Statistics for psychology Assignment
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This assignment covers topics such as quantitative and qualitative variables, frequency distribution tables, measures of central tendency, correlation, and linear regression. It includes explanations, examples, and calculations for each topic.
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Running Head: Statistics for psychology Assignment
Statistics for psychology Assignment
Student’s Name
Institution Affiliation
Statistics for psychology Assignment
Student’s Name
Institution Affiliation
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Statistics for psychology Assignment
Question One
a. Meaning of the following terms:
i. quantitative variable
This is a variable whose value can be expressed as a number for example
the age, height and size of a person (Kaps & Lamberson, 2017).
ii. qualitative variable
This is a variable whose value is expressed in terms of categories for
instance person’s gender can be male or female (Kaps & Lamberson,
2017).
iii. discrete variable
Ott & Longnecker (2015), define this as a variable that has countable
values whose numbers are either finite or infinite. These values are
normally natural numbers for example number of woman a village which
can be 1000.
iv. continuous variable
Ott & Longnecker (2015) also define continuous variable as a variable that can
take an infinite number of values within a given interval. These values are real
numbers for example the weight of a child which can 2.5 Kilograms.
v. quantitative data
This is a data that can be acquired through counting or numbering or
measuring (Punch, 2013)
Question One
a. Meaning of the following terms:
i. quantitative variable
This is a variable whose value can be expressed as a number for example
the age, height and size of a person (Kaps & Lamberson, 2017).
ii. qualitative variable
This is a variable whose value is expressed in terms of categories for
instance person’s gender can be male or female (Kaps & Lamberson,
2017).
iii. discrete variable
Ott & Longnecker (2015), define this as a variable that has countable
values whose numbers are either finite or infinite. These values are
normally natural numbers for example number of woman a village which
can be 1000.
iv. continuous variable
Ott & Longnecker (2015) also define continuous variable as a variable that can
take an infinite number of values within a given interval. These values are real
numbers for example the weight of a child which can 2.5 Kilograms.
v. quantitative data
This is a data that can be acquired through counting or numbering or
measuring (Punch, 2013)
Statistics for psychology Assignment
vi. qualitative data
This is data that can obtain by placing things, person or items into
different categories or ranks (Punch, 2013)
b. Indicating whether following variables are quantitative or qualitative and classify
the quantitative variables as discrete or continuous
i. Women’s favorite TV programs
Qualitative variable
ii. Salaries of football players
Quantitative variable, which is continuous variable
iii. Number of cars owned by families
Quantitative variable, which is discrete variable
iv. Marital status of people
Qualitative variable
v. Number of typographical errors in newspapers
Quantitative variable, which is discrete variable
vi. qualitative data
This is data that can obtain by placing things, person or items into
different categories or ranks (Punch, 2013)
b. Indicating whether following variables are quantitative or qualitative and classify
the quantitative variables as discrete or continuous
i. Women’s favorite TV programs
Qualitative variable
ii. Salaries of football players
Quantitative variable, which is continuous variable
iii. Number of cars owned by families
Quantitative variable, which is discrete variable
iv. Marital status of people
Qualitative variable
v. Number of typographical errors in newspapers
Quantitative variable, which is discrete variable
Statistics for psychology Assignment
Question Two
i. Constructing a frequency distribution table
ii. Explain why we need to group data in the
form of a frequency table.
To make the analysis and presentation of data
easier and simple
iii. Determine the relative frequency and percentage distribution for these
data.
Level of Stress Frequency Relative Frequency Percentage
Somewhat 14 0.47 47%
Very 10 0.33 33%
None 6 0.20 20%
Total 30 1.00 100%
iv. Draw a pie chart based on the results in (iii).
Level of Stress
Frequency
Somewhat 14
Very 10
None 6
Total 30
Question Two
i. Constructing a frequency distribution table
ii. Explain why we need to group data in the
form of a frequency table.
To make the analysis and presentation of data
easier and simple
iii. Determine the relative frequency and percentage distribution for these
data.
Level of Stress Frequency Relative Frequency Percentage
Somewhat 14 0.47 47%
Very 10 0.33 33%
None 6 0.20 20%
Total 30 1.00 100%
iv. Draw a pie chart based on the results in (iii).
Level of Stress
Frequency
Somewhat 14
Very 10
None 6
Total 30
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Statistics for psychology Assignment
Question Three
Explain the following terms
and using appropriate
examples to describe and
illustrate each term.
i. Central
tendency
According to Hanna and Dempster(2012). , central tendency is measure of
center or representative values. Measures of central tendency include mean,
median and mode.
ii. Measures of dispersion
Measure od dispersion is the degree to which quantitative data tend to vary
or spread around the mean (Bajpai, 2009). Measures of dispersion include
variance, Standard deviation, Range and interquartile range.Skewness
This is a statistical measure that explain how symmetric or non-symmetric data
is(Miller, 2008).
