worked examples of mathematics used in psychology260255,en.pdf · school of psychology queen‟s...

School of Psychology

Queen‟s University,

Belfast, BT7 1NN

1

Worked Examples of mathematics used in Psychology

Worked Example 1: Level 1 Introductory Psychology I (PSY1001) Laboratory

Demonstration

Worked Example 1 draws from CCEA GCSE Mathematics Module T1 and T2,

(Foundation Tier) – Statistics Component

For these exercises, describe in detail what the results presented in the tables and

graphs mean.

Answer:

This graph shows the mean extraversion and neuroticism scores for people grouped

by the number of friends they have. It shows that people with no friends have similar

levels extraversion and neuroticism, with a score of around 11 points. The trend for

extraversion shows that as the level of extraversion increases, so does the number

of close friends. This would be expected because extraversion is a personality trait

associated with outgoingness and sociability, so you would expect people who have

high levels of extraversion to have larger circles of friends. The opposite trend is

found with neuroticism, where lower levels of neuroticism are associated with having

more friends. This again is expected because neuroticism is a personality trait

associated with anxiety and worry. You would expect people with lower levels of

neuroticism to have larger circle of friends because people who are less prone to

anxiety and worry are probably able to make and maintain friendships.

NONE 1-2 3-5 5- 10+ 6

8

10

12

14

Mean EXTRA

Mean NEUROT

EXTRAVERSION AND NEUROTICISM

WITH NUMBER OF CLOSE FRIENDS

NUMBER OF CLOSE FRIENDS

Me

an

Score



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EXTRAVERSION

NEUROTICISM

MEAN S.D. MEAN S.D.

MALE 12.20 5.30 7.27 4.38

FEMALE 11.53 4.34 10.36 5.00

Answer:

This table shows males are, on average, more extraverted compared to females with

males having a mean extraversion score of 12.2 and females having a score of

11.53. There is more variance in male extraversion scores, as males have a larger

standard deviation (5.3) compared to females (4.34). Females seem to be more

neurotic on average compared to males with a mean score of 10.36 compared to

7.27 for males. Neuroticism scores varied more for females, who have a larger

standard deviation compared to males.

8 12 16 20 24 28

0

10

20

30

FREQUENCY DISTRIBUTION

LIFE SATISFACTION INDEX - Z

LSI-Z SCORES

FR

EQ

UE

NC

Y

Answer:

This graph shows the distribution of standardised scores from the life satisfaction

questionnaire from a sample of approximately 100 people. The modal score on the

questionnaire was 24 with 29 people obtaining this score. The range of scores was

20 (with scores from 8 to 28). The shape of the distribution approximates a normal

distribution, with some evidence of negative skew seen in the tail to the left hand

side of the distribution.



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Most Enjoyed Least Enjoyed

1. Time spent with spouse (32%)

2. Doing things with children (22%)

3. Recreational activities/hobbies (18%)

4. Getting together with friends (13%)

1. Time spent alone (51%)

2. Taking care of the house (48%)

3. Recreational activities/hobbies (1%)

Answer:

This table shows a list of most and least popular past-times. The most popular

activity was spending time with spouse where 32% of respondents indicated they

most enjoyed this activity. Spending time alone was the least enjoyed activity with

51% of respondents indicating this. There appears to be some missing activities

rated as most enjoyed because the percentages of this group only adds up to 85%

meaning 15% o responses are unknown.

Background to this experiment: Participants were presented with a list of twenty

words and asked to remember as many words as possible. Each word was

presented one at a time on a computer screen for 5 seconds. Every participant saw

exactly the same list of words. After all 20 words were presented, the participants

were asked to write down as many words as they could remember from the list in

any order. The graph shows the percentage of participants who remembered each

word, depending on where the word appeared in the list. E.g. the word in word-

position 1 was presented first in the list, word position 10 is the middle word of the

list, and word position 20 was the final word presented.



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Answer:

This line graph shows the accuracy of recall for words presented as part of a list,

depending on each word‟s position in the list. It seems that the group were better at

remembering words presented at the start and the end of the list, compared to words

located in the middle of the list. Words at the end of the list were remembered better

than any other word positions, which would be expected because these words would

have been most recent in the participants‟ memory. Accuracy for words in the middle

of the list (approx. 40%) were approximately half that of the accuracy of recall for

words at the end of the list (approx. 80%).

