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In 1PL model, zero ability equals average ability, or change accuracy?
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In a test evaluation scenario, latent variable analyses are used to represent the ability of the test-taker and the difficulty of the question. A 1PL model uses a latent, random log-odds threshold, $alpha$ to gauge the likelihood of correct response by a test taker. This is called a 1PL model.
In this 1PL dichotomous model, where $alpha = 1$, the zero ability people are the people who only have chance accuracy, or the average ability people (who could have very high accuracy)?
irt
edited 17 mins ago
AdamO
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up vote
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In a test evaluation scenario, latent variable analyses are used to represent the ability of the test-taker and the difficulty of the question. A 1PL model uses a latent, random log-odds threshold, $alpha$ to gauge the likelihood of correct response by a test taker. This is called a 1PL model.
In this 1PL dichotomous model, where $alpha = 1$, the zero ability people are the people who only have chance accuracy, or the average ability people (who could have very high accuracy)?
irt
edited 17 mins ago
AdamO
31.3k256132
asked 3 hours ago
Carrot
61
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Carrot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Please edit your question to make it more clear. You use some technical terms like "1PL" or "alpha" that may be interpreted differently by different people.
– Tim♦
3 hours ago
@Tim - it's always necessary to include some terminology that is specific to the area (IMHO) in a question. If someone doesn't know what 1PL is, they won't know the answer, and people searching for answers about 1PL models will typically search for that term. (Searching 1PL irt on Google takes you to the Wikipedia page for IRT, which explains it.) I edited the question a little to expand it.
– Jeremy Miles
20 mins ago
$alpha = 1$ means the log odds of correct response from a 0 ability (average performer) is `plogis(1) = 0.73. 0 ability is average performance. Average performance is always at least as good as chance accuracy unless questions are set up to solicit the wrong response more often
– AdamO
16 mins ago
@JeremyMiles I know the terminology because I worked with IRT models and had a paper on them. I'm just asking to use wording that makes it accessible for broader audience. Moreover the Greek letters have no objective meanings, so "alpha" could mean different things in some specific context.
– Tim♦
14 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In a test evaluation scenario, latent variable analyses are used to represent the ability of the test-taker and the difficulty of the question. A 1PL model uses a latent, random log-odds threshold, $alpha$ to gauge the likelihood of correct response by a test taker. This is called a 1PL model.
In this 1PL dichotomous model, where $alpha = 1$, the zero ability people are the people who only have chance accuracy, or the average ability people (who could have very high accuracy)?
irt
edited 17 mins ago
AdamO
31.3k256132
asked 3 hours ago
Carrot
61
New contributor
Carrot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In a test evaluation scenario, latent variable analyses are used to represent the ability of the test-taker and the difficulty of the question. A 1PL model uses a latent, random log-odds threshold, $alpha$ to gauge the likelihood of correct response by a test taker. This is called a 1PL model.
In this 1PL dichotomous model, where $alpha = 1$, the zero ability people are the people who only have chance accuracy, or the average ability people (who could have very high accuracy)?
irt
irt
edited 17 mins ago
AdamO
31.3k256132
asked 3 hours ago
Carrot
61
New contributor
Carrot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 mins ago
AdamO
31.3k256132
asked 3 hours ago
Carrot
61
New contributor
Carrot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 mins ago
AdamO
31.3k256132
edited 17 mins ago
AdamO
31.3k256132
edited 17 mins ago
AdamO
31.3k256132
31.3k256132
asked 3 hours ago
Carrot
61
New contributor
Carrot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Carrot
61
asked 3 hours ago
Carrot
61
61
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Carrot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Carrot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Carrot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Please edit your question to make it more clear. You use some technical terms like "1PL" or "alpha" that may be interpreted differently by different people.
– Tim♦
3 hours ago
@Tim - it's always necessary to include some terminology that is specific to the area (IMHO) in a question. If someone doesn't know what 1PL is, they won't know the answer, and people searching for answers about 1PL models will typically search for that term. (Searching 1PL irt on Google takes you to the Wikipedia page for IRT, which explains it.) I edited the question a little to expand it.
– Jeremy Miles
20 mins ago
$alpha = 1$ means the log odds of correct response from a 0 ability (average performer) is `plogis(1) = 0.73. 0 ability is average performance. Average performance is always at least as good as chance accuracy unless questions are set up to solicit the wrong response more often
– AdamO
16 mins ago
@JeremyMiles I know the terminology because I worked with IRT models and had a paper on them. I'm just asking to use wording that makes it accessible for broader audience. Moreover the Greek letters have no objective meanings, so "alpha" could mean different things in some specific context.
– Tim♦
14 mins ago
add a comment |Â
Please edit your question to make it more clear. You use some technical terms like "1PL" or "alpha" that may be interpreted differently by different people.
