To find $\\beta_0$ and $\\beta_1$ you first find $\\displaystyle \\beta_1 = \\frac{SPXY}{SSX}$ where:
\n$\\displaystyle SPXY=\\sum xy - \\frac{(\\sum x)\\times (\\sum y)}{\\var{n}}$
\n$\\displaystyle SSX=\\sum x^2 - \\frac{(\\sum x)^2}{\\var{n}}$
\nThen $\\displaystyle \\beta_0= \\frac{1}{\\var{n}}\\left[\\sum y-\\beta_1 \\sum x\\right]$
\nNow go back and fill in the values for $\\beta_0$ and $\\beta_1$.
\n", "scripts": {}, "marks": 0}], "prompt": "
Calculate the equation of the best fitting regression line:
\n\\[Y = \\beta_0 + \\beta_1 X.\\] Find $\\beta_0$ and $\\beta_1$ to 5 decimal places, then input them below to 3 decimal places. You will use these approximate values in the rest of the question.
\n$\\beta_1=\\;$[[0]], $\\beta_0=\\;$[[1]] (both to 3 decimal places.)
\nYou are given the following information:
\n| First Test$(X)$ | \n$\\sum x=\\;\\var{t[0]}$ | \n$\\sum x^2=\\;\\var{ssq[0]}$ | \n
|---|---|---|
| Later Score$(Y)$ | \n$\\sum y=\\;\\var{t[1]}$ | \n$\\sum y^2=\\;\\var{ssq[1]}$ | \n
Also you are given $\\sum xy = \\var{sxy}$.
\nClick on Show steps if you want more information on calculating $\\beta_0$ and $\\beta_1$. You will not lose any marks by doing so.
\n", "marks": 0}, {"showCorrectAnswer": true, "correctAnswerFraction": false, "showPrecisionHint": false, "scripts": {}, "allowFractions": false, "type": "numberentry", "maxValue": "ls+0.01", "minValue": "ls-0.01", "prompt": "
What is the predicted Later score for employee $\\var{obj[ch]}$ in the First test?
Use the values of $\\beta_0$ and $\\beta_1$ you input above.
\nEnter the predicted Later score here: (to 2 decimal places)
", "marks": 1}, {"stepsPenalty": 0, "scripts": {}, "gaps": [{"correctAnswerFraction": false, "showPrecisionHint": false, "scripts": {}, "allowFractions": false, "type": "numberentry", "showCorrectAnswer": true, "minValue": "res[ch]-0.01", "maxValue": "res[ch]+0.01", "marks": 1}], "type": "gapfill", "showCorrectAnswer": true, "steps": [{"type": "information", "showCorrectAnswer": true, "prompt": "\nThe residual value is given by:
\nRESIDUAL = OBSERVED - FITTED.
\nIn this case the observed value for $\\var{obj[ch]}$ is $\\var{r2[ch]}$ and you get the fitted value by feeding the First test value $\\var{r1[ch]}$ into the regression equation.
\n\n", "scripts": {}, "marks": 0}], "prompt": "
Use the result above to calculate the residual value for employee $\\var{obj[ch]}$.
\nClick on Show steps to see what is meant by the residual value if you have forgotten. You will not lose any marks by doing so.
\nResidual value = (to 2 decimal places).[[0]]
", "marks": 0}], "statement": "\nTo monitor its staff appraisal methods, a personnel department compares the results of the tests carried out on employees at their first appraisal with an assessment score of the same individuals two years later. The resulting data are as follows:
\n| Employee | $\\var{obj[0]}$ | $\\var{obj[1]}$ | $\\var{obj[2]}$ | $\\var{obj[3]}$ | $\\var{obj[4]}$ | $\\var{obj[5]}$ | $\\var{obj[6]}$ | $\\var{obj[7]}$ |
|---|---|---|---|---|---|---|---|---|
| First Test $(X)$ | \n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n$\\var{r1[6]}$ | \n$\\var{r1[7]}$ | \n
| Later Score $(Y)$ | \n$\\var{r2[0]}$ | \n$\\var{r2[1]}$ | \n$\\var{r2[2]}$ | \n$\\var{r2[3]}$ | \n$\\var{r2[4]}$ | \n$\\var{r2[5]}$ | \n$\\var{r2[6]}$ | \n$\\var{r2[7]}$ | \n
21/12/2012:
\nChecked rounding, OK. Added tag cr1.
