Two sample t-tests Practice

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Two sample t-tests Practice

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Question 1

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.

Individual	A	B	C	D	E	F	G	H	I	J	K	L
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	$\var{r2[0]}$	$\var{r2[1]}$	$\var{r2[2]}$	$\var{r2[3]}$	$\var{r2[4]}$	$\var{r2[5]}$	$\var{r2[6]}$	$\var{r2[7]}$	$\var{r2[8]}$	$\var{r2[9]}$	$\var{r2[10]}$	$\var{r2[11]}$

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.

Advice

The table of differences is given by:

Individual	A	B	C	D	E	F	G	H	I	J	K	L
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	$\var{d[0]}$	$\var{d[1]}$	$\var{d[2]}$	$\var{d[3]}$	$\var{d[4]}$	$\var{d[5]}$	$\var{d[6]}$	$\var{d[7]}$	$\var{d[8]}$	$\var{d[9]}$	$\var{d[10]}$	$\var{d[11]}$

We test the following hypothesis:

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

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

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.

The paired t-statistic is given by:

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

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

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

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

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

We see that the t-statistic lies between $\var{t95}$ and $\var{t99}$ and the table tells us that the $p$ value lies between $0.01$ and $0.05$ .

Hence we conclude that we do not reject the null hypothesis. There is slight evidence of a difference between the average scores of the two groups.

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Question 2

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:

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

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

Advice

We test the following hypothesis,

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

We find that the mean score of Group 1 is $\overline{x}_1=\var{m1}$ with standard deviation $s_1=\var{sd1}$ and the mean score of Group 2 is $\overline{x}_2=\var{m2}$ with standard deviation $s_2=\var{sd2}$.

All calculated to 3 decimal places.

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

The estimate of the pooled variance is calculated to be:

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

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

We find that the t-statistic has value:

\[\begin{eqnarray*}T&=& \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{s\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\&=&\frac{(\var{m1}-\var{m2})-(0)}{\var{s}\sqrt{\frac{1}{\var{n1}}+\frac{1}{\var{n2}}}}\\&=&\var{tvalue}\end{eqnarray*}\] to 3 decimal places.

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

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

\[\begin{array}{r|rrrrr}&0.20&0.10&0.05&0.01&0.001\\\hline18&1.330&1.734&2.101&2.878&3.922\end{array}\]

We see that the t-statistic is less than $\var{t90}$ and the table tells us that the $p$ value is greater than $0.10$.

Hence we conclude that we do not reject the null hypothesis. There is no evidence of a difference between the average scores of the two groups.

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Two sample t-tests Practice

Two sample t-tests Practice

Question 1

a)

b)

c)

Advice

Question 2

a)

b)

c)

d)

Advice