Getting accurate interpolated probability from logistic regression equation
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I have to ascertain what specific rasch item a student needs to attain to have a 70% probability of passing a future criterion test (the tests are correlated, the results below are output from the logistic regression equation).
I ran a logistic regression equation on a series of rasch items. Because the rasch items represent discrete ability scores, and the number of items was not very large (15-items per student) I have to interpolate what rasch item would be needed to have a 70% probability of passing a criterion. Below is the output and code I have tried to use to create the probability.
intercept = -0.8392
slope = 0.4120
Finding probability of 0.70 given the above intercept/slope:
#Eq1
exp((log(0.70) - intercept)/slope)
#Output: 3.225788
This output would indicate a rasch score of 3.225788 would represent a probability of 0.70. But when I use that output to assess the probability of 3.225788, it comes out to a probability of 0.62.
#Eq2
exp(-0.8392+0.4120*(3.225788)) / (1 + exp(-0.8392 + 0.4120*(3.22578)))
#Output: 0.62
I also tried repeating equation 1 by first assigning the log(0.7) to an object (p) in the hopes that this could solve a rounding error that I read about in McElreath's "Statistical Rethinking" but it didn't appear to help.
Please do let me know if you need a reproducible dataset. I thought perhaps the intercept/slope would be enough, but can put together more if needed.
r logistic
asked Nov 24 '18 at 14:10
aleksis.paulaleksis.paul
133
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migrated from stackoverflow.com Nov 25 '18 at 10:18
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
I have to ascertain what specific rasch item a student needs to attain to have a 70% probability of passing a future criterion test (the tests are correlated, the results below are output from the logistic regression equation).
I ran a logistic regression equation on a series of rasch items. Because the rasch items represent discrete ability scores, and the number of items was not very large (15-items per student) I have to interpolate what rasch item would be needed to have a 70% probability of passing a criterion. Below is the output and code I have tried to use to create the probability.
intercept = -0.8392
slope = 0.4120
Finding probability of 0.70 given the above intercept/slope:
#Eq1
exp((log(0.70) - intercept)/slope)
#Output: 3.225788
This output would indicate a rasch score of 3.225788 would represent a probability of 0.70. But when I use that output to assess the probability of 3.225788, it comes out to a probability of 0.62.
#Eq2
exp(-0.8392+0.4120*(3.225788)) / (1 + exp(-0.8392 + 0.4120*(3.22578)))
#Output: 0.62
I also tried repeating equation 1 by first assigning the log(0.7) to an object (p) in the hopes that this could solve a rounding error that I read about in McElreath's "Statistical Rethinking" but it didn't appear to help.
Please do let me know if you need a reproducible dataset. I thought perhaps the intercept/slope would be enough, but can put together more if needed.
r logistic
asked Nov 24 '18 at 14:10
aleksis.paulaleksis.paul
133
$endgroup$
migrated from stackoverflow.com Nov 25 '18 at 10:18
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
I have to ascertain what specific rasch item a student needs to attain to have a 70% probability of passing a future criterion test (the tests are correlated, the results below are output from the logistic regression equation).
I ran a logistic regression equation on a series of rasch items. Because the rasch items represent discrete ability scores, and the number of items was not very large (15-items per student) I have to interpolate what rasch item would be needed to have a 70% probability of passing a criterion. Below is the output and code I have tried to use to create the probability.
intercept = -0.8392
slope = 0.4120
Finding probability of 0.70 given the above intercept/slope:
#Eq1
exp((log(0.70) - intercept)/slope)
#Output: 3.225788
This output would indicate a rasch score of 3.225788 would represent a probability of 0.70. But when I use that output to assess the probability of 3.225788, it comes out to a probability of 0.62.
#Eq2
exp(-0.8392+0.4120*(3.225788)) / (1 + exp(-0.8392 + 0.4120*(3.22578)))
#Output: 0.62
I also tried repeating equation 1 by first assigning the log(0.7) to an object (p) in the hopes that this could solve a rounding error that I read about in McElreath's "Statistical Rethinking" but it didn't appear to help.
