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Some questions which demonstrate the adaptive marking feature.

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A quadratic function $f(x)$ goes through the points $(\\var{xs[0]},\\var{ys[0]})$, $(\\var{xs[1]},\\var{ys[1]})$ and $(\\var{xs[2]},\\var{ys[2]})$.

\n

Find the two roots of $f$.

\n

First root: [[0]]

\n

Second root: [[1]]

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Write a quadratic function $g(x)$ whose roots are the values you entered above, with $x^2$ coefficient $1$.

\n

$g(x) = $ [[0]]

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What is the midpoint of the two roots of $f$?

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(This question doesn't make a lot of pedagogic sense, it just shows off how adaptive marking works)

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Give the student three points lying on a quadratic, and ask them to find the roots.

\n

Then ask them to find the equation of the quadratic, using their roots. Error in calculating the roots is carried forward.

\n

Finally, ask them to find the midpoint of the roots (just for fun). Error is carried forward again.

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Greatest root of the quadratic

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Three given x coordinates

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Least root of the quadratic equation

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y coordinates corresponding to the given x coordinates

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Angle of rotation in radians

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Rotation matrix for the given rotation

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Angle of rotation in degrees

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Write a matrix $\\mathrm{M}$ corresponding to a rotation of $\\var{rotation}^{\\circ}$ anti-clockwise about the origin.

\n

Round numbers to 2 decimal places.

\n

$\\mathrm{M} = $ [[0]]

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$\\operatorname{det}\\mathrm{M} = $ [[0]] (Round your answer to 2 decimal places)

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$\\mathrm{M}^{-1} = $ [[0]]

\n

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Ask the student to find a matrix corresponding to a given rotation about the origin.

\n

Then ask them to find the determinant. Their answer is marked against the matrix they gave, not just the correct one.

\n

Finally, ask them to find the inverse of their matrix. Marking is against the matrix and determinant they gave.

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Two sample t-test to see if there is a difference between scores on questions between two groups when the questions are asked in a different order.

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An educational psychologist claimed that the order in which questions were asked affected the student’s ability to answer them correctly and hence their total score. In order to test this, $20$ students were randomly divided into two groups of $10$. The first group were given questions in increasing order of difficulty and the second group in decreasing order of difficulty. The ordered test scores obtained were:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Group 1{r1[0]}{r1[1]}{r1[2]}{r1[3]}{r1[4]}{r1[5]}{r1[6]}{r1[7]}{r1[8]}{r1[9]}
Group 2{r2[0]}{r2[1]}{r2[2]}{r2[3]}{r2[4]}{r2[5]}{r2[6]}{r2[7]}{r2[8]}{r2[9]}
\n

Carry out a two-sample t-test to decide if there is evidence of a difference in the average test scores for the two sets of students.

", "advice": "

We test the following hypothesis,

\n

$H_0:\\; \\mu_1=\\mu_2$ versus $H_1:\\; \\mu_1 \\neq \\mu_2$

\n

We find that the mean score of Group 1 is $\\overline{x}_1=\\var{mean1}$ with standard deviation $s_1=\\var{sd1}$ and the mean score of Group 2 is $\\overline{x}_2=\\var{mean2}$ with standard deviation $s_2=\\var{sd2}$.

\n

(All calculated to 3 decimal places.)

\n

Using the formula for the two-sample $t$-statistic as  shown above with $n_1=n_2=10$:

\n

The estimate of the pooled variance is calculated to be:

\n

\\[s^2=\\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}= \\frac{\\var{n1-1}\\times \\var{sd1}^2+\\var{n2-1}\\times \\var{sd2}^2}{\\var{n1+n2-2}}=\\var{s^2}.\\] 

\n

Hence $s = \\sqrt{\\var{s^2}}=\\var{s}$ to 3 decimal places.