Source: Questions based on level 1 laboratory handbook exercises, 2010/11.



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Worked Example 2: Level 1 Introductory Psychology 1 (PSY1001) Laboratory

Class Work

Worked Example 2 draws from CCEA GCSE Mathematics Modules:

T1 and T2, (Foundation Tier) – Mathematics & Algebra and Statistics

Components

T5 (Foundation Tier) -- Mathematics & Algebra Component

T3 (Higher Tier) – Statistics Component


T6 (Higher Tier) – Mathematics & Algebra Component

GCSE Additional Mathematics:

Statistics, Topics 1 and 3

A-Level Modules:

S1, Topic 2

S2, Topics 5 and 6

Intelligence tests provide a standardised measure of ability commonly referred to as IQ.

IQ has a normal distribution centred on a mean score of 100 with a standard deviation of 15.

We are going to calculate the mean and standard deviation of IQ scores collected from a

sample of 20 people:

Participant Number IQ Score

1 110

2 124

3 99

4 108

5 117

6 136

7 96

8 107

9 101

10 112

11 120

12 95

13 114

14 126

15 100

16 119

17 121

18 108

19 101

20 113



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Question:

We are going to use the below equation to calculate the mean IQ score of the group of

people:

Where:

is the symbol for mean

is a symbol that means „sum of/add up‟

is the symbol for individual IQ scores

N is the number of participants in the experiment

Fill in the values for and N in the equation below and calculate the mean IQ score:

Solution:

Now input the IQ scores to Microsoft excel and check your answer using the “=average”

function.

We will now calculate the standard deviation using the below equation:

Complete the table to help with the calculation:

Remember from above:



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Participant Number IQ Score ( ) 1 110

2 124

3 99

4 108

5 117

6 136

7 96

8 107

9 101

10 112

11 120

12 95

13 114

14 126

15 100

16 119

17 121

18 108

19 101

20 113

___________

Solution:

Participant Number IQ Score ( ) 1 110

110-111.35 = -1.35 -1.35*-1.35 =

1.8225

2 124 12.65 160.0225

3 99 -12.35 152.5225

4 108 -3.35 11.2225

5 117 5.65 31.9225

6 136 24.65 607.6225

7 96 -15.35 235.6225

8 107 -4.35 18.9225

9 101 -10.35 107.1225

10 112 0.65 0.4225

11 120 8.65 74.8225

12 95 -16.35 267.3225

13 114 2.65 7.0225

14 126 14.65 214.6225

15 100 -11.35 128.8225

16 119 7.65 58.5225

17 121 9.65 93.1225

18 108 -3.35 11.2225

19 101 -10.35 107.1225

20 113 1.65 2.7225



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2292.55

Now substitute the values into the standard deviation equation:

Solution:

10.985

Now check your answer using excel and/or PASW.

Question:

Compare the mean and standard deviation of this sample of 20 people with the mean and

standard deviation of the population of IQ scores ( = 100, s = 15) what could you conclude

about your sample?

Solution:

The sample of 20 people have a higher than average IQ at 111.35 This sample can‟t be

considered a representative sample of the population, where you would expect an average

IQ of around 100. The sample seems to be biased in favour of higher IQ scores. The

standard deviation is also lower than what would be expected in a representative sample at

10.985 as opposed to 15. This indicates there was less variance in IQ scores within the

group of 20 people than what you would expect if you were able to test the whole population.



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Question:

Does this histogram of the IQ data from the sample of 20 people approximate the normal

distribution that you would expect IQ data to take?

Solution:

Yes, the histogram shows that the IQ data closely fit the normal distribution.



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Worked Example 3: Level 2 Perception (PSY 2058) Laboratory Demonstration

and Coursework



T3 (Higher Tier) – Mathematics & Algebra Component and Statistics

Component


T6 (Higher Tier) Mathematics & Algebra Component


Pure Maths, Topic 4

Statistics, Topic 6 and 7

A-Level Modules:

C2, Topic 4

S2, Topics 5 and 6

The speed/accuracy trade-off is used to describe „aimed movement‟ where someone is

required to move their hand from a starting point (S) to stop within a target area (T).

