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Below is a brief snippet of R syntax. If you run it, you will see that model 2 v model 3 produces the same F ratio in each output. However, model 1 vs model 2 produces different F ratios. …

Yes, these functions appear to generate the same output. To more clearly see this, try the following code n <- 100 seed.val <- 3 set.seed(seed.val) x <- sample(c(0,1),n,replace=T,prob=c(19/37,18/37) …

The Wald $\chi^2$ test statistics are just the squares of the $z$ test statistics you have in the R output. In this case, the $P$-values will should be the same. …

Using R, the correct code would be chisq.test(v,p=freq) When I attempted to run this test, there was a problem: Chi-squared test for given probabilities data: v X-squared = 16939, df = 25, …

From what you've described, my inclination is that you have item redundancy. Look at how highly correlated the flagged polygamous items are. If they are highly correlated, then you can consider just d …

The $R^2$ values reported are indeed coefficients of determination, but they are all partial-$R^2$s (of at least the second-type). … That is to say, the partial $R^2$ values reported most likely will not sum to 100%. …

This is an issue of linearizing non-linear problems.  (A strange response, and I'll briefly elaborate before providing an answer.)  We can solve polynomial regression using linear regression because w …

The plot is providing a graphical representation of the chi-square test for independence. (In the 2x2 context, the same test as asking if two groups have the same proportion.) The null hypothesis is t …

The first value is the right-side critical value $\chi_{R,cv}^2$; the second value is the left-side critical value, $\chi_{L,cv}^2$. … , you need to take the square root of everything in the trilinear inequality: $$\sqrt{\frac{(n-1)s^2}{\chi_{R,cv}^2}} < \sigma < \sqrt{\frac{(n-1)s^2}{\chi_{L,cv}^2}}$$ If there is an R function, maybe …

If you assume that your data follows a hyperbolic model, then you can use the following approach (detailed more here https://engineerexcel.com/hyperbolic-curve-fitting-excel/): use a harmonic transfo …

You will need to look at the standardized solution. In that output, you should have an estimate for the variance for the items in the model. One of them has been flagged as having an impossible varian …

First, I will introduce yet a fourth model for the discussion in my answer: fit1.5 <- lm(y_2 ~ x_1 + x_2 + y_1) Part 0 The difference between fit1 and fit1.5 is best summarized as the difference b …

For a meta-analysis of this sort, the idea is to combine the effect size information from your disparate samples into one measure for the effect size. The conventional protocol here would be to estim …

Introduction of higher powered terms will invariably result in issues of multicollinearity.  This is essentially unavoidable.  Thus, it is best to only include the terms if theoretically justifiable.  …

In this case, you can obtain an $R^2$ for the latent variable, and you can do an F-test with the two different models (2 iv's vs. 3 iv's). …