– Tim♦
3 hours ago
@Tim - it's always necessary to include some terminology that is specific to the area (IMHO) in a question. If someone doesn't know what 1PL is, they won't know the answer, and people searching for answers about 1PL models will typically search for that term. (Searching 1PL irt on Google takes you to the Wikipedia page for IRT, which explains it.) I edited the question a little to expand it.
– Jeremy Miles
20 mins ago
$alpha = 1$ means the log odds of correct response from a 0 ability (average performer) is `plogis(1) = 0.73. 0 ability is average performance. Average performance is always at least as good as chance accuracy unless questions are set up to solicit the wrong response more often
– AdamO
16 mins ago
@JeremyMiles I know the terminology because I worked with IRT models and had a paper on them. I'm just asking to use wording that makes it accessible for broader audience. Moreover the Greek letters have no objective meanings, so "alpha" could mean different things in some specific context.
– Tim♦
14 mins ago
Please edit your question to make it more clear. You use some technical terms like "1PL" or "alpha" that may be interpreted differently by different people.
– Tim♦
3 hours ago
Please edit your question to make it more clear. You use some technical terms like "1PL" or "alpha" that may be interpreted differently by different people.
– Tim♦
3 hours ago
@Tim - it's always necessary to include some terminology that is specific to the area (IMHO) in a question. If someone doesn't know what 1PL is, they won't know the answer, and people searching for answers about 1PL models will typically search for that term. (Searching 1PL irt on Google takes you to the Wikipedia page for IRT, which explains it.) I edited the question a little to expand it.
– Jeremy Miles
20 mins ago
@Tim - it's always necessary to include some terminology that is specific to the area (IMHO) in a question. If someone doesn't know what 1PL is, they won't know the answer, and people searching for answers about 1PL models will typically search for that term. (Searching 1PL irt on Google takes you to the Wikipedia page for IRT, which explains it.) I edited the question a little to expand it.
– Jeremy Miles
20 mins ago
$alpha = 1$ means the log odds of correct response from a 0 ability (average performer) is `plogis(1) = 0.73. 0 ability is average performance. Average performance is always at least as good as chance accuracy unless questions are set up to solicit the wrong response more often
– AdamO
16 mins ago
$alpha = 1$ means the log odds of correct response from a 0 ability (average performer) is `plogis(1) = 0.73. 0 ability is average performance. Average performance is always at least as good as chance accuracy unless questions are set up to solicit the wrong response more often
– AdamO
16 mins ago
@JeremyMiles I know the terminology because I worked with IRT models and had a paper on them. I'm just asking to use wording that makes it accessible for broader audience. Moreover the Greek letters have no objective meanings, so "alpha" could mean different things in some specific context.
– Tim♦
14 mins ago
@JeremyMiles I know the terminology because I worked with IRT models and had a paper on them. I'm just asking to use wording that makes it accessible for broader audience. Moreover the Greek letters have no objective meanings, so "alpha" could mean different things in some specific context.
– Tim♦
14 mins ago
add a comment |Â
2 Answers
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up vote
3
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Assuming that your 1PL definition is
$$P(x = 1 | theta, alpha) = frac11 + exp[-1cdot (theta - alpha)]$$
then no, when $theta = 0$ and $alpha = 1$, $P(x = 1 | theta, alpha) ne 0.5$.
The form of $P(x = 1 | theta, alpha) = 0.5$, commonly referred to as the inflection point, occurs only when $theta = alpha$; in other words, when the difficulty of the item matches the ability of the participant. The is generally why the $alpha$ parameters are referred to as 'difficulty parameters', because larger $alpha$ values clearly require higher ability values before the probability of positive endorsement becomes close to 1.
answered 2 hours ago
philchalmers
2,14211021
add a comment |Â
up vote
2
down vote
The ability parameter zero corresponds to the 50% probability of a correct answer for the average difficult item.
P(item = 1) = exp(theta − beta)/ (1 + exp(theta − beta))
The probability of answering an item is the combination of two independent forces, the subject ability (theta) and item difficulty (beta).
If theta and beta are zero, the probability of getting item right is 0.5.
This happens when the location of the Item Characteristic Curve (ICC) is zero.
Picture from STATA ITEM Response THEORY REFERENCE Manual RELEASE 15
answered 3 hours ago
paoloeusebi
394
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paoloeusebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Can you define what the "average difficult item" is? I don't think this accurately reflects the question
– philchalmers
2 hours ago
Thanks for the update, but I still don't see why $beta = 0$ is the 'average difficulty'. Are you suggesting that the difficulty parameters must be constrained to sum to 0 by construction?
