. Why not repeated measures t-test but anova? How many subject groups will you have, how many repeats you have (I guess 2)) 2. 2 linear regression models: one with LoC and the other with neuroticism. Than you can compare adjusted-r scores, if both estimates are significant. 3. Linear regression...
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    Failure to calculate Kappa for a constant rater.

    Only one rater did two ratings in different time points. Correlation coefficient wouldn't work in my case since I have binary variable with only 0 and 1. If this was some kind of continuous variable I would use ICC (intra-class coefficient) and it would cause no problem. In my example one of...
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    Paired samples t test interpretation help (included image of SPSS output)

    Okay then paired-sample t-test is fine. Why do you think this is a limitation. Actually measuring the result from an unchanged sample can be a strength of your study. If the samples were different your would be less reliable at showing the change in mean_css. Different sample would introduce new...