PSYC 510
H -P -OMEWORK: TWO AIRED SAMPLES T TESTS ASSIGNMENT
INSTRUCTIONS
OVERVIEW
This Homework: Two-Paired Samples -Tests AssignmentT is designed to assess your
understanding of the concepts and applications covered thus far in this course. Concepts specific
to this module include the assumptions for a correlated groups test, how to calculate it both by t
hand and using SPSS, and how to present it using professional conventions. Its strengths and
weaknesses as compared to the independent samples test from our previous module is also t
discussed. Development of knowledge and skills for appropriate use of this popular test, as
required in this Homework: Two-Paired Samples -Tests AssignmentT, will prepare you to be a
more informed consumer and producer of research both professionally and non-professionally.
INSTRUCTIONS
Be sure you have reviewed this module’s before completing this Learn section Homework:
Two-Paired Samples -Tests Assignment Homework: Two-Paired Samples -Tests T. This T
Assignment is worth 60 points. Each question is worth 3 points. Six points are awarded for
mechanics/structure.
Part I contains general concepts from this module’s . Learn section
Part II requires use of SPSS. You will have to take screen shots and/or copy and paste from
your SPSS to place answers within this file. Make sure you only insert relevant and legible
images.
Part III is the cumulative section. These may include short answer and/or use of SPSS but
will review material from previous module(s).
Directions for each subsection are provided in the top of each table (in the blue shaded
areas).
Answers should be placed where indicated (wherever there is “ ”). ANSWER
Submit the file as a WORD document (.doc or .docx). Make sure the filename of your
submission includes your full name, course and section.
oExample: HW8_JohnDoe_510B01
Make sure to check the before you begin thisHomework Grading Rubric Homework: Two-
Paired Samples -Tests AssignmentT.
Part I: General Concepts
These questions are based on the concepts covered in this
module’s assigned readings and presentations.
Answer the following questions using your own words.
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1. Explain what counterbalancing is, how it is achieved, and which confound it helps to minimize.
Counterbalancing: In any experiment the order of the conditions wherein participants have to perform
can impact the actual performances of the participants. To deal with such conditions in order to help
improve the performance of the participants, counterbalancing technique is used. Let’s assume that the
performances of the participants have improved due to practice or experience e.g., bathroom singing,
then asking the half of the participants to sing alone and half of the participants to sing before audience
will reduce the conditions / order’s effect.
How it is Achieved: This technique is often used in repeated measures design. The simplest way to
achieve is to define the conditions e.g. singing alone is condition A and singing in front of audience is
condition B, half of the participants will now follow AB order and the remaining half will be follow BA
order counterfeit the order effect impacting their performances.
Which Confound It Helps to Minimize: Counterbalancing is usually used to help minimize the impact
of sequencing / order effects, but it takes the assumptions that participant’s’ performance has improved to
some extent let’s say through practice and there is not any other swamping effects.
2. Using your own words, discuss how a correlated-groups test has more statistical power in t
comparison to an independent-groups test.t
Correlated t test mostly have more statistical power than independent t test due to the following reasons:
Mean Difference: In correlated t test, mean difference is a comparison of subject performance in one
condition with another condition, meaning that each subject acts as its own control and doesn’t allow the
differences between subjects to enter into the analysis. This way individual differences between the
subjects is minimized, meaning that 2 subjects can’t be same, this increases the statistical power of the
correlated t test.
Size of Error: Such control also helps in keeping the standard error of the means differences smaller,
than independent t test, which ultimately helps in yielding a larger t value or more power.
3. Discuss one strength and one weakness of a within-subjects design using your own words.
(Do not include statistical power, as it is assessed in a previous question)
Strength: It requires small sample, we can obtain 20 scores in each condition by using 20 participants,
but for independent design we need 40 participants to obtain 20 scores. So with small sample it also
saves time that would be required to find 20 additional participants.
Weakness: We need a pair of participants to implement this design, just in case we lose a participant of
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first condition in second condition then we can’t make any inferences. Moreover two conditions can also
induce the order effect which can affect the results.
SCENARIO: A researcher believes exercise may affect anxiety in women, but research
appears inconclusive. She identifies a group of women (N = 30) who had not exercised
before, but are now planning to begin exercising. She gives them a 50-item anxiety
inventory before they begin exercising and administers it again after 6 months of
exercising. The anxiety inventory is measured on an interval scale and higher numbers
indicate higher anxiety. In addition, scores on the inventory are normally distributed.
The mean of the difference scores is 3.4, S = 1.8, and there were 30 in the sample. Using
D
this information, answer the following questions showing all work when calculations are
required to earn up to full credit.
4) Calculate the correlated-groups test to determine the . (Note you will first have to solve t tobt
for the standard error of the difference score).
Work:
Standard Error (SE) = Standard Deviation (S ) / Square
D
Root (SR) of N
SE = 1.8 / SR (30) = 1.8 / 5.48
SE = 0.3285
T = Difference Score / SE
T = 3.4 / 0.3284 = 10.35
ANSWER
tobt = 10.35
5) Calculate the effect size using Cohen’s . Interpret it using the appropriate conventions d
(small, medium, or large).