– philchalmers
1 hour ago
Average on a logit scale. Would you like to suggest edits on my answer?
– paoloeusebi
32 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Assuming that your 1PL definition is
$$P(x = 1 | theta, alpha) = frac11 + exp[-1cdot (theta - alpha)]$$
then no, when $theta = 0$ and $alpha = 1$, $P(x = 1 | theta, alpha) ne 0.5$.
The form of $P(x = 1 | theta, alpha) = 0.5$, commonly referred to as the inflection point, occurs only when $theta = alpha$; in other words, when the difficulty of the item matches the ability of the participant. The is generally why the $alpha$ parameters are referred to as 'difficulty parameters', because larger $alpha$ values clearly require higher ability values before the probability of positive endorsement becomes close to 1.
answered 2 hours ago
philchalmers
2,14211021
add a comment |Â
up vote
3
down vote
Assuming that your 1PL definition is
$$P(x = 1 | theta, alpha) = frac11 + exp[-1cdot (theta - alpha)]$$
then no, when $theta = 0$ and $alpha = 1$, $P(x = 1 | theta, alpha) ne 0.5$.
The form of $P(x = 1 | theta, alpha) = 0.5$, commonly referred to as the inflection point, occurs only when $theta = alpha$; in other words, when the difficulty of the item matches the ability of the participant. The is generally why the $alpha$ parameters are referred to as 'difficulty parameters', because larger $alpha$ values clearly require higher ability values before the probability of positive endorsement becomes close to 1.
answered 2 hours ago
philchalmers
2,14211021
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Assuming that your 1PL definition is
$$P(x = 1 | theta, alpha) = frac11 + exp[-1cdot (theta - alpha)]$$
then no, when $theta = 0$ and $alpha = 1$, $P(x = 1 | theta, alpha) ne 0.5$.
The form of $P(x = 1 | theta, alpha) = 0.5$, commonly referred to as the inflection point, occurs only when $theta = alpha$; in other words, when the difficulty of the item matches the ability of the participant. The is generally why the $alpha$ parameters are referred to as 'difficulty parameters', because larger $alpha$ values clearly require higher ability values before the probability of positive endorsement becomes close to 1.
answered 2 hours ago
philchalmers
2,14211021
Assuming that your 1PL definition is
$$P(x = 1 | theta, alpha) = frac11 + exp[-1cdot (theta - alpha)]$$
then no, when $theta = 0$ and $alpha = 1$, $P(x = 1 | theta, alpha) ne 0.5$.
The form of $P(x = 1 | theta, alpha) = 0.5$, commonly referred to as the inflection point, occurs only when $theta = alpha$; in other words, when the difficulty of the item matches the ability of the participant. The is generally why the $alpha$ parameters are referred to as 'difficulty parameters', because larger $alpha$ values clearly require higher ability values before the probability of positive endorsement becomes close to 1.
answered 2 hours ago
philchalmers
2,14211021
answered 2 hours ago
philchalmers
2,14211021
answered 2 hours ago
philchalmers
2,14211021
answered 2 hours ago
philchalmers
2,14211021
2,14211021
add a comment |Â
add a comment |Â
up vote
2
down vote
The ability parameter zero corresponds to the 50% probability of a correct answer for the average difficult item.
P(item = 1) = exp(theta − beta)/ (1 + exp(theta − beta))
The probability of answering an item is the combination of two independent forces, the subject ability (theta) and item difficulty (beta).
If theta and beta are zero, the probability of getting item right is 0.5.
This happens when the location of the Item Characteristic Curve (ICC) is zero.
Picture from STATA ITEM Response THEORY REFERENCE Manual RELEASE 15
answered 3 hours ago
paoloeusebi
394
New contributor
paoloeusebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can you define what the "average difficult item" is? I don't think this accurately reflects the question
– philchalmers
2 hours ago
Thanks for the update, but I still don't see why $beta = 0$ is the 'average difficulty'. Are you suggesting that the difficulty parameters must be constrained to sum to 0 by construction?
– philchalmers
1 hour ago
Average on a logit scale. Would you like to suggest edits on my answer?
– paoloeusebi
32 mins ago
add a comment |Â
up vote
2
down vote
The ability parameter zero corresponds to the 50% probability of a correct answer for the average difficult item.
P(item = 1) = exp(theta − beta)/ (1 + exp(theta − beta))
The probability of answering an item is the combination of two independent forces, the subject ability (theta) and item difficulty (beta).
If theta and beta are zero, the probability of getting item right is 0.5.
This happens when the location of the Item Characteristic Curve (ICC) is zero.