\nPossible use of scenarios, so added tag sc.
\nThe array r2 generates the regression data which has an inbuilt noise via r1[x]+random(-9..9).
\n", "licence": "None specified", "description": "
Find a regression equation.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": ""}, {"variable_groups": [], "variables": {"tol": {"templateType": "anything", "group": "Ungrouped variables", "definition": "0.001", "name": "tol", "description": ""}, "r1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "repeat(round(normalsample(67,8)),10)", "name": "r1", "description": ""}, "spxy": {"templateType": "anything", "group": "Ungrouped variables", "definition": "sxy-t[0]*t[1]/n", "name": "spxy", "description": ""}, "b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "precround(spxy/ss[0],3)", "name": "b", "description": ""}, "sc": {"templateType": "anything", "group": "Ungrouped variables", "definition": "r1[ch]", "name": "sc", "description": ""}, "ssq": {"templateType": "anything", "group": "Ungrouped variables", "definition": "[sum(map(x^2,x,r1)),sum(map(x^2,x,r2))]", "name": "ssq", "description": ""}, "ch": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(0..9)", "name": "ch", "description": ""}, "b1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(0.25..0.45#0.05)", "name": "b1", "description": ""}, "t": {"templateType": "anything", "group": "Ungrouped variables", "definition": "[sum(r1),sum(r2)]", "name": "t", "description": ""}, "a1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(10..20)", "name": "a1", "description": ""}, "sxy": {"templateType": "anything", "group": "Ungrouped variables", "definition": "sum(map(r1[x]*r2[x],x,0..n-1))", "name": "sxy", "description": ""}, "ss": {"templateType": "anything", "group": "Ungrouped variables", "definition": "[ssq[0]-t[0]^2/n,ssq[1]-t[1]^2/n]", "name": "ss", "description": ""}, "n": {"templateType": "anything", "group": "Ungrouped variables", "definition": "10", "name": "n", "description": ""}, "obj": {"templateType": "anything", "group": "Ungrouped variables", "definition": "['A','B','C','D','E','F','G','H','I','J']", "name": "obj", "description": ""}, "r2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "map(round(a1+b1*x+random(-9..9)),x,r1)", "name": "r2", "description": ""}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "precround(1/n*(t[1]-spxy/ss[0]*t[0]),3)", "name": "a", "description": ""}, "ls": {"templateType": "anything", "group": "Ungrouped variables", "definition": "precround(a+b*sc,2)", "name": "ls", "description": ""}, "res": {"templateType": "anything", "group": "Ungrouped variables", "definition": "map(precround(r2[x]-(a+b*r1[x]),2),x,0..n-1)", "name": "res", "description": ""}, "tsqovern": {"templateType": "anything", "group": "Ungrouped variables", "definition": "[t[0]^2/n,t[1]^2/n]", "name": "tsqovern", "description": ""}}, "ungrouped_variables": ["tsqovern", "a", "b", "obj", "r1", "r2", "ss", "res", "ssq", "n", "a1", "ch", "spxy", "ls", "tol", "t", "sc", "sxy", "b1"], "name": "Find and use linear regression equation for a small sample - PSY2010", "functions": {"pstdev": {"type": "number", "language": "jme", "definition": "sqrt(abs(l)/(abs(l)-1))*stdev(l)", "parameters": [["l", "list"]]}}, "parts": [{"stepsPenalty": 0, "scripts": {}, "gaps": [{"precisionPartialCredit": 0, "allowFractions": false, "showCorrectAnswer": true, "minValue": "b-tol", "maxValue": "b+tol", "precision": "3", "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "correctAnswerFraction": false, "marks": 1}, {"precisionPartialCredit": 0, "allowFractions": false, "showCorrectAnswer": true, "minValue": "a-tol", "maxValue": "a+tol", "precision": "3", "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "correctAnswerFraction": false, "marks": 1}], "type": "gapfill", "showCorrectAnswer": true, "steps": [{"type": "information", "showCorrectAnswer": true, "prompt": "To find $a$ and $b$ you first find $\\displaystyle b = \\frac{SPXY}{SSX}$ where:
\n$\\displaystyle SPXY=\\sum xy - \\frac{(\\sum x)\\times (\\sum y)}{10}$
\n$\\displaystyle SSX=\\sum x^2 - \\frac{(\\sum x)^2}{10}$
\nThen $\\displaystyle a = \\frac{1}{10}\\left[\\sum y-b \\sum x\\right]$
\nNow go back and fill in the values for $a$ and $b$.