It describes how faster movements result in lower accuracy and higher accuracy can be

achieved by lowering speed of movement. Depending on the difficulty of a task, humans can

adjust their movement speed in order to achieve respectable accuracy.



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When the target area T is thin, the task is more difficult because movement requires higher

precision so as not to miss or overshoot the target. In this case movement must be slowed

down to hit the target accurately.

When the start-point and target are farther apart, the task difficulty is increased as the brain

needs to co-ordinate movement across a longer distance. In this case the longer distance

allows the arm to be accelerated to higher speeds, resulting in faster performance, however

accuracy may decrease as the extra deceleration of the arm may result in the target being

missed. This is why emergency buttons in aircraft cockpits are larger than other buttons. In

an emergency pilots will be working hard and quickly to avert disaster, so making the buttons

larger will make them easier to press with „fast and furious‟ hand movements.

The task difficulty index (Id) is a logarithmic function of the distance between the start-point

and target (amplitude of movement, A) and the width of the target (W) where:

Fitt‟s law states that there is a linear relationship between movement time (MT) and task

difficulty expressed as:

where m and c are constants.

An experiment was run where the amplitude and width of a target were manipulated across

16 trials. Movement Time was measured in milliseconds. Amplitude was measured at 4

levels (3, 6, 12 and 24 cm) and width was measured at 4 levels (1, 2, 4 and 8 cm). The

results of each trial are in the table below:

Trial Number Amplitude (cm) Width (cm) Movement Time (ms)

1 3 1 360

2 3 2 212

3 3 4 182

4 3 8 166

5 6 1 217

6 6 2 353

7 6 4 179

8 6 8 172

9 12 1 234

10 12 2 383

11 12 4 325

12 12 8 229

13 24 1 481

14 24 2 404

15 24 4 338

16 24 8 298



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Question: Input this table into Excel and calculate, in a new column, the Index of Task

Difficulty (Id) for each trial. Use the log function where „Number‟ is the expression and

„Base‟ is 2.

Solution:

Assuming the above table is inputted in Excel where the top-left cell “Trial Number” is A1;

the excel expression for item difficulty for trial 1 will be: =log(2*B2/C2,2). This formula can be

filled down to calculate Id for the other rows, which are given in the following table.

Trial Number Movement Time (ms) Task Difficulty

1 360 2.584963

2 212 1.584963

3 182 0.584963

4 166 -0.41504

5 217 3.584963

6 353 2.584963

7 179 1.584963

8 172 0.584963

9 234 4.584963

10 383 3.584963

11 325 2.584963

12 229 1.584963

13 481 5.584963

14 404 4.584963

15 338 3.584963

16 298 2.584963

Question:

Does the data support the hypothesis that Fitt‟s Law provides an adequate description of the

role assumed by movement amplitude and precision in determining movement time?

Solution:

Since Fitt‟s law is a linear equation, correlation can be used to calculate the extent of the

linear relationship, using the square of the correlation coefficient as an indication of common

variance.

∵

[PASW or Excel can be used to calculate the correlation coefficient.]

There is a strong correlation between movement time and difficulty index (r=.7698) where

59.2% of the variance in movement time is accounted for by the difficulty index variance.



Belfast, BT7 1NN

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Question:

What is the correspondence of the slope and intercept values obtained from a regression

analysis of MT and Id to the parameters m and c in the equation:

Solution:

Using PASW, you can perform a regression analysis with a predictor of Id and a criterion

variable of MT. The relevant analysis output is given the table:

Coefficientsa

Model

Unstandardized Coefficients

Standardized

Coefficients

t Sig. B Std. Error Beta

1 (Constant) 164.986 30.735 5.368 .000

Id 45.775 10.143 .770 4.513 .000

a. Dependent Variable: MT

The unstandardised beta coefficient for Id is the slope of the regression line which

corresponds to the gradient (m) parameter of the Fitt‟s law equation. The unstandardised

beta coefficient of the constant is the intercept of the regression line, which corresponds to

the intercept (c) parameter of the Fitt‟s law equation.

Substitution of the constants gives the solution:



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Question:

Draw a scatter plot showing Movement Time (y-axis) and Index of Difficulty (x-axis); include

the line of best fit as described in the regression.

Solution:

Source: Questions based on level 2 laboratory demonstration and coursework from 2008.



