1) In a planned study, here is a known population with a normal distribution, pop
mean = 50, pop standard dev = 5. What is the predicted (standardized) effects
size (d) if the researchers predict that those given an experimental treatment
have a mean of
a) 50

 null 50  50
d  treatment

 0 ; this is no effect

5
b) 52

 null 52  50
d  treatment

 0.4 ; this is a small effect

5
c) 54

 null 54  50
d  treatment

 0.8 ; this is a large effect

5
d) 56

 null 56  50
d  treatment

 1.2 ; this is a large effect

5
e) 47

 null 47  50
d  treatment

 0.6 ; this is medium (negative) effect

5
For each part, also indicate whether the effect is approximately small, medium,
or large.
2) For each of the following studies, make a chart of the four possible correct
and incorrect decisions, and explain what each would mean.
a) A study of whether increasing the amount of recess time improves
school children's in-class behavior.
Reject H0
Fail to reject
H0
H0 is true
Incorrectly concluding that
increasing recess time does
improve children’s in-class
behavior, when it really
doesn’t
Correctly concluding that
increasing recess time does
not improve children’s in-class
behavior, when it really
doesn’t
H0 is false
Correctly concluding that
increasing recess time does
improve children’s in-class
behavior, when it really does
Incorrectly concluding that
increasing recess time does
not improve children’s in-class
behavior, when it really does
b) A study of whether color-blind individuals can distinguish gray shades
better than the population at large.
Reject H0
Fail to reject
H0
H0 is true
Incorrectly concluding that
color blind individuals can
distinguish gray shades better
than the general population,
when they really can’t
Correctly concluding that color
blind individuals can not
distinguish gray shades better
than the general population,
when they really can’t
H0 is false
Correctly concluding that color
blind individuals can
distinguish gray shades better
than the general population,
when they really can
Incorrectly concluding that
color blind individuals can not
distinguish gray shades better
than the general population,
when they really can
c) A study comparing individuals who have ever been in psychotherapy to
the general public to see if they are more tolerant of other people's upsets
than is the general population.
Reject H0
Fail to reject
H0
H0 is true
Incorrectly concluding that
individuals who have been in
psychotherapy are more
tolerant of others, when they
really aren’t
Correctly concluding that
individuals who have been in
psychotherapy are not more
tolerant of others, when they
really aren’t
H0 is false
Correctly concluding that
individuals who have been in
psychotherapy are more
tolerant of others, when they
really are
Incorrectly concluding that
individuals who have been in
psychotherapy are not more
tolerant of others, when they
really are
3) Here is information about several possible versions of a planned experiment.
Figure effect size and power for each; sketch the distributions involved, showing
the area for alpha, beta and power (Assume all populations have a normal
distribution). (I haven’t included sketches because that takes a lot longer to do
in the digital realm)
Population
A
Mean
90
Std dev
4
Predicted
Mean
treatment
91
n
alpha-level
1 or 2
tailed
100
0.05
1
B
C
D
E
F
90
90
90
90
90
4
2
4
4
4
92
91
91
91
91
100
100
16
100
100
0.05
0.05
0.05
0.01
0.05
1
1
1
1
2
NOTE: These problems use distributions of sample means rather than just
populations to compute these answers. This means that you’ll need to use the
standard error rather than the population standard deviation in the compuations
of statistical power.

 null 91  90

 0.25 ; this is a small effect
a) d  treatment

4
 critical Z for  = 0.05 & 1-tailed test is 1.645
 convert this z-score into a raw score (using the information from the
null distribution)
  90  90.66
X  Z  X    1.645  4

100 
 convert this raw score into a z-score (using the information from the
treatment distribution)
X   90.66  91
Z

 0.85
4
X
100
 using the unit normal table look up the area under the curve in the tail
for that z-score: z(-0.85) = 0.1977 = 
 Power = 1- = 1 – 0.1977 = 0.8123 or 81.23%
treatment  null 92  90

 0.50 ; this is a medium effect

4
 critical Z for  = 0.05 & 1-tailed test is 1.645
 convert this z-score into a raw score (using the information from the
null distribution)
  90  90.66
X  Z  X    1.645  4

100 
 convert this raw score into a z-score (using the information from the
treatment distribution)
X   90.66  92
Z