Question Four
In this case t-test will be employed as the small size is less than 30 and the population
standard deviation is unknown.
Hypotheses:
47%
33%
20%
Somewhat
Very
None
Question Three
Explain the following terms
and using appropriate
examples to describe and
illustrate each term.
i. Central
tendency
According to Hanna and Dempster(2012). , central tendency is measure of
center or representative values. Measures of central tendency include mean,
median and mode.
ii. Measures of dispersion
Measure od dispersion is the degree to which quantitative data tend to vary
or spread around the mean (Bajpai, 2009). Measures of dispersion include
variance, Standard deviation, Range and interquartile range.Skewness
This is a statistical measure that explain how symmetric or non-symmetric data
is(Miller, 2008).
Question Four
In this case t-test will be employed as the small size is less than 30 and the population
standard deviation is unknown.
Hypotheses:
47%
33%
20%
Somewhat
Very
None
Statistics for psychology Assignment
H0=RM 100
H1 ≠ RM 100
Computation of t-statistics
t= x−μ
s
√n
, where n=sample ¿25 , s=sampkle standard deviation=Rm 15 ,
μ= population mean=Rm 100∧x=sample mean=Rm 130
t= 130−100
15
√ 25
= 30
3 =10
t- critical at 0.05 significance level and degree of freedom=25-1=24
tα =0.05 ,df =24=2.064
Decision and conclusion
The computed t-statistic(10) is greater than t-critical(2.064). This implies that the
null hypothesis will be rejected, hence the training did work as the level of sales has
significantly increased.
b. Determination of whether your teaching improves students’ knowledge/skills (i.e.
test scores). Draw conclusions about the impact of this module in general.
H0=RM 100
H1 ≠ RM 100
Computation of t-statistics
t= x−μ
s
√n
, where n=sample ¿25 , s=sampkle standard deviation=Rm 15 ,
μ= population mean=Rm 100∧x=sample mean=Rm 130
t= 130−100
15
√ 25
= 30
3 =10
t- critical at 0.05 significance level and degree of freedom=25-1=24
tα =0.05 ,df =24=2.064
Decision and conclusion
The computed t-statistic(10) is greater than t-critical(2.064). This implies that the
null hypothesis will be rejected, hence the training did work as the level of sales has
significantly increased.
b. Determination of whether your teaching improves students’ knowledge/skills (i.e.
test scores). Draw conclusions about the impact of this module in general.
Statistics for psychology Assignment
t-Test: Paired Two Sample for Means
Pre-module
Score
Post Module
Score
Mean 18.4 20.45
Variance 9.9368 16.4711
Observations 20 20
Pearson Correlation 0.7175
Hypothesized Mean
Difference
0
df 19
t Stat -3.2313
P(T<=t) one-tail 0.0022
t Critical one-tail 1.7291
P(T<=t) two-tail 0.0044
t Critical two-tail 2.0930
Conclusion
From the tablet-statistic(-3.2313) is absolutely greater than the t-critical two-tail(2.093).
this indicate that null hypothesis will be rejected, hence the teaching has improved the
students’ knowledge and skills.
Question Five
a. What do you understand by the statistical procedure called correlation?
Correlation is a statistical procedure that’s used to measure the level of association
between two variables, say variable a and b. (Gupta, 2016).
b. Do you expect the ages of husbands and wives to be positively or negatively related?
The two variables are expected to be positively related.
t-Test: Paired Two Sample for Means
Pre-module
Score
Post Module
Score
Mean 18.4 20.45
Variance 9.9368 16.4711
Observations 20 20
Pearson Correlation 0.7175
Hypothesized Mean
Difference
0
df 19
t Stat -3.2313
P(T<=t) one-tail 0.0022
t Critical one-tail 1.7291
P(T<=t) two-tail 0.0044
t Critical two-tail 2.0930
Conclusion
From the tablet-statistic(-3.2313) is absolutely greater than the t-critical two-tail(2.093).
this indicate that null hypothesis will be rejected, hence the teaching has improved the
students’ knowledge and skills.
Question Five
a. What do you understand by the statistical procedure called correlation?
Correlation is a statistical procedure that’s used to measure the level of association
between two variables, say variable a and b. (Gupta, 2016).
b. Do you expect the ages of husbands and wives to be positively or negatively related?
The two variables are expected to be positively related.
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Statistics for psychology Assignment
c. Plot a scatter diagram.
15 20 25 30 35 40 45 50 55 60
0
10
20
30
40
50
60
Wife’s age Against Husband's Age
Husband's Age
Wife's Age
The correlation coefficient is expected to be close to +1 as the data dots are
forming a linear pattern with a positive slope due its left-right upward trend.
d. Finding the correlation coefficient. Is the value of r consistent with you expected in
parts (b) and (c)?