Cohen’s d = Mean Difference / SD of Difference
D = 3.4 / 1.8 = 1.89
The effect size for this analysis (d = 1.89) was found to
exceed Cohen's (1988) convention for a (d = large effect
0.8).
ANSWER
Cohen’s d = 1.89
6) Is this a directional or non-directional study, and what is the ? (Note you will have t cv
to first calculate the df)
Work:
Nondirectional for two reasons:
1. Any change in the anxiety levels (pre / post
exercise) is of interest, not specifically increase or
decrease only.
2. Two-tailed t is asked to be used to determine 95%
cv
ANSWER
Nondirectional
Df = 29
tcv = + 2.045 (two tailed)
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PSYC 510
CI.
Df = n – 1 = 30 – 1
Df = 29
tcv = + 2.045 (two tailed)
7) Calculate the 95% confidence intervals (make sure you use the two-tailed ).t cv
Work:
CI95% = Mean Difference + tcv(SE)
CI95% = 3.4 + 2.045 (0.3285)
CI95% = 3.4 + 0.6718
CI95% = 2.7282, 4.0718
Lower Level CI = 2.7282
95%
Upper Level CI = 4.0718
95%
ANSWER
CI95% = 2.7282, 4.0718
8) Write an APA-style Results section based on your analyses. All homework “Results
sections” should follow the examples provided in the videos. Don’t forget to include the
effect size, confidence interval, and a decision about the null hypothesis.
ANSWER
A paired-samples t-test was conducted to compare the levels of anxiety in women in non-
exercise and exercise conditions with a random sample of 30 women in both conditions. There
was a significant difference in the anxiety levels for exercise and non-exercise (M = 3.4, SD =
1.8) conditions; t(29) = 10.35, p < .001, 95% CI (2.7282, 4.0718), d = 1.89. The effect size for
this analysis (d = 1.89) was found to exceed Cohen’s (1988) convention for a large effect (d =
0.8). So, our null hypothesis was rejected and the anxiety levels in women in both exercise
conditions were statistically different. These results suggest that women in non-exercise
condition had lower levels of anxiety than in exercising condition.
Part II: SPSS Application
These questions require the use of SPSS. Remember you must
submit all of your work within this word document. You will need
to take a screen shot of your data view if necessary, or copy and
paste your output into the spaces below. Remember to report the
exact p value provided by SPSS output – simply reporting p<.05 or
p>.05 is not acceptable (unless SPSS output states p=.000 – in
that case you can report p<.001).
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Research Scenario: Do you cope better than peers when facing difficult situations? In one
study, researchers reported that most individuals believe that they can cope better than their
fellow students (Igou, 2008). A recent researcher wanted to replicate the study. In this design,
participants read a scenario of a negative event and were asked to use a 10 point scale to rate
how it would affect their immediate well being (1 – 10, the higher the score, the better the
mood). They they were asked to imagine the event from the perspective of an ordinary fellow
student and rate how it would affect that person using the same scale.
Using the information provided, create an SPSS data file and conduct the appropriate
statistical test to determine whether people believe they can cope than their peers. better
Answer the following questions based on your analyses.
9) Conduct a correlated groups t-test. Paste the appropriate SPSS output below.
ANSWER
Paired Samples Statistics
Mean N Std. Deviation Std. Error Mean
Pair 1 Self_Rating 5.5833 12 1.92865 .55675
Peer_Rating 4.8333 12 1.69670 .48979
Paired Samples Correlations
N Correlation Sig.
Pair 1 Self_Rating & Peer_Rating 12 .949 .000
Paired Samples Test
Paired Differences
t df
Sig. (2-
tailed)Mean
Std.
Deviation
Std.
Error
Mean
95% Confidence
Interval of the
Difference
Lower Upper
Pair 1 Self_Rating - Peer_Rating .75000 .62158 .17944 .35507 1.14493 4.180 11 .002
10) Use the output to calculate r (show your work in the space provided and remember if
2
a number is negative, when squaring it will lose its sign and be positive). Interpret it
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Self-rating Rating of peer
4 4
8 6
7 6
5 5
7 6
5 4
3 2
5 4
2 2
7 7
8 7
6 5
PSYC 510
(small, medium, large) based on the conventions for this effect size calculation.
Work:
R2 = t / t + df
2 2
R2 = (4.18) / (4.18) + 11 = 17.47 / 28.47
2 2
R2 = 0.614
R2 has exceeded the Cohen’s (1988)
convention large (R = 0.25)
2
ANSWER
R2 = 0.614
11) Paste an appropriate SPSS graph (make sure you use the procedures outlined in this
module’s SPSS video– and don’t forget to label your y axis “Coping Ability”.
ANSWER
12) Write an APA-style Results section based on your analyses. All homework “Results
sections” should follow the examples provided in the videos. Don’t forget to include the
effect size, confidence interval, and a decision about the null hypothesis. If significant,
make sure you interpret how the conditions differ (refer to the Figure or report the
means and standard deviations).
