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):

$$

logbigg(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.

- Log Odds for $X_1 = 3$ for each group.
- Odds for $X_1 = 3$ for each group.
- 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.*

**Contents**hide

#### Best Answer

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 + 0.27times X_1 + 0.732times X_2 \ &= 0.3211 + 0.27times 3 + 0.732times 1 \ &= 0.3211 + 0.81quadquad + 0.732 \ &= 1.8631 \ \ text{log odds}(X_1 = 3; B) &= 0.3211 + 0.27times X_1 + 0.732times X_2 \ &= 0.3211 + 0.27times 3 + 0.732times 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.*)

### Similar Posts:

- Solved – Looking for a layman’s explanation of how to manually calculate log odds
- Solved – Looking for a layman’s explanation of how to manually calculate log odds
- Solved – Connection between Hazard Ratios, Survival, and Probability
- Solved – P(A|B) and P(A|C) known, what is P(A|BC)
- Solved – P(A|B) and P(A|C) known, what is P(A|BC)