Correlation Matrix
Husband’s age Wife’s age
Husband’s age 1
Wife’s age 0.97156725 1
c. Plot a scatter diagram.
15 20 25 30 35 40 45 50 55 60
0
10
20
30
40
50
60
Wife’s age Against Husband's Age
Husband's Age
Wife's Age
The correlation coefficient is expected to be close to +1 as the data dots are
forming a linear pattern with a positive slope due its left-right upward trend.
d. Finding the correlation coefficient. Is the value of r consistent with you expected in
parts (b) and (c)?
Correlation Matrix
Husband’s age Wife’s age
Husband’s age 1
Wife’s age 0.97156725 1
Statistics for psychology Assignment
The computed r is 0.9716 which is close to +1, therefore, value of r is consistent with
part (b) and (c) as 0.9716 indicates a positive relationship between the two variables.
e. Conclusion from this result?
From the investigation, it’s clear that the Husband’s age and Wife’s age has a high
positive linear relationship.
Question Six
Table 4
a. Does the insurance premium
depend on the driving experience or
does the driving experience depend on
the insurance premium? Do you expect
a positive or a negative relationship
between these two variables?
From information in the above,
it’s clear that insurance premium depend on driving experience. A high driving
experience leads to a low premium. Therefore, driving experience and insurance
premium will be negatively related.
b. What statistical model do you think you can apply to confirm your answer in (a)?
Linear regression model
c. Compute Sxx S, SSyy, and SSxy.
Driving
Experience
(years)
Monthly Auto
Insurance Premium
(RM)
5 64
2 87
12 50
9 71
15 44
6 56
25 42
16 60
The computed r is 0.9716 which is close to +1, therefore, value of r is consistent with
part (b) and (c) as 0.9716 indicates a positive relationship between the two variables.
e. Conclusion from this result?
From the investigation, it’s clear that the Husband’s age and Wife’s age has a high
positive linear relationship.
Question Six
Table 4
a. Does the insurance premium
depend on the driving experience or
does the driving experience depend on
the insurance premium? Do you expect
a positive or a negative relationship
between these two variables?
From information in the above,
it’s clear that insurance premium depend on driving experience. A high driving
experience leads to a low premium. Therefore, driving experience and insurance
premium will be negatively related.
b. What statistical model do you think you can apply to confirm your answer in (a)?
Linear regression model
c. Compute Sxx S, SSyy, and SSxy.
Driving
Experience
(years)
Monthly Auto
Insurance Premium
(RM)
5 64
2 87
12 50
9 71
15 44
6 56
25 42
16 60
Statistics for psychology Assignment
Driving experience is denoted by X and Auto insurance premium by Y
X Y x−x Sxx=( x−x)2 y− y Syy =( y − y)2 Sxy= ( x−x ) ( y− y )
5 64 -6.25 39.0625 4.75 22.5625 -29.6875
2 87 -9.25 85.5625 27.75 770.0625 -256.6875
12 50 0.75 0.5625 -9.25 85.5625 -6.9375
9 71 -2.25 5.0625 11.75 138.0625 -26.4375
15 44 3.75 14.0625 -15.25 232.5625 -57.1875
6 56 -5.25 27.5625 -3.25 10.5625 17.0625
25 42 13.75 189.0625 -17.25 297.5625 -237.1875
16 60 4.75 22.5625 0.75 0.5625 3.5625
Sum 90 474 383.5 1557.5 -593.50
n 8 8
Mean 90
8 =11.2
5
474
8 =59.25
Therefore,
Sxx=383.5 , S yy=1557.50 , S xy=−593.50
Standard deviation(Sx)= √ Sxx
n−1 = √ 383.5
8−1 =7.402
Standard deviation(Sy )= √ S yy
n−1 = √ 1557.50
8−1 =14.916
d. Find the least squares regression line by choosing appropriate dependent and
independent variables based on your answer in (c).
The least squares regression line is given by y=bx+ a
b= Sxy
Sxx
=−593.50
383.5 =−1.548
Driving experience is denoted by X and Auto insurance premium by Y
X Y x−x Sxx=( x−x)2 y− y Syy =( y − y)2 Sxy= ( x−x ) ( y− y )
5 64 -6.25 39.0625 4.75 22.5625 -29.6875
2 87 -9.25 85.5625 27.75 770.0625 -256.6875
12 50 0.75 0.5625 -9.25 85.5625 -6.9375
9 71 -2.25 5.0625 11.75 138.0625 -26.4375
15 44 3.75 14.0625 -15.25 232.5625 -57.1875
6 56 -5.25 27.5625 -3.25 10.5625 17.0625
25 42 13.75 189.0625 -17.25 297.5625 -237.1875
16 60 4.75 22.5625 0.75 0.5625 3.5625
Sum 90 474 383.5 1557.5 -593.50
n 8 8
Mean 90
8 =11.2
5
474
8 =59.25
Therefore,
Sxx=383.5 , S yy=1557.50 , S xy=−593.50
Standard deviation(Sx)= √ Sxx
n−1 = √ 383.5
8−1 =7.402
Standard deviation(Sy )= √ S yy
n−1 = √ 1557.50
8−1 =14.916
d. Find the least squares regression line by choosing appropriate dependent and
independent variables based on your answer in (c).