", "scripts": {}, "marks": 0}], "prompt": "Calculate the equation of the best fitting regression line:
\n\\[Y = a + b \\times X.\\] Find $a$ and $b$ to 5 decimal places, then input them below to 3 decimal places. You will use these approximate values in the rest of the question.
\n$b=\\;$[[0]], $a=\\;$[[1]] (enter both to 3 decimal places).
\nYou are given the following information:
\n| First Test$(X)$ | \n$\\sum x=\\;\\var{t[0]}$ | \n$\\sum x^2=\\;\\var{ssq[0]}$ | \n
|---|---|---|
| Later Score$(Y)$ | \n$\\sum y=\\;\\var{t[1]}$ | \n$\\sum y^2=\\;\\var{ssq[1]}$ | \n
Also you are given $\\sum xy = \\var{sxy}$.
\nClick on Show steps if you want more information on calculating $a$ and $b$. You will not lose any marks by doing so.
\n", "marks": 0}, {"precisionPartialCredit": 0, "allowFractions": false, "correctAnswerFraction": false, "minValue": "ls-0.01", "prompt": "
What is the predicted Later score for employee $\\var{obj[ch]}$?
Use the values of $a$ and $b$ you input above.
\nEnter the predicted Later score here: (to 2 decimal places)
", "precision": "2", "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "showCorrectAnswer": true, "marks": 1, "maxValue": "ls+0.01"}, {"stepsPenalty": 0, "scripts": {}, "gaps": [{"precisionPartialCredit": 0, "allowFractions": false, "showCorrectAnswer": true, "minValue": "res[ch]-0.01", "maxValue": "res[ch]+0.01", "precision": "2", "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "correctAnswerFraction": false, "marks": 1}], "type": "gapfill", "showCorrectAnswer": true, "steps": [{"type": "information", "showCorrectAnswer": true, "prompt": "The residual value is given by:
\nRESIDUAL = OBSERVED - FITTED.
\nIn this case the observed value for $\\var{obj[ch]}$ is $\\var{r2[ch]}$ and you get the fitted value by feeding the First test value $\\var{r1[ch]}$ into the regression equation.
\n", "scripts": {}, "marks": 0}], "prompt": "
Use the result above to calculate the residual value for employee $\\var{obj[ch]}$.
\nClick on Show steps to see what is meant by the residual value if you have forgotten. You will not lose any marks by doing so.
