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For a standard linear regression, you assume the error term is normally distributed. Here, the variance parameter has to be estimated - hence the use of the $t$-distribution for the test statistic. …

Yeah, how about a nonlinear regression model. There are a few models that are used all of the time in biostats. …

You could use a Poisson regression and add an offset for the size of each group. See When to use an offset in a Poisson regression? and poisson vs logistic regression for more explanation. …

I've spent a lot of time looking through R manuals. The best one that I've seen is Grant Farnsworth's "Econometrics in R". Check it out: http://cran.r-project.org/doc/contrib/Farnsworth-EconometricsIn …

A nonlinear regression function is never going to decide the functional form for you. …

Normality of the residuals is an assumption of running a linear model. So, if your residuals are normal, it means that your assumption is valid and model inference (confidence intervals, model predict …

Also, I think this is more of a time series problem, not a cross sectional logistic regression problem. …

It's not true or false. It's just saying that more children is correlated with households having lower income. Check the correlation between the two variables. There's certainly no causation here. We …

You can do this (put a continuous variable into bins) but it's generally considered a loss of information. It would be appropriate if there is clearly a different effect when moving from one bin to a …

So maybe it's not a good idea, but here's a start: #first, some simulated data y = seq(from = 1, to = 10, by = 0.5) x1 = y + rnorm(length(y)) #not a significant variable x2 = y * run …

I think I found the problem: when you went from the gradient to solving for $\beta$, you didn't reverse the order of the terms. The inversion should have been $$(x_nx^T_n)^{-1}=\frac{1}{x^T_nx_n},$$ …