Mixing
Devising a grading scheme that satisfies two tests
Suppose two students take a $10$-question-long True/False quiz. One point is awarded for each correct answer.
Student $1$'s ten answers are given by $(F, T, F, T, T, T, F, T, T, F)$, where $F$ denotes false and $T$ denotes true. He receives a grade of $2/3$ (as a percentage, that is, $66.66%$).
Student $2$'s ten answers are given by $(T, T, F, T, F, F, T, F, F, T)$. They receive a score of $1/6$ (as a percentage, that is, $0.16666%$).
What's a possible grading scheme that allows this to happen? Obviously, there has to be some sort of penalty for wrong answers, or maybe a penalty for wrong true answers only (and no penalty on wrong false answers)? I'm pretty sure the penalty will be $-1/3$ too, just from messing around.
I can't fit any grading scheme to satisfy this. Can someone please help me come up with one?
algebra-precalculus puzzle
asked Nov 20 '18 at 5:35
joseph
4329
add a comment |
Suppose two students take a $10$-question-long True/False quiz. One point is awarded for each correct answer.
Student $1$'s ten answers are given by $(F, T, F, T, T, T, F, T, T, F)$, where $F$ denotes false and $T$ denotes true. He receives a grade of $2/3$ (as a percentage, that is, $66.66%$).
Student $2$'s ten answers are given by $(T, T, F, T, F, F, T, F, F, T)$. They receive a score of $1/6$ (as a percentage, that is, $0.16666%$).
What's a possible grading scheme that allows this to happen? Obviously, there has to be some sort of penalty for wrong answers, or maybe a penalty for wrong true answers only (and no penalty on wrong false answers)? I'm pretty sure the penalty will be $-1/3$ too, just from messing around.
I can't fit any grading scheme to satisfy this. Can someone please help me come up with one?
algebra-precalculus puzzle
asked Nov 20 '18 at 5:35
joseph
4329
Student one gets 2/3, student 2 gets 1/6 and from what you've told us, other students get a random rational number in [0,1].
– William Elliot
Nov 20 '18 at 10:38
add a comment |
Suppose two students take a $10$-question-long True/False quiz. One point is awarded for each correct answer.
Student $1$'s ten answers are given by $(F, T, F, T, T, T, F, T, T, F)$, where $F$ denotes false and $T$ denotes true. He receives a grade of $2/3$ (as a percentage, that is, $66.66%$).
Student $2$'s ten answers are given by $(T, T, F, T, F, F, T, F, F, T)$. They receive a score of $1/6$ (as a percentage, that is, $0.16666%$).
What's a possible grading scheme that allows this to happen? Obviously, there has to be some sort of penalty for wrong answers, or maybe a penalty for wrong true answers only (and no penalty on wrong false answers)? I'm pretty sure the penalty will be $-1/3$ too, just from messing around.
I can't fit any grading scheme to satisfy this. Can someone please help me come up with one?
algebra-precalculus puzzle
asked Nov 20 '18 at 5:35
joseph
4329
Suppose two students take a $10$-question-long True/False quiz. One point is awarded for each correct answer.
Student $1$'s ten answers are given by $(F, T, F, T, T, T, F, T, T, F)$, where $F$ denotes false and $T$ denotes true. He receives a grade of $2/3$ (as a percentage, that is, $66.66%$).
Student $2$'s ten answers are given by $(T, T, F, T, F, F, T, F, F, T)$. They receive a score of $1/6$ (as a percentage, that is, $0.16666%$).
What's a possible grading scheme that allows this to happen? Obviously, there has to be some sort of penalty for wrong answers, or maybe a penalty for wrong true answers only (and no penalty on wrong false answers)? I'm pretty sure the penalty will be $-1/3$ too, just from messing around.
I can't fit any grading scheme to satisfy this. Can someone please help me come up with one?
algebra-precalculus puzzle
algebra-precalculus puzzle
asked Nov 20 '18 at 5:35
joseph
4329
asked Nov 20 '18 at 5:35
joseph
4329
asked Nov 20 '18 at 5:35
joseph
4329
asked Nov 20 '18 at 5:35
joseph
4329
asked Nov 20 '18 at 5:35
joseph
4329
4329
Student one gets 2/3, student 2 gets 1/6 and from what you've told us, other students get a random rational number in [0,1].
