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Logistic Regression Info

The above is all the info I was given to work with.

For the b column, are higher bs indicative of stronger association? What could the constant mean?

What does the exp (b) column mean (how do I interpret it)?

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  • $\begingroup$ It is simply the exponential function. $\endgroup$ Mar 30, 2017 at 18:14
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    $\begingroup$ What is an exponential function? $\endgroup$
    – Ria
    Mar 30, 2017 at 18:16
  • $\begingroup$ For each b it computes the base function of the natural logarithm. So for b=0 you get 1. For b>0 you get numbers greater than 1. When b is negative exp(b)=1/exp(c) where c = -b. $\endgroup$ Mar 30, 2017 at 18:22

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Before we start with the interpretation of the $b$s make sure you know what logarithms and exponentials are.

$exp(b)$ means nothing else than $e^b$. $e$ is a constant with a never ending number of digits. Approximately, $e=2.718$. $e$ is very important constant in mathematics as it shows up in many different places (just like $\pi$). So lets say we have an equation like $e^x=12$ we can solve it using the logarithm: $ln(12)=x$ (In general it is very useful to look deeper into exponential functions and logarithms. There are excellent explanations from elementary mid-school explanations to university grade explanations on khanacademy.org for free).

A logistic regression model has like any other regression model a regression equation at its core. Using logistic regression you model values in the dependent variable that are between 0 and 1, lets call them $p$ (like probability).

The regression equation of logistic regression is:

$\hat{p}=\frac{1}{1+e^{-(b_0+b_1X)}}$

This is the same thing as stating:

$ln\left(\frac{\hat{p}}{1-\hat{p}}\right)=b_0+b_1X$.

this term on the left side is called logit:

$ln\left(\frac{\hat{p}}{1-\hat{p}}\right)$

Something similar but without the ln() we call odds ratio, that is the probability of something decided by the counter probability (e.g. probability of getting cancer divided by probability of not getting cancer).

$\frac{\hat{p}}{1-\hat{p}}$

In your case the equations would look like:

$ln\left(\frac{\hat{p}}{1-\hat{p}}\right)=-5.847-0.366*age-2.018*children+0.39*education+0.487*health$

or

$$\hat{p}=\frac{1}{1+e^{-(-5.847-0.366*age-2.018*N+0.39*education+0.487*health)}}$$

So why am I telling you all this? $b$ and $exp(b)$ give you the same information but in a different way.

Generally speaking: the further away $b$ from $0$ and the $exp(b)$ from $1$ the greater the association between this predictor (IV) and the criterion (DV).

Let us look at "age of wife": $b=-0.366$ means that if you fix all other predictors one extra year of age leads to 0.366 less in the logit. For example: let us say the logit of the DV is 5 (our criterion) and the age of the wife is 30. We can say that if the wife was 31 the logit would be 5-0.366 as long as we don't change any other varialbe (like children, education...). So the $b$ gives us some additive effect. $exp(b)=0.693$ means that the odds ratio of age 31 will be 5*0.693. So this is a multiplicative effect of aging 1 year on the odds ratio of the DV.

I hope this helps. One remark though: It seems as you are a statistics and math beginner. However, when it comes to statistical inference from logistic regression (Which will be the next logical step) things become even more complex when you have to deal with null deviance and Likelihood ratio tests instead of F tests and $R^2$ as in "regular" linear regression. I would therefore recommend to ask a friend or college who may know this stuff and help you a bit.

EDIT: the constant is the intercept. So if all values Age, children, health... are ALL 0, then the predicted logit would be -5.847 and the predicted odds ratio 0.003.

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    $\begingroup$ +1 Nice job, Carol! Here's a little tip: setting your $\TeX$ between paired dollar signs ($\$\$$ delimiters) will center the formulas in separate paragraphs. Complex formulas, such as your last one for $\hat p$, will be more readable. $\endgroup$
    – whuber
    Mar 30, 2017 at 21:12
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    $\begingroup$ Terrific (+1). One quibble: Your last phrase should say the predicted probability would be .003. $\endgroup$
    – rolando2
    Mar 31, 2017 at 2:12

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