\nResidual value = (to 2 decimal places).[[0]]
", "marks": 0}], "statement": "To monitor its staff appraisal methods, a personnel department compares the results of the tests carried out on employees at their first appraisal with an assessment score of the same individuals two years later. The resulting data are as follows:
\n| Employee | $\\var{obj[0]}$ | $\\var{obj[1]}$ | $\\var{obj[2]}$ | $\\var{obj[3]}$ | $\\var{obj[4]}$ | $\\var{obj[5]}$ | $\\var{obj[6]}$ | $\\var{obj[7]}$ | $\\var{obj[8]}$ | $\\var{obj[9]}$ |
|---|---|---|---|---|---|---|---|---|---|---|
| First Test $(X)$ | \n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n$\\var{r1[6]}$ | \n$\\var{r1[7]}$ | \n$\\var{r1[8]}$ | \n$\\var{r1[9]}$ | \n
| Later Score $(Y)$ | \n$\\var{r2[0]}$ | \n$\\var{r2[1]}$ | \n$\\var{r2[2]}$ | \n$\\var{r2[3]}$ | \n$\\var{r2[4]}$ | \n$\\var{r2[5]}$ | \n$\\var{r2[6]}$ | \n$\\var{r2[7]}$ | \n$\\var{r2[8]}$ | \n$\\var{r2[9]}$ | \n
30/09/2102:
\n \t\tIntroduced three functions:
\n \t\t1. To produce the ranking of a list of 8 numbers.
\n \t\t2. To produce a list of 8 numbers from a scale of 1..20 which are all distinct.
\n \t\t3. To produce the maximum of the numbers in a list.
\n \t\t4. Given an array such as in 2. to find another such array which has max diff between any two corresponding entries less than a given number. This is to ensure that the two array produced do not differ too much, as the point of the exercise is to show that there is a positive high correlation.
\n \t\t\n \t\t", "licence": "None specified", "description": "
Find a regression equation.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": ""}], "duration": 0.0, "name": "Regression Practice 1", "showQuestionGroupNames": false, "allQuestions": true, "percentpass": 0.0, "shufflequestions": false, "resources": [], "navigation": {"onleave": {"message": "", "action": "none"}, "allowregen": true, "reverse": true, "preventleave": true, "browse": true, "showresultspage": "never", "showfrontpage": true}, "feedback": {"showanswerstate": true, "advicethreshold": 0.0, "showactualmark": true, "allowrevealanswer": true, "showtotalmark": true}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Find and use linear regression equation for a small sample", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"tol": {"group": "Ungrouped variables", "templateType": "anything", "definition": "0.001", "name": "tol", "description": ""}, "r1": {"group": "Ungrouped variables", "templateType": "anything", "definition": "repeat(round(normalsample(67,8)),10)", "name": "r1", "description": ""}, "spxy": {"group": "Ungrouped variables", "templateType": "anything", "definition": "sxy-t[0]*t[1]/n", "name": "spxy", "description": ""}, "a": {"group": "Ungrouped variables", "templateType": "anything", "definition": "precround(1/n*(t[1]-spxy/ss[0]*t[0]),3)", "name": "a", "description": ""}, "ssq": {"group": "Ungrouped variables", "templateType": "anything", "definition": "[sum(map(x^2,x,r1)),sum(map(x^2,x,r2))]", "name": "ssq", "description": ""}, "sc": {"group": "Ungrouped variables", "templateType": "anything", "definition": "r1[ch]", "name": "sc", "description": ""}, "ch": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(0..9)", "name": "ch", "description": ""}, "b1": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(0.25..0.45#0.05)", "name": "b1", "description": ""}, "t": {"group": "Ungrouped variables", "templateType": "anything", "definition": "[sum(r1),sum(r2)]", "name": "t", "description": ""}, "a1": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(10..20)", "name": "a1", "description": ""}, "sxy": {"group": "Ungrouped