Perform a paired t-test

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Chris Graham

said

Needs to be tested

6 years, 5 months ago

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Tom Lowe commented 4 years, 5 months ago

Question could use some adaptive marking as each part requires an answer from the previous part. Also it seems slightly strange that 12/18 of the marks on offer are for calculating the differences between two values - could these be awarded half marks?

○

Chris Graham 6 years, 5 months ago

Gave some feedback: Needs to be tested

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Chris Graham 6 years, 5 months ago

Created this.

Name	Status	Author	Last Modified
Perform a paired t-test	Needs to be tested	Chris Graham	23/11/2022 15:44
Charlie's copy of Perform a paired t-test	draft	Charlie Earle	09/12/2018 12:48

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Question statement

The following data was obtained from $\var{n}$ right-handed participants. The observations consist of the time taken in seconds for each participant to complete a dexterity task by firstly using their right hand and secondly using their left hand.

{object}	A	B	C	D	E	F	G	H	I	J	K	L
Right hand	$\var{r1[0]}$	$\var{r1[1]}$	$\var{r1[2]}$	$\var{r1[3]}$	$\var{r1[4]}$	$\var{r1[5]}$	$\var{r1[6]}$	$\var{r1[7]}$	$\var{r1[8]}$	$\var{r1[9]}$	$\var{r1[10]}$	$\var{r1[11]}$
Left hand	$\var{r2[0]}$	$\var{r2[1]}$	$\var{r2[2]}$	$\var{r2[3]}$	$\var{r2[4]}$	$\var{r2[5]}$	$\var{r2[6]}$	$\var{r2[7]}$	$\var{r2[8]}$	$\var{r2[9]}$	$\var{r2[10]}$	$\var{r2[11]}$

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

Give any introductory information the student needs.

			Name	Type	Generated Value

t90	number	1.796
t95	number	2.201
t99	number	3.106
t999	number	4.437

			Name	Type	Generated Value

mu1	number	19
mu2	number	21.4
n	integer	12
r1	list	List of 12 items
r2	list	List of 12 items
sig1	number	2.6
sig2	number	2.4

			Name	Type	Generated Value

d	list	List of 12 items
meandiff	number	4.000
stdiff	number	4.452
tvalue	number	3.112400373

			Name	Type	Generated Value

cmsg1	list	List of 5 items
cmsg	list	List of 5 items
msg	list	List of 5 items
pmsg	list	List of 5 items

			Name	Type	Generated Value

object	string	Participant
t	integer	1
v	list	[ 0, 1, 0, 0, 0 ]
pvalue	number	0.010

Automatically regenerate variables?

Name

Data type

Value

(a number)

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: `number`

1.796

This variable doesn't seem to be used anywhere.

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Part e) Choose one from a list

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

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

{object}

Differences

$(d_i)$

Gap 0.

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Gap 11.

Advice

(a)

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 $(d_i)$	$\var{d[0]}$	$\var{d[1]}$	$\var{d[2]}$	$\var{d[3]}$	$\var{d[4]}$	$\var{d[5]}$	$\var{d[6]}$	$\var{d[7]}$	$\var{d[8]}$	$\var{d[9]}$	$\var{d[10]}$	$\var{d[11]}$

(b)

We test the following hypothesis:

$H_0:\;\mu_d=0$

$H_1:\;\mu_d\neq 0$

The mean of the differences is given by

\begin{align}
\bar{x}_d
&= \frac{\sum{d_i}}{n} \\
&= \frac{\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]}}{\var{n}} \\
&=\var{dpformat(meandiff,3)}.
\end{align}

The standard deviation of the differences is given by

\begin{align}
s_d&= \sqrt{\frac{\sum(d_i-\bar{x}_d)^2}{n-1}}\\
&= \sqrt{\frac{(\var{d[0]}-\var{dpformat(meandiff,3)})^2+(\var{d[1]}-\var{dpformat(meandiff,3)})^2+...+(\var{d[11]}-\var{dpformat(meandiff,3)})^2}{\var{n}-1}}\\
&=\var{dpformat(stdiff,3)}.
\end{align}

(c)

The paired test statistic is given by:

\begin{align}
T &=\frac{\bar{x}_{d}-\mu_d}{s_d/\sqrt{n}}\\
&=\frac{\var{dpformat(meandiff,3)}-0}{\var{dpformat(stdiff,3)}/\sqrt{\var{n}}}\\
&=\var{dpformat(tvalue,3)}.
\end{align}

(d)

For a paired sample $t$-test, we look up $t$ values on, $n-1 = \var{n}-1 = 11$, degrees of freedom.

From looking at $t$ tables, we obtain the following values

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

We see that the test statistic {msg[t]} and from the table we can see that the $p$ value {pmsg[t]}.

(e)

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

Give a worked solution to the whole question.

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Tom Lowe commented 4 years, 5 months ago

Chris Graham 6 years, 5 months ago

Chris Graham 6 years, 5 months ago

Error in variable testing condition

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Warning

Generated value: `number`

Parts

Gap-fill

(a)

(b)

(c)

(d)

(e)

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Perform a paired t-test

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Tom Lowe commented 4 years, 5 months ago

Chris Graham 6 years, 5 months ago

Chris Graham 6 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

(a)

(b)

(c)

(d)

(e)

Generated value: `number`