Looking for a layman's explanation of how to manually calculate log odds?

Question

I will start that I am not as math oriented as I would like to and could use a layman's / non-staticians explanation walk through of how to calculate the log odds.

I am reading Hosmer, Lemeshow, and Sturdivant's Appliged Logistical Regression which is helpful, but I could use a primer / tutorial to help solidify the concepts for me.

Given something like the following equation (values randomly chosen): $$ \log\bigg(\frac \pi {(1-\pi)}\bigg) = 0.3211 + 0.27 X_1 + 0.732 X_2 $$ And the following information:

• Indicator Variable Group A $X_2 = 1$, Group B $X_2 = 0$.
• Sample Size 1000
• Likelihood Value for Model = 0.0598

How would I compute the following from the above information manually without using R or another application.

1. Log Odds for $X_1 = 3$ for each group.
2. Odds for $X_1 = 3$ for each group.
3. Probability $X_1 = 3$ for each group.

If I understand correctly the Log Odds is the $\ln(p/(1-p))$ and log-odds and odds are different; but I am NOT clear on how to apply the above information to calculate the above information and am looking for a step-by-step walk through that covers most of the steps required of how to apply this information and perform the calculation.

Note: While I am utilizing this to assist me in a class it is not part of an assignment or test and the equation is made up by me purely for example purposes as I feel I am in need of a starting point example as I've spent some time reading the book (particularly chapter 3) but it is not clicking like I need it to.

Answer 1

You are right that the "$\ln(p/(1−p))$" is the log odds. That means all you have to do to calculate the log odds is plug in the values you want for your $X$'s and do the arithmetic. Here is the calculation for #1:
\begin{align} \text{log odds}(X_1 = 3; A) &= 0.3211 + 0.27\times X_1 + 0.732\times X_2 \\ &= 0.3211 + 0.27\times 3\ \ \ + 0.732\times 1 \\ &= 0.3211 + 0.81\quad\quad\ \ + 0.732 \\ &= 1.8631 \\ \ \\ \text{log odds}(X_1 = 3; B) &= 0.3211 + 0.27\times X_1 + 0.732\times X_2 \\ &= 0.3211 + 0.27\times 3\ \ \ + 0.732\times 0 \\ &= 0.3211 + 0.81 \\ &= 1.1311 \\ \end{align} To get the odds from the log odds, you exponentiate the log odds. That means you raise the special number e ($\approx 2.718281828$) to the power of the log odds. Here is the calculation for #2:
\begin{align} \text{odds}(X_1 = 3; A) &= e^{1.8631} \\ &= 6.443681 \\ \ \\ \text{odds}(X_1 = 3; B) &= e^{1.1311} \\ &= 3.099064 \end{align} (Full disclosure: I used R to do the arithmetic on that one.)

To get the probability, you divide the odds by the quantity odds plus one. Here is the calculation for #3:
\begin{align} \text{probability}(X_1 = 3; A) &= \frac{6.443681}{6.443681 + 1} \\ \ \\ &= 0.8656579 \\ \ \\ \text{probability}(X_1 = 3; B) &= \frac{3.099064}{3.099064 + 1} \\ \ \\ &= 0.7560419 \end{align} (I used R for this one, too.)