The least squares regression line is given by y=bx+ a
b= Sxy
Sxx
=−593.50
383.5 =−1.548
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Statistics for psychology Assignment
a= y−b x
¿ 59.25− (−1.548 ) ( 11.25 )
¿ 76.665
Hence, the regression line will be
y=bx+ a
y=−1.548 x +76.665
e. Interpret the meaning of the values of a and b calculated in part (c).
The value of a is 76.665, this indicates the value of premium when there’s no driving
experience
The value of b is -1.548 this indicates that when driving experience increase by 1 unit,
the value of premium will decrease by 1.548.
f. Plot the scatter diagram and the regression line.
g. Calculate r and r2
and explain what
they mean.
r =Cov ( x , y )
Sx∗SY
Cov ( x , y )= S xy
n−1 =−593.50
8−1 =−84.786
r = −84.786
7.402∗14.916
0 5 10 15 20 25 30
0
10
20
30
40
50
60
70
80
90
100
f(x) = − 1.54758800521512 x + 76.6603650586702
R² = 0.589722941313115
Monthly Auto Insurance Premium (RM) Y Against Driving Experience
X
Driving Experience(X)
Monthly Auto Premium(Y)
a= y−b x
¿ 59.25− (−1.548 ) ( 11.25 )
¿ 76.665
Hence, the regression line will be
y=bx+ a
y=−1.548 x +76.665
e. Interpret the meaning of the values of a and b calculated in part (c).
The value of a is 76.665, this indicates the value of premium when there’s no driving
experience
The value of b is -1.548 this indicates that when driving experience increase by 1 unit,
the value of premium will decrease by 1.548.
f. Plot the scatter diagram and the regression line.
g. Calculate r and r2
and explain what
they mean.
r =Cov ( x , y )
Sx∗SY
Cov ( x , y )= S xy
n−1 =−593.50
8−1 =−84.786
r = −84.786
7.402∗14.916
0 5 10 15 20 25 30
0
10
20
30
40
50
60
70
80
90
100
f(x) = − 1.54758800521512 x + 76.6603650586702
R² = 0.589722941313115
Monthly Auto Insurance Premium (RM) Y Against Driving Experience
X
Driving Experience(X)
Monthly Auto Premium(Y)
Statistics for psychology Assignment
¿−0.7679
r2= ( −0.7679 ) 2
¿ 0.5897
h. Predict the monthly auto insurance premium for a driver with 10 years of driving
experience.
y=−1.548 ( 10 )+76.665
¿ RM 61.19
Reference
Bajpai, N. (2009). Business Statistics. India: Pearson Education India.
Gupta, A. (2016). Risk Management and Simulation. United Kingdom: CRC Press.
Hanna, D., & Dempster, M. (2012). Psychology Statistics For Dummies. United Kingdom: John
Wiley & Sons.
Kaps, M., & R., W. (2017). Biostatistics for Animal Science. United Stated: CABI.
Miller, M. (2013). Mathematics and Statistics for Financial Risk Management. Canada: John
Wiley & Sons.
Ott, L., & Longnecker , M. (2015). An Introduction to Statistical Methods and Data Analysis.
United States of America: Cengage Learning.
Punch, K. F. (2013). Introduction to social research: Quantitative and qualitative approaches.
sage.
¿−0.7679
r2= ( −0.7679 ) 2
¿ 0.5897
h. Predict the monthly auto insurance premium for a driver with 10 years of driving
experience.
y=−1.548 ( 10 )+76.665
¿ RM 61.19
Reference
Bajpai, N. (2009). Business Statistics. India: Pearson Education India.
Gupta, A. (2016). Risk Management and Simulation. United Kingdom: CRC Press.
Hanna, D., & Dempster, M. (2012). Psychology Statistics For Dummies. United Kingdom: John
Wiley & Sons.
Kaps, M., & R., W. (2017). Biostatistics for Animal Science. United Stated: CABI.
Miller, M. (2013). Mathematics and Statistics for Financial Risk Management. Canada: John
Wiley & Sons.
Ott, L., & Longnecker , M. (2015). An Introduction to Statistical Methods and Data Analysis.
United States of America: Cengage Learning.
Punch, K. F. (2013). Introduction to social research: Quantitative and qualitative approaches.
sage.
Statistics for psychology Assignment
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