Timeline for How to test proportion of each factor level against mean proportion across all levels (binary outcome)
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 13 at 19:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 14 at 19:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 15 at 21:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 15, 2023 at 2:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 16, 2023 at 0:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 17, 2023 at 22:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 12, 2022 at 8:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 13, 2022 at 20:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 10, 2022 at 16:48 | answer | added | kjetil b halvorsen♦ | timeline score: 0 | |
| Jun 10, 2022 at 16:48 | history | edited | kjetil b halvorsen♦ |
edited tags
|
|
| Nov 9, 2015 at 13:44 | comment | added | tristan |
OK, in that case you would probably do a two-tailed exact binomial proportion test. In R you have binom.test which can perform this. I would also adjust the significance level for multiple testing (from 0.05 to 0.0125). I think this is what you are saying in your "Idea", and I think it would be sound.
|
|
| Nov 9, 2015 at 13:06 | comment | added | Mark Heckmann | @tristan The Chi-sqaure test is an overall test performed on a contingency table. It will not tell me which groups differ from the mean propability. I need four p-Values, not one. | |
| Nov 9, 2015 at 11:51 | comment | added | tristan | Is there any reason you couldn't do a chi-squared test? You would have a 4x2 matrix (each row is a region, the columns are the counts for outcome 0 and outcome 1). The expected value in each cell is calculated as $E_{ij} = n_i \frac{X_j}{N}$ where $n_i$ is the population size of region $i$, $X_j$ is total number with outcome $j$ and $N$ is total population size. Degrees of freedom = 3 (I think). The main reason I can imagine is if the outcomes in each row are dependent? | |
| Nov 9, 2015 at 11:38 | history | asked | Mark Heckmann | CC BY-SA 3.0 |