Please do let me know if you need a reproducible dataset. I thought perhaps the intercept/slope would be enough, but can put together more if needed.
r logistic
asked Nov 24 '18 at 14:10
aleksis.paulaleksis.paul
133
$endgroup$
I have to ascertain what specific rasch item a student needs to attain to have a 70% probability of passing a future criterion test (the tests are correlated, the results below are output from the logistic regression equation).
I ran a logistic regression equation on a series of rasch items. Because the rasch items represent discrete ability scores, and the number of items was not very large (15-items per student) I have to interpolate what rasch item would be needed to have a 70% probability of passing a criterion. Below is the output and code I have tried to use to create the probability.
intercept = -0.8392
slope = 0.4120
Finding probability of 0.70 given the above intercept/slope:
#Eq1
exp((log(0.70) - intercept)/slope)
#Output: 3.225788
This output would indicate a rasch score of 3.225788 would represent a probability of 0.70. But when I use that output to assess the probability of 3.225788, it comes out to a probability of 0.62.
#Eq2
exp(-0.8392+0.4120*(3.225788)) / (1 + exp(-0.8392 + 0.4120*(3.22578)))
#Output: 0.62
I also tried repeating equation 1 by first assigning the log(0.7) to an object (p) in the hopes that this could solve a rounding error that I read about in McElreath's "Statistical Rethinking" but it didn't appear to help.
Please do let me know if you need a reproducible dataset. I thought perhaps the intercept/slope would be enough, but can put together more if needed.
r logistic
r logistic
asked Nov 24 '18 at 14:10
aleksis.paulaleksis.paul
133
asked Nov 24 '18 at 14:10
aleksis.paulaleksis.paul
133
asked Nov 24 '18 at 14:10
aleksis.paulaleksis.paul
133
asked Nov 24 '18 at 14:10
aleksis.paulaleksis.paul
133
asked Nov 24 '18 at 14:10
aleksis.paulaleksis.paul
133
133
migrated from stackoverflow.com Nov 25 '18 at 10:18
This question came from our site for professional and enthusiast programmers.
migrated from stackoverflow.com Nov 25 '18 at 10:18
This question came from our site for professional and enthusiast programmers.
add a comment |
add a comment |
1 Answer
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I think your arithmetic is wrong ...
- back-transform 0.7 to the log-odds (logit) scale:
log(x/(1-x))orplogis()in R:
p <- 0.7
log(p/(1-p)) ## 0.847
logit.p <- qlogis(p) ## same
- Solve for the desired value (
a+b*x=p->x = (p-a)/b):
int <- -0.8392
slope <- 0.4120
val <- (logit.p-int)/slope ## 4.093
- Checking: the logistic function (
exp(x)/(1+exp(x)), or1/(1+exp(-x))) is also available asplogis()in R:
plogis(int+slope*val) ## 0.7
There's a dose.p function in the MASS package that will do this automatically (and compute approximate standard errors) when provided with a glm object:
answered Nov 24 '18 at 17:41
Ben BolkerBen Bolker
23.8k16494
$endgroup$
$begingroup$
Thank you! I’m a bit embarrassed that I didn’t notice I wasn’t properly transforming the probability into odds before trying to solve the equation.
$endgroup$
– aleksis.paul
Nov 25 '18 at 1:19
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think your arithmetic is wrong ...
- back-transform 0.7 to the log-odds (logit) scale:
log(x/(1-x))orplogis()in R:
p <- 0.7
log(p/(1-p)) ## 0.847
logit.p <- qlogis(p) ## same
- Solve for the desired value (
a+b*x=p->x = (p-a)/b):
int <- -0.8392
slope <- 0.4120
val <- (logit.p-int)/slope ## 4.093
- Checking: the logistic function (
exp(x)/(1+exp(x)), or1/(1+exp(-x))) is also available asplogis()in R:
plogis(int+slope*val) ## 0.7
There's a dose.p function in the MASS package that will do this automatically (and compute approximate standard errors) when provided with a glm object:
answered Nov 24 '18 at 17:41
Ben BolkerBen Bolker
23.8k16494
$endgroup$
$begingroup$
Thank you! I’m a bit embarrassed that I didn’t notice I wasn’t properly transforming the probability into odds before trying to solve the equation.
