The table of differences is given by:
\n| {object} | A | B | C | D | E | F | G | H | I | J | K | L | M | N | O |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Before | \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$\\var{r1[10]}$ | \n$\\var{r1[11]}$ | \n$\\var{r1[12]}$ | \n$\\var{r1[13]}$ | \n$\\var{r1[14]}$ | \n
| After | \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$\\var{r2[10]}$ | \n$\\var{r2[11]}$ | \n$\\var{r2[12]}$ | \n$\\var{r2[13]}$ | \n$\\var{r2[14]}$ | \n
| Differences | \n$\\var{d[0]}$ | \n$\\var{d[1]}$ | \n$\\var{d[2]}$ | \n$\\var{d[3]}$ | \n$\\var{d[4]}$ | \n$\\var{d[5]}$ | \n$\\var{d[6]}$ | \n$\\var{d[7]}$ | \n$\\var{d[8]}$ | \n$\\var{d[9]}$ | \n$\\var{d[10]}$ | \n$\\var{d[11]}$ | \n$\\var{d[12]}$ | \n$\\var{d[13]}$ | \n$\\var{d[14]}$ | \n
The mean of the differences is $\\overline{x_d}=\\var{meandiff}$.
\nThe variance $V$ of the differences is
\\[\\begin{eqnarray*} V&=&\\frac{1}{14}\\left(\\simplify[]{{d[0]^2}+{d[1]^2}+{d[2]^2}+{d[3]^2}+{d[4]^2}+{d[5]^2}+{d[6]^2}+{d[7]^2}+{d[8]^2}+{d[9]^2}+{d[10]^2}+{d[11]^2}+{d[12]^2}+{d[13]^2}+{d[14]^2}}-15\\times \\var{meandiff}^2\\right)\\\\ &=&\\var{variance(d)} \\end{eqnarray*} \\]
Hence we have the standard deviation $s_d= \\sqrt{V}=\\var{stdiff}$ to 3 decimal places.
The paired t-statistic is given by:
\n\\[t_d=\\frac{\\overline{x_d}-\\mu_d}{\\frac{s_d}{\\sqrt{n}}}\\]
\nIn this example $n=15$ and the null hypothesis is that the means are the same i.e. $\\mu_d=0$.
\nOn calculating we find $t_d=\\var{tvalue}$.
\nLooking up this value on the T-distribution table for $t_{14}$
\n\\[\\begin{array}{r|rrrrr}&0.20&0.10&0.05&0.01&0.001\\\\\\hline14&1.345&1.761&2.145&2.977&4.140\\end{array}\\]
\nWe see that the t-statistic {msg[t]} and the table tells us that the $p$ value {pmsg[t]}.
\nHence we conclude that {cmsg[t]}
\n\n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n
Find the mean and standard deviations of the difference between the before and after responses
\nCalculate differences for after response– before response.
\nMean of difference = [[0]] (input to 3 decimal places )
\nStandard deviation of difference = [[1]] (input to 3 decimal places)
\nNow find the paired t-test statistic using the values you have just calculated =[[2]] (3 decimal places)
\n\n ", "gaps": [{"minvalue": "{meandiff}", "type": "numberentry", "maxvalue": "{meandiff}", "marks": 0.5, "showPrecisionHint": false}, {"minvalue": "{stdiff}", "type": "numberentry", "maxvalue": "{stdiff}", "marks": 0.5, "showPrecisionHint": false}, {"minvalue": "{tvalue}", "type": "numberentry", "maxvalue": "{tvalue}", "marks": 1.0, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}, {"maxanswers": 0.0, "displaycolumns": 0.0, "prompt": "
Give the value of the t-statistic you have found, choose the range for the $p$ value by looking up the t-statistic tables:
", "matrix": "v", "shufflechoices": false, "minanswers": 0.0, "choices": ["$p$ is greater than $10\\%$
", "$p$ lies between $5 \\%$ and $10\\%$
", "$p$ lies between $1 \\%$ and $5\\%$
", "$p$ lies between $0.1 \\%$ and $1\\%$
", "$p$ is less than $0.1 \\%$
"], "displaytype": "radiogroup", "maxmarks": 0.0, "distractors": ["", "", "", "", ""], "marks": 0.0, "type": "1_n_2", "minmarks": 0.0}, {"maxanswers": 0.0, "displaycolumns": 0.0, "prompt": "Given the $p$-value and the range you have found, what is the strength of evidence against the null hypothesis that there is a difference in the average responses of the two groups?