– William Elliot
Nov 20 '18 at 10:38
add a comment |
Student one gets 2/3, student 2 gets 1/6 and from what you've told us, other students get a random rational number in [0,1].
– William Elliot
Nov 20 '18 at 10:38
Student one gets 2/3, student 2 gets 1/6 and from what you've told us, other students get a random rational number in [0,1].
– William Elliot
Nov 20 '18 at 10:38
Student one gets 2/3, student 2 gets 1/6 and from what you've told us, other students get a random rational number in [0,1].
– William Elliot
Nov 20 '18 at 10:38
add a comment |
2 Answers
2
active
oldest
votes
If student A got one wrong, the wrong answer would have to score $-8/3$. But in that case student B can have at most four right (they disagree on seven), and so would score negative.
If student A got two wrong, the wrong answers count $-2/3$ each. This is consistent with student B getting five right and five wrong, for an overall score of $5/3$. For example, this would be the case if the correct answers were F,T,F,T,T,T,F,T,F,T.
answered Nov 20 '18 at 15:10
Especially Lime
21.6k22858
add a comment |
As a linear system, we do not have enough information to obtain a unique answer**. Let $x_i$ be the number correct for student $i$. Then assuming a penalty $p$ for incorrect answers, (and a linear penalty relationship,) the knowledge we have is $$x_1 - p(10-x_1) = 1/6 \x_2 - p(10-x_2) = 4/6$$The goal is to solve for $p$, but we have two equations and three unknowns.
The other piece of information we have is that the second through fourth answers are the same between both students, and all other questions are opposites. Maybe someone can derive a relationship from that, finding combinations where the coefficients evenly divide the numerator(s)?
**If this is a homework problem, it should be sufficient to determine one of several possible models; however, if a student is considering arguing a grade with the professor, and maybe missed the review session where the scoring was discussed but is too embarrassed to admit that, it's important that we find one grading rubric. :-)
answered Nov 20 '18 at 15:00
Russ
37919
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005982%2fdevising-a-grading-scheme-that-satisfies-two-tests%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If student A got one wrong, the wrong answer would have to score $-8/3$. But in that case student B can have at most four right (they disagree on seven), and so would score negative.
If student A got two wrong, the wrong answers count $-2/3$ each. This is consistent with student B getting five right and five wrong, for an overall score of $5/3$. For example, this would be the case if the correct answers were F,T,F,T,T,T,F,T,F,T.
answered Nov 20 '18 at 15:10
Especially Lime
21.6k22858
add a comment |
If student A got one wrong, the wrong answer would have to score $-8/3$. But in that case student B can have at most four right (they disagree on seven), and so would score negative.
If student A got two wrong, the wrong answers count $-2/3$ each. This is consistent with student B getting five right and five wrong, for an overall score of $5/3$. For example, this would be the case if the correct answers were F,T,F,T,T,T,F,T,F,T.
answered Nov 20 '18 at 15:10
Especially Lime
21.6k22858
add a comment |
If student A got one wrong, the wrong answer would have to score $-8/3$. But in that case student B can have at most four right (they disagree on seven), and so would score negative.
If student A got two wrong, the wrong answers count $-2/3$ each. This is consistent with student B getting five right and five wrong, for an overall score of $5/3$. For example, this would be the case if the correct answers were F,T,F,T,T,T,F,T,F,T.
answered Nov 20 '18 at 15:10
Especially Lime
21.6k22858
If student A got one wrong, the wrong answer would have to score $-8/3$. But in that case student B can have at most four right (they disagree on seven), and so would score negative.