$endgroup$
– aleksis.paul
Nov 25 '18 at 1:19
add a comment |
$begingroup$
I think your arithmetic is wrong ...
- back-transform 0.7 to the log-odds (logit) scale:
log(x/(1-x))orplogis()in R:
p <- 0.7
log(p/(1-p)) ## 0.847
logit.p <- qlogis(p) ## same
- Solve for the desired value (
a+b*x=p->x = (p-a)/b):
int <- -0.8392
slope <- 0.4120
val <- (logit.p-int)/slope ## 4.093
- Checking: the logistic function (
exp(x)/(1+exp(x)), or1/(1+exp(-x))) is also available asplogis()in R:
plogis(int+slope*val) ## 0.7
There's a dose.p function in the MASS package that will do this automatically (and compute approximate standard errors) when provided with a glm object:
answered Nov 24 '18 at 17:41
Ben BolkerBen Bolker
23.8k16494
$endgroup$
$begingroup$
Thank you! I’m a bit embarrassed that I didn’t notice I wasn’t properly transforming the probability into odds before trying to solve the equation.
$endgroup$
– aleksis.paul
Nov 25 '18 at 1:19
add a comment |
$begingroup$
I think your arithmetic is wrong ...
- back-transform 0.7 to the log-odds (logit) scale:
log(x/(1-x))orplogis()in R:
p <- 0.7
log(p/(1-p)) ## 0.847
logit.p <- qlogis(p) ## same
- Solve for the desired value (
a+b*x=p->x = (p-a)/b):
int <- -0.8392
slope <- 0.4120
val <- (logit.p-int)/slope ## 4.093
- Checking: the logistic function (
exp(x)/(1+exp(x)), or1/(1+exp(-x))) is also available asplogis()in R:
plogis(int+slope*val) ## 0.7
There's a dose.p function in the MASS package that will do this automatically (and compute approximate standard errors) when provided with a glm object:
answered Nov 24 '18 at 17:41
Ben BolkerBen Bolker
23.8k16494
$endgroup$
I think your arithmetic is wrong ...
- back-transform 0.7 to the log-odds (logit) scale:
log(x/(1-x))orplogis()in R:
p <- 0.7
log(p/(1-p)) ## 0.847
logit.p <- qlogis(p) ## same
- Solve for the desired value (
a+b*x=p->x = (p-a)/b):
int <- -0.8392
slope <- 0.4120
val <- (logit.p-int)/slope ## 4.093
- Checking: the logistic function (
exp(x)/(1+exp(x)), or1/(1+exp(-x))) is also available asplogis()in R:
plogis(int+slope*val) ## 0.7
There's a dose.p function in the MASS package that will do this automatically (and compute approximate standard errors) when provided with a glm object:
answered Nov 24 '18 at 17:41
Ben BolkerBen Bolker
23.8k16494
answered Nov 24 '18 at 17:41
Ben BolkerBen Bolker
23.8k16494
answered Nov 24 '18 at 17:41
Ben BolkerBen Bolker
23.8k16494
answered Nov 24 '18 at 17:41
Ben BolkerBen Bolker
23.8k16494
23.8k16494
$begingroup$
Thank you! I’m a bit embarrassed that I didn’t notice I wasn’t properly transforming the probability into odds before trying to solve the equation.
$endgroup$
– aleksis.paul
Nov 25 '18 at 1:19
add a comment |
$begingroup$
Thank you! I’m a bit embarrassed that I didn’t notice I wasn’t properly transforming the probability into odds before trying to solve the equation.
$endgroup$
– aleksis.paul
Nov 25 '18 at 1:19
$begingroup$
Thank you! I’m a bit embarrassed that I didn’t notice I wasn’t properly transforming the probability into odds before trying to solve the equation.
$endgroup$
– aleksis.paul
Nov 25 '18 at 1:19
$begingroup$
Thank you! I’m a bit embarrassed that I didn’t notice I wasn’t properly transforming the probability into odds before trying to solve the equation.
$endgroup$
– aleksis.paul
Nov 25 '18 at 1:19
add a comment |
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