\n

We find that the t-statistic has value:

\n

\\begin{align}
T &= \\frac{(\\overline{x}_1-\\overline{x}_2)-(\\mu_1-\\mu_2)}{s\\sqrt{\\frac{1}{n_1}+\\frac{1}{n_2}}} \\\\
&= \\frac{(\\var{mean1}-\\var{mean2})-(0)}{\\var{s}\\sqrt{\\frac{1}{\\var{n1}}+\\frac{1}{\\var{n2}}}} \\\\
&= \\var{t_statistic}
\\end{align}

\n

Our test statistic is $|T|=\\var{abs(t_statistic)}$.

\n

Given that we have $n_1+n_2-2=18$ degrees of freedom, we look up this value on the T-distribution table for $t_{18}$

\n

\\[\\begin{array}{r|rrrrr}&0.10&0.05&0.01&0.001\\\\\\hline18&1.734&2.101&2.878&3.922\\end{array}\\]

\n

We see that the t-statistic {t_statistic_range} and the table tells us that the $p$ value {p_value_range}.

\n

Hence we conclude that we {reject} the null hypothesis. There is {evidence_strength} evidence of a difference between the average scores of the two groups.

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Describe where the p-value lies in relation to the critical values

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Population standard deviation of sample 2

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Size of sample 2

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Describe where the t-statistic lies in relation to the critical values

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Population mean of sample 2

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How much evidence is there against the null hypothesis?

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Size of sample 1

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Population mean of sample 1 (we'll generate samples from different distributions to produce different outcomes)

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Do we reject the null hypothesis?

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p-value corresponding to the t-statistic

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Marking matrix for the multiple choice questions

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Sample standard deviation of sample 2

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Sample 1

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Sample mean of sample 1

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Sample mean of sample 1

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Sample 2

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Population standard deviation of sample 1

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Sample standard deviation of sample 1

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Used in the formula for the t statistic

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Find the mean and standard deviations of the scores of the two groups. Round your answers to 3 decimal places.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
MeanStandard deviation
Group 1[[0]][[1]]
Group 2[[2]][[3]]
\n

Now find the two sample t-test statistic $T$ using the values you have just calculated and enter it here: [[4]]

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The two-sample t-statistic for two independent sets of data where one set has $n_1$ data points and the other set $n_2$ data points is calculated as follows:

\n

\\[T = \\frac{(\\overline{x}_1-\\overline{x}_2)-(\\mu_1-\\mu_2)}{s\\times\\sqrt{\\frac{1}{n_1}+\\frac{1}{n_2}}}\\;\\;\\;\\]

\n

where $\\overline{x}_1,\\;\\overline{x}_2$ are the sample means and 

\n

\\[s^2=\\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}\\]

\n

where $s_1,\\;s_2$ are the sample standard deviations.

\n

Use the values you calculated to 3 decimal places in order to find $T$.

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You have not given your answer to the correct precision.

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Given the value $|T|$ of the t-statistic you have found, choose the range for the $p$ value by looking up the t tables:

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$p$ is less than $0.1\\%$

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$p$ lies between $0.1\\%$ and $1\\%$

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$p$ lies between $1 \\%$ and $5\\%$

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$p$ lies between $5 \\%$ and $10\\%$

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$p$ is greater than $10\\%$

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Given the $p$-value and the range you have found, what is the strength of evidence against the null hypothesis that there is no difference in the average times for the left and right hands?

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Very Strong Evidence

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Strong Evidence

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Evidence

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Weak Evidence

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No Evidence

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What do you decide based on the above analysis?

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We reject the null hypothesis at the $0.1\\%$ level

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We reject the null hypothesis at the $1\\%$ level.

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We reject the null hypothesis at the $5\\%$ level.

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We do not reject the null hypothesis but consider further investigation.

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We do not reject the null hypothesis.

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Given the formula for BMI, students are asked to determine a patient's BMI given their height in cm (as people usually do) and their mass in kg.

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Calculate the following with the use of a calculator.

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\n

The formula for BMI (Body Mass Index) is

\n

\\[\\text{BMI}=\\frac{\\text{mass (kg)}}{\\text{height (m)}^2}.\\] 

\n

The patient's BMI to two decimal places is [[0]].

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The equation for BMI is

\n

\\[\\text{BMI}=\\frac{\\text{mass (kg)}}{\\text{height (m)}^2}\\] 

\n

Notice the units in the formula are kilograms and metres. We convert the $\\var{height}$ cm into metres by moving the decimal point twice to the left to get $\\var{height/100}$ m. 