Picture from STATA ITEM Response THEORY REFERENCE Manual RELEASE 15
answered 3 hours ago
paoloeusebi
394
New contributor
paoloeusebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can you define what the "average difficult item" is? I don't think this accurately reflects the question
– philchalmers
2 hours ago
Thanks for the update, but I still don't see why $beta = 0$ is the 'average difficulty'. Are you suggesting that the difficulty parameters must be constrained to sum to 0 by construction?
– philchalmers
1 hour ago
Average on a logit scale. Would you like to suggest edits on my answer?
– paoloeusebi
32 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The ability parameter zero corresponds to the 50% probability of a correct answer for the average difficult item.
P(item = 1) = exp(theta − beta)/ (1 + exp(theta − beta))
The probability of answering an item is the combination of two independent forces, the subject ability (theta) and item difficulty (beta).
If theta and beta are zero, the probability of getting item right is 0.5.
This happens when the location of the Item Characteristic Curve (ICC) is zero.
Picture from STATA ITEM Response THEORY REFERENCE Manual RELEASE 15
answered 3 hours ago
paoloeusebi
394
New contributor
paoloeusebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The ability parameter zero corresponds to the 50% probability of a correct answer for the average difficult item.
P(item = 1) = exp(theta − beta)/ (1 + exp(theta − beta))
The probability of answering an item is the combination of two independent forces, the subject ability (theta) and item difficulty (beta).
If theta and beta are zero, the probability of getting item right is 0.5.
This happens when the location of the Item Characteristic Curve (ICC) is zero.
Picture from STATA ITEM Response THEORY REFERENCE Manual RELEASE 15
answered 3 hours ago
paoloeusebi
394
New contributor
paoloeusebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 30 mins ago
answered 3 hours ago
paoloeusebi
394
New contributor
paoloeusebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 3 hours ago
paoloeusebi
394
answered 3 hours ago
paoloeusebi
394
394
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paoloeusebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
paoloeusebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
paoloeusebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can you define what the "average difficult item" is? I don't think this accurately reflects the question
– philchalmers
2 hours ago
Thanks for the update, but I still don't see why $beta = 0$ is the 'average difficulty'. Are you suggesting that the difficulty parameters must be constrained to sum to 0 by construction?
– philchalmers
1 hour ago
Average on a logit scale. Would you like to suggest edits on my answer?
– paoloeusebi
32 mins ago
add a comment |Â
Can you define what the "average difficult item" is? I don't think this accurately reflects the question
– philchalmers
2 hours ago
Thanks for the update, but I still don't see why $beta = 0$ is the 'average difficulty'. Are you suggesting that the difficulty parameters must be constrained to sum to 0 by construction?
– philchalmers
1 hour ago
Average on a logit scale. Would you like to suggest edits on my answer?
– paoloeusebi
32 mins ago
Can you define what the "average difficult item" is? I don't think this accurately reflects the question
– philchalmers
2 hours ago
Can you define what the "average difficult item" is? I don't think this accurately reflects the question
– philchalmers
2 hours ago
Thanks for the update, but I still don't see why $beta = 0$ is the 'average difficulty'. Are you suggesting that the difficulty parameters must be constrained to sum to 0 by construction?
– philchalmers
1 hour ago
Thanks for the update, but I still don't see why $beta = 0$ is the 'average difficulty'. Are you suggesting that the difficulty parameters must be constrained to sum to 0 by construction?
– philchalmers
1 hour ago
Average on a logit scale. Would you like to suggest edits on my answer?
– paoloeusebi
32 mins ago
Average on a logit scale. Would you like to suggest edits on my answer?
– paoloeusebi
32 mins ago
add a comment |Â
Carrot is a new contributor. Be nice, and check out our Code of Conduct.
Carrot is a new contributor. Be nice, and check out our Code of Conduct.
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Please edit your question to make it more clear. You use some technical terms like "1PL" or "alpha" that may be interpreted differently by different people.
– Tim♦
3 hours ago
@Tim - it's always necessary to include some terminology that is specific to the area (IMHO) in a question. If someone doesn't know what 1PL is, they won't know the answer, and people searching for answers about 1PL models will typically search for that term. (Searching 1PL irt on Google takes you to the Wikipedia page for IRT, which explains it.) I edited the question a little to expand it.
– Jeremy Miles
20 mins ago
$alpha = 1$ means the log odds of correct response from a 0 ability (average performer) is `plogis(1) = 0.73. 0 ability is average performance. Average performance is always at least as good as chance accuracy unless questions are set up to solicit the wrong response more often
– AdamO
16 mins ago
@JeremyMiles I know the terminology because I worked with IRT models and had a paper on them. I'm just asking to use wording that makes it accessible for broader audience. Moreover the Greek letters have no objective meanings, so "alpha" could mean different things in some specific context.
– Tim♦
14 mins ago