Chapter 2
11) For the following scores, find the (a) mean, (b) median,
(c) sum of squared deviations, (d) variance, and (e) standard deviation:
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2,
2, 0, 5, 1, 4, 1, 3, 0, 0, 1, 4, 4, 0, 1, 4, 3, 4, 2, 1, 0
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a)
(2+2+0+5+1+4+1+3+0+0+1+4+4+0+1+4+3+4+2+1+0)/21=2
b) median value: 2
c) (2-2)2+(2-2)2+(0-2)2+(5-2)2+(1-2)2+(4-2)2+(1-2)2+(3-2)2+(0-2)2+(0-2)2+(1-2)2+(4-2)2+(4-2)2+(0-2)2+(1-2)2+(4-2)2+(3-2)2+(4-2)2+(2-2)2+(1-2)2+(0-2)2 = 56
d) 56/21 = 2.667
e) SQRT (2.667) = 1.633
12) For the following scores, find the (a) mean, (b) median,
(c) sum of squared deviations, (d) variance, and (e) standard deviation:
1,112;
1,245; 1,361; 1,372; 1,472
a)
(1112+1245+1361+1372+1472)/5=1312.4
b) 1361
c) (1112-1312.4)² +
(1245-1312.4)² + (1361-1312.4)² + (1372- 1312.4)² + (1472-1312.4)² = 76089.2
d)
76089.2/5=15217.84
e)
SQRT (15217.84) = 123.361
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3.0,
3.4, 2.6, 3.3, 3.5, 3.2
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a)
(3+3.4+2.6+3.3+3.5+3.2)/6=3.167
b) 3.25
c) (3.0-3.167)² +
(3.4-3.167)² + (2.6-3.167)² + (3.3- 3.167)² + (3.5-3.167)² + (3.2-3.167)² = 0.533
d) 0.53/6=0.089
e) SQRT (0.089) = 0.298
16)
A
psychologist interested in political behavior measured the square footage of
the desks in the official office of four U.S. governors and of four chief
executive officers (CEOs) of major U.S. corporations. The figures for the
governors were 44, 36, 52, and 40 square feet. The figures for the CEOs were
32, 60, 48, and 36 square feet. (a) Figure the means and standard deviations
for the governors and for the CEOs. (b) Explain, to a person who has never had
a course in statistics, what you have done. (c) Note the ways in which the
means and standard deviations differ, and speculate on the possible meaning of
these differences, presuming that they are representative of U.S. governors and
large corporations’ CEOs in general.
a) Governors: 36; 44; 52; 40
Mean: 43
Standard deviation: 6.83
CEOs: 32; 60; 48; 36
Mean: 44
Standard deviation: 12.65
b) We have just found the
average measurements of the desks of the governors and CEOs in major US
corporations. We also found the spread of the data.
c) The average, or mean of
the measurements are almost the same in both cases. However, the standard
deviation of the CEOs’ measurements is almost double of the governors’, which
means that the CEO’s desks can be either much bigger or much smaller than the
governors’.
21)
Payne
(2001)
gave participants a computerized task in which they first see a face and then a
picture of either a gun or a tool. The task was to press one button if it was a
tool and a different one if it was a gun. Unknown to the participants while
they were doing the study, the faces served as a “prime” (something that starts
you thinking a particular way); half the time they were of a black person and
half the time of a white person. Table 2–9
shows the means and standard deviations for reaction times (the time to decide
if the picture is of a gun or a tool) after either a black or white prime. (In
Experiment 2, participants were told to decide as fast as possible.) Explain
the results to a person who has never had a course in statistics. (Be sure to
explain some specific numbers as well as the general principle of the mean and
standard deviation.)
The mean reaction time is the average time
it took for the participants to press a button, either for gun or for tool. The
mean for the gun button being pressed after the computer showing a black face
is less than that for the tool button after the black face. This shows that the participants were more
prompt to press the gun button. Also, the fact that it took longer for the
participants to push the tool button after a black face was shown, shows that
the participants did not expect the tool to show, and had to think longer about
that. In the second experiment, the mean decreased as the participants had to
press button as fast as possible, which decreased the response times. The
standard deviation shows the variability of reaction times within a given
group, which was also decreased in the second experiment.
Chapter 3
14) On a standard measure of
hearing ability, the mean is 300 and the standard deviation is 20. Give the Z
scores for persons who score (a) 340, (b) 310, and (c) 260. Give the raw scores
for persons whose Z scores on this test are (d) 2.4, (e) 1.5, (f) 0, and
(g) –4.5.
Mean = 300
Standard deviation = 20
z = (x - mean) / standard
deviation
a) z (340) = (340-300)/20=2
b) z (310) = (310-300)/20=0.5
c) z (260) = (260-300)/20=-2
x = z * (standard deviation)
+ mean
d) x = 2.4*20+300 = 348
e) x = 1 .5*20+300 = 330
f) x = 0*20+300 = 300
g) x = - 4.5*20+300 = 210
15)
A
person scores 81 on a test of verbal ability and 6.4 on a test of quantitative
ability. For the verbal ability test, the mean for people in general is 50 and
the standard deviation is 20. For the quantitative ability test, the mean for
people in general is 0 and the standard deviation is 5. Which is this person’s
stronger ability: verbal or quantitative? Explain your answer to a person who
has never had a course in statistics.
Z-scores:
Verbal ability:
z (81) = (81-50)/20 = 31/20
= 1.55
Quantitative ability:
z (6.4) = (6.4-0)/5 = 1.28
We can see that the verbal
ability score is higher than the quantitative ability score, which shows that
this person is stronger with verbal skills.
22)
Suppose
you want to conduct a survey of the attitude of psychology graduate students
studying clinical psychology toward psychoanalytic methods of psychotherapy.
One approach would be to contact every psychology graduate student you know and
ask them to fill out a questionnaire about it. (a) What kind of sampling method
is this? (b) What is a major limitation of this kind of approach?
a) This is considered a
convenience sampling.
b) This method is not always
very effective because the sample does not necessarily represent the
population, since it was not randomly selected. Because of this, the results
will not be accurately transferrable to the population.
25)
You
are conducting a survey at a college with 800 students, 50 faculty members, and
150 administrators. Each of these 1,000 individuals has a single listing in the
campus phone directory. Suppose you were to cut up the directory and pull out
one listing at random to contact. What is the probability it would be (a) a
student, (b) a faculty member, (c) an administrator, (d) a faculty member or
administrator, and (e) anyone except an administrator? (f) Explain your answers
to someone who has never had a course in statistics.
a)
P (student) = 800/1000 = 0.8
b)
P (faculty) = 50/1000 = 0.05
c)
P (administrator) = 150/1000 = 0.15
d)
P (faculty member or administrator) = 0.05+0.15 = 0.2
e)
P (anyone except an administrator) = 1-0.15 = 0.85
f) To find the probability
of the occurrence of an event, we must divide the number of successes by the
number of possible outcomes. By doing that we can find the probability of an
event happening. For example, dividing the number of students by the total
number of individuals, we get the probability of a student being selected, and
so on.
References
Aron,
A., Aron, E. N., & Coups, E. J. (2009). Statistics for psychology (5th
ed.). Upper Saddle River, NJ: Pearson/Prentice Hall.
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