 3.35
4
X
100
 using the unit normal table look up the area under the curve in the tail
for that z-score: z(-3.35) < 0.001
 1977 = 
 Power = 1- = 1 – 0.001 = 0.999 or approximately 100%
b) d 
treatment  null 91  90

 0.50 ; this is a medium effect

2
 critical Z for  = 0.05 & 1-tailed test is 1.645
 convert this z-score into a raw score (using the information from the
null distribution)
  90  90.33
X  Z  X    1.645  2

100 
 convert this raw score into a z-score (using the information from the
treatment distribution)
X   90.66  91
Z

 1.7
2
X
100
 using the unit normal table look up the area under the curve in the tail
for that z-score: z(-1.7) = 0.0446
 1977 = 
 Power = 1- = 1 – 0.0446 = 0.9554 or 95.54%
c) d 
treatment  null 91  90

 0.25 ; this is a small effect

4
 critical Z for  = 0.05 & 1-tailed test is 1.645
 convert this z-score into a raw score (using the information from the
null distribution)
  90  91.65
X  Z  X    1.645  4

16 
 convert this raw score into a z-score (using the information from the
treatment distribution)
X   90.66  91
Z

 0.34
4
X
16
 using the unit normal table look up the area under the curve in the tail
for that z-score: z(-0.34) = 0.3669
 1977 = 
 Power = 1- = 1 – 0.3669 = 0.6331 or 63.31%
d) d 
treatment  null 91  90

 0.25 ; this is a small effect

4
 critical Z for  = 0.01 & 1-tailed test is 2.33
 convert this z-score into a raw score (using the information from the
null distribution)
  90  90.93
X  Z  X    2.33 4

100 
 convert this raw score into a z-score (using the information from the
treatment distribution)
e) d 
X
90.93  91
 0.175
4
X
100
 using the unit normal table look up the area under the curve in the tail
for that z-score: z(-0.175) = 0.4286
 1977 = 
 Power = 1- = 1 – 0.4286 = 0.5714 or 57.14%
Z

treatment  null 91  90

 0.25 ; this is a small effect

4
 critical Z for  = 0.05 & 2-tailed test is ±1.96
 convert this z-score into a raw score (using the information from the
null distribution)
  90  90.78
X  Z  X    1.96  4

100 
 convert this raw score into a z-score (using the information from the
treatment distribution)
X   90.78  91
Z

 0.55
4
X
f) d 
100
 using the unit normal table look up the area under the curve in the tail
for that z-score: z(-0.55) = 0.2912
 1977 = 
 Power = 1- = 1 – 0.2912 = 0.7088 or 70.88%
4) On a particular memory task in which words are learned in a random order, it
is known that people can recall a mean of 11 words with a standard deviation of
4, and the distribution follows a normal curve. A cognitive psychologist, to test a
particular theory, modifies that task so that the words are presented in a way in
which words that have a related meaning are presented together. The cognitive
psychologist predicts that under these conditions, people will recall so many
more words that there will be a large effect size. She plans to test this with a
sample of 20 people, using the alpha level = 0.01 and a two tailed test.
a) What is the power of this study?
 she is expecting a large effect size, so we will assume Cohen’s d = 0.8

 null
d  treatment
 treatment  d    null  0.8 4   11  14.2

 critical Z for  = 0.01 & 2-tailed test is ±2.58
 convert this z-score into a raw score (using the information from the
null distribution)
  11  13.31
X  Z  X    2.58  4

20 
 convert this raw score into a z-score (using the information from the
treatment distribution)
X   13.31  14.2 2.68
Z


 1.00
4
X
0.89
20
 using the unit normal table look up the area under the curve in the tail
for that z-score: z(-1.00) = 0.1587
 1977 = 
 Power = 1- = 1 – 0.1587 = 0.8413 or 84.13%
b) Sketch the distributions involved, showing the area for alpha, beta, and
power.
(I haven’t included sketches because that takes a lot longer to do in the
digital realm)
c) Explain your answer to someone who understands hypothesis testing
involving means of samples but has never learned about effect size or
power.
Using her experience with the research area, the researcher expects to
have a large effect. Using this expectation, she is able to predict the
approximate raw score mean of the treatment group. Using this
information (and information about the population standard deviation and
the number of participants in her sample), she was able to predict the
likelihood of being able to detect an effect of that size if it is present.