If student A got two wrong, the wrong answers count $-2/3$ each. This is consistent with student B getting five right and five wrong, for an overall score of $5/3$. For example, this would be the case if the correct answers were F,T,F,T,T,T,F,T,F,T.
answered Nov 20 '18 at 15:10
Especially Lime
21.6k22858
answered Nov 20 '18 at 15:10
Especially Lime
21.6k22858
answered Nov 20 '18 at 15:10
Especially Lime
21.6k22858
answered Nov 20 '18 at 15:10
Especially Lime
21.6k22858
21.6k22858
add a comment |
add a comment |
As a linear system, we do not have enough information to obtain a unique answer**. Let $x_i$ be the number correct for student $i$. Then assuming a penalty $p$ for incorrect answers, (and a linear penalty relationship,) the knowledge we have is $$x_1 - p(10-x_1) = 1/6 \x_2 - p(10-x_2) = 4/6$$The goal is to solve for $p$, but we have two equations and three unknowns.
The other piece of information we have is that the second through fourth answers are the same between both students, and all other questions are opposites. Maybe someone can derive a relationship from that, finding combinations where the coefficients evenly divide the numerator(s)?
**If this is a homework problem, it should be sufficient to determine one of several possible models; however, if a student is considering arguing a grade with the professor, and maybe missed the review session where the scoring was discussed but is too embarrassed to admit that, it's important that we find one grading rubric. :-)
answered Nov 20 '18 at 15:00
Russ
37919
add a comment |
As a linear system, we do not have enough information to obtain a unique answer**. Let $x_i$ be the number correct for student $i$. Then assuming a penalty $p$ for incorrect answers, (and a linear penalty relationship,) the knowledge we have is $$x_1 - p(10-x_1) = 1/6 \x_2 - p(10-x_2) = 4/6$$The goal is to solve for $p$, but we have two equations and three unknowns.
The other piece of information we have is that the second through fourth answers are the same between both students, and all other questions are opposites. Maybe someone can derive a relationship from that, finding combinations where the coefficients evenly divide the numerator(s)?
**If this is a homework problem, it should be sufficient to determine one of several possible models; however, if a student is considering arguing a grade with the professor, and maybe missed the review session where the scoring was discussed but is too embarrassed to admit that, it's important that we find one grading rubric. :-)
answered Nov 20 '18 at 15:00
Russ
37919
add a comment |
As a linear system, we do not have enough information to obtain a unique answer**. Let $x_i$ be the number correct for student $i$. Then assuming a penalty $p$ for incorrect answers, (and a linear penalty relationship,) the knowledge we have is $$x_1 - p(10-x_1) = 1/6 \x_2 - p(10-x_2) = 4/6$$The goal is to solve for $p$, but we have two equations and three unknowns.
The other piece of information we have is that the second through fourth answers are the same between both students, and all other questions are opposites. Maybe someone can derive a relationship from that, finding combinations where the coefficients evenly divide the numerator(s)?
**If this is a homework problem, it should be sufficient to determine one of several possible models; however, if a student is considering arguing a grade with the professor, and maybe missed the review session where the scoring was discussed but is too embarrassed to admit that, it's important that we find one grading rubric. :-)
answered Nov 20 '18 at 15:00
Russ
37919
As a linear system, we do not have enough information to obtain a unique answer**. Let $x_i$ be the number correct for student $i$. Then assuming a penalty $p$ for incorrect answers, (and a linear penalty relationship,) the knowledge we have is $$x_1 - p(10-x_1) = 1/6 \x_2 - p(10-x_2) = 4/6$$The goal is to solve for $p$, but we have two equations and three unknowns.
The other piece of information we have is that the second through fourth answers are the same between both students, and all other questions are opposites. Maybe someone can derive a relationship from that, finding combinations where the coefficients evenly divide the numerator(s)?
**If this is a homework problem, it should be sufficient to determine one of several possible models; however, if a student is considering arguing a grade with the professor, and maybe missed the review session where the scoring was discussed but is too embarrassed to admit that, it's important that we find one grading rubric. :-)
answered Nov 20 '18 at 15:00
Russ
37919
edited Nov 20 '18 at 15:06
answered Nov 20 '18 at 15:00
Russ
37919
answered Nov 20 '18 at 15:00
Russ
37919
answered Nov 20 '18 at 15:00
Russ
37919
37919
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005982%2fdevising-a-grading-scheme-that-satisfies-two-tests%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Student one gets 2/3, student 2 gets 1/6 and from what you've told us, other students get a random rational number in [0,1].
– William Elliot
Nov 20 '18 at 10:38