variables", "templateType": "anything", "definition": "sum(map(r1[x]*r2[x],x,0..n-1))", "name": "sxy", "description": ""}, "ss": {"group": "Ungrouped variables", "templateType": "anything", "definition": "[ssq[0]-t[0]^2/n,ssq[1]-t[1]^2/n]", "name": "ss", "description": ""}, "n": {"group": "Ungrouped variables", "templateType": "anything", "definition": "10", "name": "n", "description": ""}, "obj": {"group": "Ungrouped variables", "templateType": "anything", "definition": "['A','B','C','D','E','F','G','H','I','J']", "name": "obj", "description": ""}, "r2": {"group": "Ungrouped variables", "templateType": "anything", "definition": "map(round(a1+b1*x+random(-9..9)),x,r1)", "name": "r2", "description": ""}, "b": {"group": "Ungrouped variables", "templateType": "anything", "definition": "precround(spxy/ss[0],3)", "name": "b", "description": ""}, "ls": {"group": "Ungrouped variables", "templateType": "anything", "definition": "precround(a+b*sc,2)", "name": "ls", "description": ""}, "res": {"group": "Ungrouped variables", "templateType": "anything", "definition": "map(precround(r2[x]-(a+b*r1[x]),2),x,0..n-1)", "name": "res", "description": ""}, "tsqovern": {"group": "Ungrouped variables", "templateType": "anything", "definition": "[t[0]^2/n,t[1]^2/n]", "name": "tsqovern", "description": ""}}, "ungrouped_variables": ["tsqovern", "a", "b", "obj", "r1", "r2", "ss", "res", "ssq", "n", "a1", "ch", "spxy", "ls", "tol", "t", "sc", "sxy", "b1"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "showQuestionGroupNames": false, "functions": {"pstdev": {"type": "number", "language": "jme", "definition": "sqrt(abs(l)/(abs(l)-1))*stdev(l)", "parameters": [["l", "list"]]}}, "parts": [{"marks": 0, "scripts": {}, "gaps": [{"precisionPartialCredit": 0, "allowFractions": false, "showCorrectAnswer": true, "minValue": "b-tol", "maxValue": "b+tol", "precision": "3", "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "correctAnswerFraction": false, "marks": 1}, {"precisionPartialCredit": 0, "allowFractions": false, "showCorrectAnswer": true, "minValue": "a-tol", "maxValue": "a+tol", "precision": "3", "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "correctAnswerFraction": false, "marks": 1}], "type": "gapfill", "showCorrectAnswer": true, "steps": [{"type": "information", "showCorrectAnswer": true, "prompt": "To find $a$ and $b$ you first find $\\displaystyle b = \\frac{SPXY}{SSX}$ where:
\n$\\displaystyle SPXY=\\sum xy - \\frac{(\\sum x)\\times (\\sum y)}{10}$
\n$\\displaystyle SSX=\\sum x^2 - \\frac{(\\sum x)^2}{10}$
\nThen $\\displaystyle a = \\frac{1}{10}\\left[\\sum y-b \\sum x\\right]$
\nNow go back and fill in the values for $a$ and $b$.
", "scripts": {}, "marks": 0}], "prompt": "Calculate the equation of the best fitting regression line:
\n\\[Y = a + b \\times X.\\] Find $a$ and $b$ to 5 decimal places, then input them below to 3 decimal places. You will use these approximate values in the rest of the question.
\n$b=\\;$[[0]], $a=\\;$[[1]] (enter both to 3 decimal places).
\nYou are given the following information:
\n| First Test$(X)$ | \n$\\sum x=\\;\\var{t[0]}$ | \n$\\sum x^2=\\;\\var{ssq[0]}$ | \n
|---|---|---|
| Later Score$(Y)$ | \n$\\sum y=\\;\\var{t[1]}$ | \n$\\sum y^2=\\;\\var{ssq[1]}$ | \n
Also you are given $\\sum xy = \\var{sxy}$.
\nClick on Show steps if you want more information on calculating $a$ and $b$. You will not lose any marks by doing so.
\n", "stepsPenalty": 0}, {"precisionPartialCredit": 0, "allowFractions": false, "showCorrectAnswer": true, "minValue": "ls-0.01", "prompt": "
What is the predicted Later score for employee $\\var{obj[ch]}$?
Use the values of $a$ and $b$ you input above.
