Paired sample t-test.

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Bill Foster 12 years, 3 months ago

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Question statement

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.

{object}	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.

Give any introductory information the student needs.

			Name	Type	Generated Value

pmsg	list	List of 5 items
msg	list	List of 5 items
cmsg1	list	List of 5 items
t999	number	4.437
meandiff	number	1.000
object	string	Individual
sig1	number	2.6
tvalue	number	0.721
stdiff	number	4.805
sig2	number	3.8
t99	number	3.106
t95	number	2.201
t90	number	1.796
cmsg	list	List of 5 items
d	list	List of 12 items
mu1	number	24.5
r1	list	List of 12 items
r2	list	List of 12 items
mu2	number	27.3
n	integer	12
t	integer	4
v	list	[ 0, 0, 0, 0, 1 ]
pvalue	number	0.486
attempt		Definition of variable `attempt` is empty.

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
[' 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 $0.10$']

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `list`

[ " 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 $0.10$" ]

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add
3. Gap Gap 2. Number entry
  Add
Part b) Choose one from a list
Part c) Choose one from a list

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

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

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

Mean of difference = Gap 0. (input to 3 decimal places )

Standard deviation of difference = Gap 1. (input to 3 decimal places)

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

Advice

The table of differences is given by:

{object}	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 {msg[t]} and the table tells us that the $p$ value {pmsg[t]}.

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.

Give a worked solution to the whole question.

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Part	Test	Passed?
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Number entry
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Taxonomy: mathcentre

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Bill Foster 12 years, 3 months ago

Error in variable testing condition

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Generated value: `list`

Parts

Gap-fill

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Paired sample t-test.

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Bill Foster 12 years, 3 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

Generated value: `list`