", "matrix": "v", "shufflechoices": false, "minanswers": 0.0, "choices": ["None", "Slight", "Moderate", "Strong", "Very strong"], "displaytype": "radiogroup", "maxmarks": 0.0, "distractors": ["", "", "", "", ""], "marks": 0.0, "type": "1_n_2", "minmarks": 0.0}], "extensions": ["stats"], "statement": "\nSuppose that 15 individuals, diagnosed with bipolar disorder take part in an experiment that grades their happiness on a scale from 1 to 25. They take the test before treatment and then again after a specific drug has been prescribed. The data is shown below:
\n| {object} | A | B | C | D | E | F | G | H | I | J | K | L | M | N | O |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Before | \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$\\var{r1[10]}$ | \n$\\var{r1[11]}$ | \n$\\var{r1[12]}$ | \n$\\var{r1[13]}$ | \n$\\var{r1[14]}$ | \n
| After | \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$\\var{r2[10]}$ | \n$\\var{r2[11]}$ | \n$\\var{r2[12]}$ | \n$\\var{r2[13]}$ | \n$\\var{r2[14]}$ | \n
QUESTION: Is there a difference between the average responses of the two groups? In order to do this answer the following questions:
\n ", "variable_groups": [], "progress": "in-progress", "type": "question", "variables": {"t95": {"definition": 2.145, "name": "t95"}, "tvalue": {"definition": "precround(abs(tscore(0,d)),3)", "name": "tvalue"}, "msg": {"definition": "['is greater than $\\\\var{t999}$','lies between $\\\\var{t99}$ and $\\\\var{t999}$','lies between $\\\\var{t95}$ and $\\\\var{t99}$','lies between $\\\\var{t90}$ and $\\\\var{t95}$','is less than $\\\\var{t90}$']", "name": "msg"}, "t999": {"definition": 4.14, "name": "t999"}, "meandiff": {"definition": "precround(mean(d),3)", "name": "meandiff"}, "object": {"definition": "'Individual'", "name": "object"}, "sig1": {"definition": "random(2..3#0.2)", "name": "sig1"}, "thismany": {"definition": 15.0, "name": "thismany"}, "stdiff": {"definition": "precround(pstdev(d),3)", "name": "stdiff"}, "sig2": {"definition": "random(2..3#0.2)", "name": "sig2"}, "t99": {"definition": 2.977, "name": "t99"}, "attempt": {"definition": "'hand'", "name": "attempt"}, "t90": {"definition": 1.761, "name": "t90"}, "cmsg": {"definition": "['A','B','C','D','E']", "name": "cmsg"}, "r1": {"definition": "repeat(min(round(normalsample(mu1,sig1)),25),15)", "name": "r1"}, "mu1": {"definition": "random(10..14#0.5)", "name": "mu1"}, "d": {"definition": "list(vector(r2)-vector(r1))", "name": "d"}, "r2": {"definition": "repeat(min(round(normalsample(mu2,sig2)),25),15)", "name": "r2"}, "mu2": {"definition": "mu1+random(2..4#0.1)", "name": "mu2"}, "pmsg": {"definition": "[' is less than $0.001$',' lies between $0.001$ and $0.01$',' lies between $0.01$ and $0.05$',' lies between $0.05$ and $0.10$',' is greater than $0.10$']", "name": "pmsg"}, "objects": {"definition": "'individuals'", "name": "objects"}, "t": {"definition": "switch(v[0]=1,0,v[1]=1,1,v[2]=1,2,v[3]=1,3,4)", "name": "t"}, "v": {"definition": "if(tvalue>=t999,[1,0,0,0,0],if(tvalue>=t99,[0,1,0,0,0],if(tvalue>=t95,[0,0,1,0,0],if(tvalue>=t90,[0,0,0,1,0],[0,0,0,0,1]))))", "name": "v"}, "pvalue": {"definition": "precround(ttest(0,d,2),3)", "name": "pvalue"}}, "metadata": {"notes": "11/07/2012:
\n
Added tags.
Calculation not yet tested.
\n23/07/2012:
\nAdded description.
\nChecked calculation.
\nChanged display slightly in Advice.
\n3/08/2012:
\nAdded tags.
\nQuestion appears to be working correctly.
\n26/01/2013:
\nAdvice needs to be finished.
", "description": "Paired t-test to see if there is a difference between responses after treatment.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}]}