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Paired t-test to see if there is a difference between times take in a task.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "\n

The following data was obtained from $12$ individuals. The observations consist of the time taken to complete a dexterity task using their left and right hands.

\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 \n \n \n \n \n
{object}ABCDEFGHIJKL
Right$\\var{r1[0]}$$\\var{r1[1]}$$\\var{r1[2]}$$\\var{r1[3]}$$\\var{r1[4]}$$\\var{r1[5]}$$\\var{r1[6]}$$\\var{r1[7]}$$\\var{r1[8]}$$\\var{r1[9]}$$\\var{r1[10]}$$\\var{r1[11]}$
Left$\\var{r2[0]}$$\\var{r2[1]}$$\\var{r2[2]}$$\\var{r2[3]}$$\\var{r2[4]}$$\\var{r2[5]}$$\\var{r2[6]}$$\\var{r2[7]}$$\\var{r2[8]}$$\\var{r2[9]}$$\\var{r2[10]}$$\\var{r2[11]}$
\n

Carry out by hand a paired t-test to test whether there is evidence of a difference in the average times for the left and right hands.

\n ", "advice": "\n

The table of differences is given by:

\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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
{object}ABCDEFGHIJKL
Right$\\var{r1[0]}$$\\var{r1[1]}$$\\var{r1[2]}$$\\var{r1[3]}$$\\var{r1[4]}$$\\var{r1[5]}$$\\var{r1[6]}$$\\var{r1[7]}$$\\var{r1[8]}$$\\var{r1[9]}$$\\var{r1[10]}$$\\var{r1[11]}$
Left$\\var{r2[0]}$$\\var{r2[1]}$$\\var{r2[2]}$$\\var{r2[3]}$$\\var{r2[4]}$$\\var{r2[5]}$$\\var{r2[6]}$$\\var{r2[7]}$$\\var{r2[8]}$$\\var{r2[9]}$$\\var{r2[10]}$$\\var{r2[11]}$
Differences$\\var{d[0]}$$\\var{d[1]}$$\\var{d[2]}$$\\var{d[3]}$$\\var{d[4]}$$\\var{d[5]}$$\\var{d[6]}$$\\var{d[7]}$$\\var{d[8]}$$\\var{d[9]}$$\\var{d[10]}$$\\var{d[11]}$
\n

We test the following hypothesis:

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$H_0:\\;\\mu_d=0$ versus $H_1:\\;\\mu_d\\neq 0$

\n

$n=\\var{n}$ and the mean of the differences is $\\overline{d}=\\var{meandiff}$.

\n

The variance $V$ of the differences is calculated to be $\\var{pstdev(d)^2}$

Hence we have the standard deviation $s_d= \\sqrt{V}=\\var{stdiff}$ to 3 decimal places.

\n

The paired t-statistic is given by:

\n

\\[\\begin{eqnarray*} T&=&\\frac{\\overline{d}-\\mu_d}{\\frac{s_d}{\\sqrt{n}}}\\\\&=&\\frac{\\var{meandiff}-0}{\\frac{\\var{stdiff}}{\\sqrt{\\var{n}}}}\\\\&=&\\var{tvalue}\\end{eqnarray*}\\]

\n

(Using the null hypothesis that the means are the same i.e. $\\mu_d=0$.)

\n

Hence our test statistic  $|T|=\\var{tvalue}$.

\n

Looking up this value on the T-distribution table for $t_{11}$

\n

\\[\\begin{array}{r|rrrrr}&0.20&0.10&0.05&0.01&0.001\\\\\\hline11&1.363&1.796&2.201&3.106&4.437\\end{array}\\]

\n

We see that the t-statistic {msg[t]} and the table tells us that the $p$ value {pmsg[t]}.

\n

Hence we conclude that we {cmsg[t]} the null hypothesis. There is {cmsg1[t]} evidence of a difference between the average scores of the two groups.

\n

 

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Find the mean and standard deviations of the difference between left and right {attempt}s.

\n

Calculate differences for left {attempt} times – right {attempt} times. Make sure you take the differences this way round.

\n

Mean of difference = [[0]] (input  to 3 decimal places )

\n

Standard deviation of difference = [[1]] (input to 3 decimal places)

\n

Now find the t-test statistic $T$ using the values you have just calculated and  input the absolute value $|T|$ here: [[2]] (3 decimal places). 

\n

 

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

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$p$ less than $0.1 \\%$

", "

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

", "

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

", "

$p$ 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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