\nEnter the predicted Later score here: (to 2 decimal places)
", "precision": "2", "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "correctAnswerFraction": false, "marks": 1, "maxValue": "ls+0.01"}, {"marks": 0, "scripts": {}, "gaps": [{"precisionPartialCredit": 0, "allowFractions": false, "showCorrectAnswer": true, "minValue": "res[ch]-0.01", "maxValue": "res[ch]+0.01", "precision": "2", "type": "numberentry", "precisionType": "dp", "showPrecisionHint": false, "strictPrecision": false, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "correctAnswerFraction": false, "marks": 1}], "type": "gapfill", "showCorrectAnswer": true, "steps": [{"type": "information", "showCorrectAnswer": true, "prompt": "The residual value is given by:
\nRESIDUAL = OBSERVED - FITTED.
\nIn this case the observed value for $\\var{obj[ch]}$ is $\\var{r2[ch]}$ and you get the fitted value by feeding the First test value $\\var{r1[ch]}$ into the regression equation.
\n", "scripts": {}, "marks": 0}], "prompt": "
Use the result above to calculate the residual value for employee $\\var{obj[ch]}$.
\nClick on Show steps to see what is meant by the residual value if you have forgotten. You will not lose any marks by doing so.
\nResidual value = (to 2 decimal places).[[0]]
", "stepsPenalty": 0}], "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "To monitor its staff appraisal methods, a personnel department compares the results of the tests carried out on employees at their first appraisal with an assessment score of the same individuals two years later. The resulting data are as follows:
\n| Employee | $\\var{obj[0]}$ | $\\var{obj[1]}$ | $\\var{obj[2]}$ | $\\var{obj[3]}$ | $\\var{obj[4]}$ | $\\var{obj[5]}$ | $\\var{obj[6]}$ | $\\var{obj[7]}$ | $\\var{obj[8]}$ | $\\var{obj[9]}$ |
|---|---|---|---|---|---|---|---|---|---|---|
| First Test $(X)$ | \n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n$\\var{r1[6]}$ | \n$\\var{r1[7]}$ | \n$\\var{r1[8]}$ | \n$\\var{r1[9]}$ | \n
| Later Score $(Y)$ | \n$\\var{r2[0]}$ | \n$\\var{r2[1]}$ | \n$\\var{r2[2]}$ | \n$\\var{r2[3]}$ | \n$\\var{r2[4]}$ | \n$\\var{r2[5]}$ | \n$\\var{r2[6]}$ | \n$\\var{r2[7]}$ | \n$\\var{r2[8]}$ | \n$\\var{r2[9]}$ | \n
30/09/2102:
\n \t\tIntroduced three functions:
\n \t\t1. To produce the ranking of a list of 8 numbers.
\n \t\t2. To produce a list of 8 numbers from a scale of 1..20 which are all distinct.
\n \t\t3. To produce the maximum of the numbers in a list.
\n \t\t4. Given an array such as in 2. to find another such array which has max diff between any two corresponding entries less than a given number. This is to ensure that the two array produced do not differ too much, as the point of the exercise is to show that there is a positive high correlation.
\n \t\t\n \t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "
Find a regression equation.