ANSWER
A paired-samples t-test was conducted to compare the coping ability of students based on self-
rating and peer-rating (scale range: 1-10) with a random sample of 12 students in both
conditions. There was a significant difference (0.75 score) in the coping ability of students
between self-rating (M = 5.58, SD = 1.93) and peer-rating (M = 4.83, SD = 1.70); t(11) = 4.18,
p = .002, 95% CI (0.3551, 1.1449), d = 0.41. The effect size for this analysis (d = 0.41) was
found to exceed Cohen’s (1988) convention for a small effect (d = 0.2). So, our null hypothesis
was rejected as the scores of coping ability in self-rating mode were significantly higher than
the scores of peer-rating. These results suggest that people can cope better than their peers.
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PSYC 510
Part III: Cumulative
These questions can be related to anything covered thus far in the
course.
Does amount of information retained differ when participating in meetings in-person verses
virtually? In this study, participants got to choose whether they attend a quarterly meeting
in person or via Google Meet. After the meeting, they were given a brief assessment of
information covered – percent correct for each condition is presented in the table below.
Using the information provided, create an SPSS data file and conduct the appropriate
statistical test to determine whether there is a statistically significant in amount difference
of information retained based on whether attendance is in-person or virtual. Answer the
following questions based on your analysis.
13) Clearly identify the independent and dependent variables in this scenario.
ANSWER
Independent Variables
1. In-Person Meetings
2. Virtual Meetings
Dependent Variable
Information Covered / Retained
14) Paste the output of your statistical analysis below.
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In-Person Virtual
78 67
65 80
79 72
84 69
91 92
88 75
79 79
89 66
90 71
85 69
PSYC 510
ANSWER
Paired Samples Statistics
Mean N Std. Deviation Std. Error Mean
Pair 1 In_Person 82.8000 10 7.88529 2.49355
Virtual 74.0000 10 7.90218 2.49889
Paired Samples Correlations
N Correlation Sig.
Pair 1 In_Person & Virtual 10 -.036 .922
Paired Samples Test
Paired Differences
t df
Sig. (2-
tailed)Mean
Std.
Deviation
Std.
Error
Mean
95% Confidence Interval
of the Difference
Lower Upper
Pair 1 In_Person - Virtual 8.8000 11.36075 3.59258 .67301 16.92699 2.449 9 .037
15) Create an appropriate graph based on the data and paste it below (make sure
to have axes labels and error bars).
ANSWER
16) Write an APA-style Results section based on your analyses. Remember to use
complete sentences, include the statistical notation, effect size, confidence
interval, and include a decision about the null hypothesis. If it is significant,
state how by reporting the means and standard deviations or refer the reader
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to the previous Figure.
ANSWER
A paired-samples t-test was conducted to compare the amount of information retained in
in-person meetings and virtual meetings with a random sample of 10 meetings in both
types. There was a significant difference (8.8 units of information) in the in-person
meetings of students between in-person meetings (M = 82.8, SD = 7.89) and virtual
meetings (M = 74.0, SD = 7.90); t(9) = 2.45, p = .037, 95% CI (0.673, 16.927), d = 1.11.
The effect size for this analysis (d = 1.11) was found to exceed Cohen’s (1988) convention
for a large effect (d = 0.8). So, our null hypothesis was rejected as the amount of
information retained in in-person meetings was higher than in the virtual meetings.
These results suggest that in-person meetings are better than virtual meeting in term of
retaining information.
17) Discuss one threat to internal validity as it specifically relates to this research
scenario.
One of the threats to internal validity is the selection or sampling bias, such bias arises
when a certain type of sample is chosen for certain type of task instead of random
assignment of the certain task across the entire sample. This bias can also arise when
people volunteer them to participate in one condition of an experiment and withdraw from
the other / next condition. This can be explained through an example. Let’s say a teacher
gives an assignment of organic chemistry to his female students and assignment of
inorganic chemistry to his male students. Females solve the organic chemistry assignment
quickly than male do the inorganic, based on that teacher decides that organic chemistry is
an easy subject. While there is a possible confounding part that is females were a certain
task and males were given another certain task. It's possible that females are good at
organic chemistry, yet this can’t be concluded until a random selection is done or both are
given the same assignment. Random allocation may not be practical in many cases but that
doesn’t rule out this potential bias of selection / sampling.
Use this information to answer the following questions:
Age at onset of dementia was determined in the general population to be = 70 and
= 7.0.
18) Based on the data above, what is the z score for someone being diagnosed with
dementia at age 65 (can round to two decimal places)? What percentage of
people might start to show signs of dementia before this age ( DO NOT
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ROUND).
ANSWER
Z score = -0.714
ANSWER
% = 23.89
Work:
z = (X − μ) / σ
z = 65 – 70 / 7
z = -5 / 7
z = -.714
Percentile = 0.2389 from z-table
So, 23.89% people would start showing signs of dementia before the age of
65.
Done!
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