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Using a scientific calculator you can type in \\[\\var{mass}\\div \\var{height/100}^2\\] or by using the fraction button you may be able to type in \\[\\frac{\\var{mass}}{\\var{height/100}^2}\\] and it will give you something like $\\var{bmi6dec}$ which you need to round {upordown} to two decimal places and get $\\var{bmi}$ $\\var{bmi}0$ $\\var{bmi}.00$

\n

\n

\n

A few things to remark:

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false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "tags": ["concentration", "ephlth", "nursing"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "", "rulesets": {}, "variables": {"marking": {"name": "marking", "group": "b", "definition": "[if(mult=1,1,0),if(mult=2,1,0),0,if(mult=10,1,0)]", "description": "", "templateType": "anything"}, "ans2": {"name": "ans2", "group": "b", "definition": "mult*100", "description": "", "templateType": "anything"}, "gramsb": {"name": "gramsb", "group": "whole numbers that result in simple decimal", "definition": "ansb*2*k", "description": "", "templateType": "anything"}, "grams": {"name": "grams", "group": "Ungrouped variables", "definition": "random(0.1..20#0.1)", "description": "", "templateType": "anything"}, "k": {"name": "k", "group": 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"extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

A {millnice} mL solution is made by mixing {gramsnice} g of {chem[2]} and water. The percentage weight per volume of the solution is [[0]] % w/v of {chem[2]}.

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Percentage weight per volume is the weight of the solute in grams divided by the amount of solution in millilitres expressed as a percentage. That is, we do the division and multiply by 100.

\n

\n

In particular, if a {millnice} mL solution is made by mixing {gramsnice} g of {chem[2]} and water. The percentage weight per volume of the solution is:

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$\\dfrac{\\text{grams}}{\\text{millilitres}}\\times 100$$=$$\\dfrac{\\var{gramsnice}}{\\var{millnice}}\\times 100$ % w/v
$=$$\\var{ansnice}$ % w/v 
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A {millb} mL solution is made by mixing {gramsb} g of {chem[3]} and water. The percentage weight per volume of the solution is [[0]] % w/v of {chem[3]}.

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Percentage weight per volume is the weight of the solute in grams divided by the amount of solution in millilitres expressed as a percentage. That is, we do the division and multiply by 100.

\n

\n

In particular, if a {millb} mL solution is made by mixing {gramsb} g of {chem[3]} and water. The percentage weight per volume of the solution is:

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$\\dfrac{\\text{grams}}{\\text{millilitres}}\\times 100$$=$$\\dfrac{\\var{gramsb}}{\\var{millb}}\\times 100$ % w/v
$=$$\\var{ansb}$ % w/v 
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A {mill} mL solution is made by mixing {grams} g of {chem[0]} and water. The percentage weight per volume of the solution is [[0]] % w/v of {chem[0]} (to 2 decimal places).

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Percentage weight per volume is the weight of the solute in grams divided by the amount of solution in millilitres expressed as a percentage. That is, we do the division and multiply by 100.

\n

\n

In particular, if a {mill} mL solution is made by mixing {grams} g of {chem[0]} and water. The percentage weight per volume of the solution is:

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$\\dfrac{\\text{grams}}{\\text{millilitres}}\\times 100$$=$$\\dfrac{\\var{grams}}{\\var{mill}}\\times 100$ % w/v
$=$$\\var{ans}$ % w/v  (to 2 dec. pl)
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A {per} % {chem[1]} solution contains {amount} g of {chem[1]} in every...

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It is important to know {per} % {chem[1]} solution contains {per} g of {chem[1]} in every 100 mL of solution.

\n

\n

Given this, since we have {amount} g of {chem[1]} we have two times the standard amount and so we must have two times the solution, that is 200 mL.

\n

\n

Given this, since we have {amount} g of {chem[1]} we have ten times the standard amount and so we must have ten times the solution, that is 1000 mL.

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100 mL.

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200 mL.

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500 mL.

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1000 mL.

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