"}, "advice": ""}, {"name": "Linear regression - find line of best fit given summary statistics, ", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"tol": {"group": "Ungrouped variables", "templateType": "anything", "definition": "0.001", "name": "tol", "description": ""}, "r1": {"group": "Ungrouped variables", "templateType": "anything", "definition": "repeat(round(normalsample(67,8)),n)", "name": "r1", "description": ""}, "spxy": {"group": "Ungrouped variables", "templateType": "anything", "definition": "sxy-t[0]*t[1]/n", "name": "spxy", "description": ""}, "a": {"group": "Ungrouped variables", "templateType": "anything", "definition": "precround(1/n*(t[1]-spxy/ss[0]*t[0]),3)", "name": "a", "description": ""}, "ssq": {"group": "Ungrouped variables", "templateType": "anything", "definition": "[sum(map(x^2,x,r1)),sum(map(x^2,x,r2))]", "name": "ssq", "description": ""}, "sc": {"group": "Ungrouped variables", "templateType": "anything", "definition": "r1[ch]", "name": "sc", "description": ""}, "ch": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(0..7)", "name": "ch", "description": ""}, "b1": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(0.25..0.45#0.05)", "name": "b1", "description": ""}, "t": {"group": "Ungrouped variables", "templateType": "anything", "definition": "[sum(r1),sum(r2)]", "name": "t", "description": ""}, "a1": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(10..20)", "name": "a1", "description": ""}, "sxy": {"group": "Ungrouped variables", "templateType": "anything", "definition": "sum(map(r1[x]*r2[x],x,0..n-1))", "name": "sxy", "description": ""}, "ss": {"group": "Ungrouped variables", "templateType": "anything", "definition": "[ssq[0]-t[0]^2/n,ssq[1]-t[1]^2/n]", "name": "ss", "description": ""}, "n": {"group": "Ungrouped variables", "templateType": "anything", "definition": "8", "name": "n", "description": ""}, "obj": {"group": "Ungrouped variables", "templateType": "anything", "definition": "['A','B','C','D','E','F','G','H']", "name": "obj", "description": ""}, "r2": {"group": "Ungrouped variables", "templateType": "anything", "definition": "map(round(a1+b1*x+random(-9..9)),x,r1)", "name": "r2", "description": ""}, "b": {"group": "Ungrouped variables", "templateType": "anything", "definition": "precround(spxy/ss[0],3)", "name": "b", "description": ""}, "ls": {"group": "Ungrouped variables", "templateType": "anything", "definition": "precround(a+b*sc,2)", "name": "ls", "description": ""}, "res": {"group": "Ungrouped variables", "templateType": "anything", "definition": "map(precround(r2[x]-(a+b*r1[x]),2),x,0..n-1)", "name": "res", "description": ""}, "tsqovern": {"group": "Ungrouped variables", "templateType": "anything", "definition": "[t[0]^2/n,t[1]^2/n]", "name": "tsqovern", "description": ""}}, "ungrouped_variables": ["tsqovern", "a", "b", "obj", "r1", "r2", "ss", "res", "ssq", "ls", "n", "a1", "ch", "spxy", "t", "tol", "sc", "sxy", "b1"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "showQuestionGroupNames": false, "functions": {"pstdev": {"type": "number", "language": "jme", "definition": "sqrt(abs(l)/(abs(l)-1))*stdev(l)", "parameters": [["l", "list"]]}}, "parts": [{"marks": 0, "scripts": {}, "gaps": [{"correctAnswerFraction": false, "showPrecisionHint": false, "scripts": {}, "allowFractions": false, "type": "numberentry", "showCorrectAnswer": true, "minValue": "b-tol", "maxValue": "b+tol", "marks": 1}, {"correctAnswerFraction": false, "showPrecisionHint": false, "scripts": {}, "allowFractions": false, "type": "numberentry", "showCorrectAnswer": true, "minValue": "a-tol", "maxValue": "a+tol", "marks": 1}], "type": "gapfill", "showCorrectAnswer": true, "steps": [{"type": "information", "showCorrectAnswer": true, "prompt": "To find $\\beta_0$ and $\\beta_1$ you first find $\\displaystyle \\beta_1 = \\frac{SPXY}{SSX}$ where:
\n$\\displaystyle SPXY=\\sum xy - \\frac{(\\sum x)\\times (\\sum y)}{\\var{n}}$
\n$\\displaystyle SSX=\\sum x^2 - \\frac{(\\sum x)^2}{\\var{n}}$
\nThen $\\displaystyle \\beta_0= \\frac{1}{\\var{n}}\\left[\\sum y-\\beta_1 \\sum x\\right]$
\nNow go back and fill in the values for $\\beta_0$ and $\\beta_1$.
\n", "scripts": {}, "marks": 0}], "prompt": "
Calculate the equation of the best fitting regression line:
\n\\[Y = \\beta_0 + \\beta_1 X.\\] Find $\\beta_0$ and $\\beta_1$ to 5 decimal places, then input them below to 3 decimal places. You will use these approximate values in the rest of the question.