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Worked Example 4: Level 2 Methods and Conceptual Issues (PSY 2056)

Research Methods and Statistics


T1 and T2, (Foundation Tier) – Mathematics & Algebra and Statistics

Components




T6 (Higher Tier) – Mathematics & Algebra Component


Statistics, Topic 3

A-Level Modules:

S1, Topic 1 and 2

S2, Topics 3 and 4

Question:

As a clinical psychologist, you want to investigate the effectiveness of a new type of therapy

for depression. Before starting the new therapy, you ask clients to complete the Beck‟s

Depression Inventory. This is self-report questionnaire that provides a measure of

depression on a scale from 0-63 where 0 is minimal depression and 63 is severe

depression. After two months of therapy, you ask clients to complete the depression

inventory again to see if levels of depression have improved during the course of the

therapy. Twenty clients completed the therapy course, their inventory scores are below.



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Client Identification Number

Beck‟s Depression Inventory Score BEFORE Therapy

Beck‟s Depression Inventory Score AFTER two months

1 44 23

2 13 17

3 58 55

4 7 7

5 32 19

6 33 45

7 28 30

8 37 28

9 41 30

10 23 18

11 52 42

12 11 8

13 32 40

14 59 54

15 8 13

16 60 34

17 29 25

18 46 42

19 15 11

20 33 28

Assuming the therapy is effective; state your one-tailed hypothesis for the depression

scores.

Solution:

It is hypothesised that clients will score significantly lower in Depression scores after two

months of therapy, compared to their scores just before starting the therapy.

Question:

Looking at the table, do you think the therapy is effective?

Solution:

Generally, there seems to be lower scores after the therapy. The mean depression score for

the whole group before treatment was 33.05, whereas after treatment the mean for the

group reduced to 28.45. This implies that depression scores reduced over the course of the

treatment. However, there are a few clients whose scores increased after the therapy

(clients 2, 6, 7, 13, 15). The difference between scores also varied across clients, for

example, client 16‟s scores reduced by 26, client 7 only reduced by 2. In conclusion, the

treatment seems to work as hypothesised for some people but it is still unclear if the effect of

the treatment is large enough to be considered effective.



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You can perform a t-test to compare the mean depression scores of the group before and

after therapy. By using a t-test you can also decide if the difference in scores is statistically

significant. The formula for a t-test is:

Where is the average difference between the depression scores before and after

treatment and is the standard error of the mean difference scores which is given by:

Where is the standard deviation of the difference scores and N is the number of clients

tested.

Question:

Calculate the t-test using excel to complete column D in following table:

Column A Column B Column C Column D

Client Identification

Number

Beck‟s Depression Inventory Score

BEFORE Therapy


AFTER two months

Difference in depression score:

Column B – Column C

1 44 23

2 13 17

3 58 55

4 7 7

5 32 19

6 33 45

7 28 30

8 37 28

9 41 30

10 23 18

11 52 42

12 11 8

13 32 40

14 59 54

15 8 13

16 60 34

17 29 25

18 46 42

19 15 11

20 33 28

Mean of Difference Scores :

Standard Deviation of Difference Scores :



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Solution:

Column A Column B Column C Column D

Client Identification

Number


BEFORE Therapy


AFTER two months

Difference in depression score:

Column B – Column C

1 44 23 44-23 = 21

2 13 17 -4

3 58 55 3

4 7 7 0

5 32 19 13

6 33 45 -12

7 28 30 -2

8 37 28 9

9 41 30 11

10 23 18 5

11 52 42 10

12 11 8 3

13 32 40 -8

14 59 54 5

15 8 13 -5

16 60 34 26

17 29 25 4

18 46 42 4

19 15 11 4

20 33 28 5

Mean of Difference Scores : 4.6

Standard Deviation of Difference Scores : 9.058

Substitute the values and calculate SEM from the above equation:

Solution:



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Now, calculate t by substituting the values to the equation:

Solution:

Values of t greater than 1.96 are statistically significant at p<.05 level. I.E. there is a 5%

chance or less that the difference between the mean of the two groups is due to chance. Is

the difference in mean depression scores before and after therapy significant at the p<.05

level?

Solution:

Yes, t=2.27 which is greater than the threshold value of 1.96

worked examples of mathematics used in psychology260255,en.pdf · school of psychology queen‟s...

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