\n$\\beta_1=\\;$[[0]], $\\beta_0=\\;$[[1]] (both to 3 decimal places.)
\nYou are given the following information:
\n| First Test$(X)$ | \n$\\sum x=\\;\\var{t[0]}$ | \n$\\sum x^2=\\;\\var{ssq[0]}$ | \n
|---|---|---|
| Later Score$(Y)$ | \n$\\sum y=\\;\\var{t[1]}$ | \n$\\sum y^2=\\;\\var{ssq[1]}$ | \n
Also you are given $\\sum xy = \\var{sxy}$.
\nClick on Show steps if you want more information on calculating $\\beta_0$ and $\\beta_1$. You will not lose any marks by doing so.
\n", "stepsPenalty": 0}, {"showCorrectAnswer": true, "showPrecisionHint": false, "scripts": {}, "allowFractions": false, "type": "numberentry", "prompt": "
What is the predicted Later score for employee $\\var{obj[ch]}$ in the First test?
Use the values of $\\beta_0$ and $\\beta_1$ you input above.
\nEnter the predicted Later score here: (to 2 decimal places)
", "minValue": "ls-0.01", "correctAnswerFraction": false, "marks": 1, "maxValue": "ls+0.01"}, {"marks": 0, "scripts": {}, "gaps": [{"correctAnswerFraction": false, "showPrecisionHint": false, "scripts": {}, "allowFractions": false, "type": "numberentry", "showCorrectAnswer": true, "minValue": "res[ch]-0.01", "maxValue": "res[ch]+0.01", "marks": 1}], "type": "gapfill", "showCorrectAnswer": true, "steps": [{"type": "information", "showCorrectAnswer": true, "prompt": "\nThe residual value is given by:
\nRESIDUAL = OBSERVED - FITTED.
\nIn this case the observed value for $\\var{obj[ch]}$ is $\\var{r2[ch]}$ and you get the fitted value by feeding the First test value $\\var{r1[ch]}$ into the regression equation.
\n\n", "scripts": {}, "marks": 0}], "prompt": "
Use the result above to calculate the residual value for employee $\\var{obj[ch]}$.
\nClick on Show steps to see what is meant by the residual value if you have forgotten. You will not lose any marks by doing so.
\nResidual value = (to 2 decimal places).[[0]]
", "stepsPenalty": 0}], "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "\nTo monitor its staff appraisal methods, a personnel department compares the results of the tests carried out on employees at their first appraisal with an assessment score of the same individuals two years later. The resulting data are as follows:
\n| Employee | $\\var{obj[0]}$ | $\\var{obj[1]}$ | $\\var{obj[2]}$ | $\\var{obj[3]}$ | $\\var{obj[4]}$ | $\\var{obj[5]}$ | $\\var{obj[6]}$ | $\\var{obj[7]}$ |
|---|---|---|---|---|---|---|---|---|
| First Test $(X)$ | \n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n$\\var{r1[6]}$ | \n$\\var{r1[7]}$ | \n
| Later Score $(Y)$ | \n$\\var{r2[0]}$ | \n$\\var{r2[1]}$ | \n$\\var{r2[2]}$ | \n$\\var{r2[3]}$ | \n$\\var{r2[4]}$ | \n$\\var{r2[5]}$ | \n$\\var{r2[6]}$ | \n$\\var{r2[7]}$ | \n
21/12/2012:
\nChecked rounding, OK. Added tag cr1.
\nPossible use of scenarios, so added tag sc.
\nThe array r2 generates the regression data which has an inbuilt noise via r1[x]+random(-9..9).
\n", "licence": "Creative Commons Attribution 4.0 International", "description": "
Find a regression equation.
"}, "advice": ""}], "name": "", "pickQuestions": 0}], "metadata": {"notes": "", "licence": "Creative Commons Attribution 4.0 International", "description": "Statistics and probability. 2 questions. Both simple regression. First with 8 data points, second with 10. Find $a$ and $b$ such that $Y=a+bX$. Then find the residual value for one of the data points.
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