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University of Toronto Mississauga

Psychology

PSY201H5

Dax Urbszat

Fall

Description

Name____________________________ 1
Psychology 201F
Practice Questions (Answers)
Put you name on the top of each page. Answer all questions. Show all of your work. Below are
some formulas that you might want to use.
The binomial formula is
Aids: Calculator. Name____________________________ 2
1. (30 points). Following the September 11 terrorist attack in New York, Air Canada wanted to
find out how apprehensive frequent flyers were about flying. They examined their data base to
identify all of those individuals who flew more the 30,000 km last year. They took a random
sample of size 100 from this list of frequent flyers and told them about their new security
procedures. They then asked them to rate how safe they would feel about taking a flight to the
U.S. next week on a scale from 0 to 100 where 0 meant that they did not feel at all safe, and 100
meant that they were perfectly safe. The data are presented below.
A. Plot a frequency histogram of these data using 10 class intervals. Be sure to specify how you
arrived at these class interval boundaries and show all of your work. The safeness scores have
been ordered for your convenience.
SAFENESS SCORE
18, 25, 26, 26, 26, 27, 27, 28, 29, 29, 30, 32, 33, 34, 35
35, 36, 36, 38, 39, 39, 39, 39, 40, 41, 42, 43, 43, 44, 44
44, 44, 45, 45, 45, 46, 47, 48, 48, 49, 49, 50, 52, 52, 52
53, 54, 55, 55, 56, 57, 63, 64, 68, 68, 70, 71, 71, 72, 73
75, 75, 75, 75, 77, 77, 78, 78, 79, 80, 80, 80, 81, 81, 82
82, 82, 82, 83, 83, 84, 85, 85, 87, 87, 87, 89, 89, 91, 92
92, 92, 93, 94, 95, 97, 97, 98, 99, 400
Accuracy of measurement: first digit to left of decimal point.
Outlier: The score 400 is not only an outlier but also violates the instructions and can be
discarded.
Range: 99-18 = 81
Category width 81/10 = 8.1, round up to 9.
Lower boundary of lowest category: 18 - ½ possible difference between scores = 18-0.5 = 17.5
Categories Frequency
17.5-26.5 5
26.5-35.5 11
35.5-44.5 16
44.5-53.5 14
53.5-62.5 5
62.5-71.5 7
71.5-80.5 14
80.5-89.5 16
89.5-98.5 10
98.5-107.5 1
B. Describe the distribution of safeness scores. The distribution is bimodal. The distribution Name____________________________ 3
e 22 31 40 49 58 67 76 85 94 103
o
c
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s12.5
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a
S s7.5
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y
c 2.5
n
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q
r 22 31 40 49 58 67 76 85 94 103
F f Safeness Scores
around each of the modes looks roughly symmetrical.
C. What measures of Central Tendency and Variability (if any) would be appropriate for these
data. Why? Because the distribution is bimodal we need to specify the location of each mode.
These locations are at 40 and 85. Because of the bimodal nature of the distribution, none of the
measures of variability are appropriate and need not be calculated.
D. Compute these measures. The two modes occur at 40 and 85.
2. (20 points) A clothes manufacturer won a contract to provide soccer shorts to all of the
recreational soccer players in a particular district. Because short size depends primarily on the
circumference of a player’s waist, he randomly selected 500 players from the district and
measured the circumference of their waists in inches. The frequency histogram of waist length is
presented below, along with each of the categories used in constructing the histogram.
This manufacturer is geared up to manufacturer 6 sizes of soccer shorts. Size one will fit
those with waist lengths between 25 and 28 inches, size two between 28 and 31 inches, size 3
between 31 and 34 inches, size 4 between 34 and 37 inches, size 5 between 37 and 40 inches,
size 6 between 40 and 43 inches. Name____________________________ 4
If the contract calls for 10,000 pairs of soccer shorts, how many pairs of each sizes 2, 3 and 4
should he plan on producing. How many individuals will he not be able to fit?
23 24.5 26 27.5 29 30.5 32 33.5 35 36.5 38 39.5 41 42.5
80
z
i60
S
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o40
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23 24.5 26 27.5 29 30.5 32 33.5 35 36.5 38 39.5 41 42.5
Waist Since in inches
Class Interval in Inches Frequency Cumulative Frequency Cum. %
22.25-23.75 1 1 0.2
23.75-25.25 8 9 1.8
25.25-26.75 20 29 5.8
26.75-28.25 39 68 13.6
28.25-29.75 87 155 31.0
29.75-31.25 59 214 42.8
31.25-32.75 73 287 57.4
32.75-34.25 60 347 69.4
34.25-35.75 54 401 80.2
35.75-37.25 49 450 90.0
37.25-38.75 29 479 95.8
38.75-40.25 15 494 98.8
40.25-41.75 4 498 99.6
41.75-43.25 2 500 100.0 Name____________________________ 5
Size two fits 28 to 31. Therefore we need to compute the percentage of scores between 28 and
31. We can do this by determining the percentile ranks of the scores 31 and 28 and then
subtracting. For size 3 it is the difference between the percentile ranks of 34 and 31, and size 4
is the difference between the percentile ranks of 37 and 34. The percentage that he can’t fit is
given by the percentile rank of the lowest size namely, 25.
y −5.8 13.6. −58 y −31 42.8 −31
28−26.75 = 28.2.5−2675 =12.3 , 31−29.75 = 31..5−2975 = 40.83
Size 2 should fit 28.53% of the population, and he should manufacture 2853 of size 2 shorts.
y −57.4 69.. −574
34 −32.75 = 34.2.5−3275 = 67.4
Size 3 should fit 67.4-40.83 = 26.57% of the population. Manufacture 2657 of size 3
y −80.2 90−80.2
37 −35.75 = 37..5−3575 = 88.37
Size 4 should fit 88.37 - 67.40 = 20.97, manufacture 2097 of size 4
y −0.2 1.. −02 , He will not be able to fit 1.53% of the population or
25−23.75 = 25..5−2375 =1.53
153 individuals. Name____________________________ 6
3. (20 points). Suppose you have a jar containing 25 red (R) marbles, 35 green (G) marbles, and
40 blue (B) marbles. Assume that you are drawing from this jar without replacement. Compute
the follow probabilities.
A. The probability of drawing 3 G marbles in a row.
B. The probability that you will not get a Red marble in three draws.
B. The probability that out of 3 draws, you will only get Red or Green marbles.
C. The probability that in 4 draws, exactly 2 of the marbles are blue and 2 are green.
A. The probability of drawing 3 G marbles in a row. Only 1 sequence gives 3 Greens in 3
draws.
p(GGG) = (35/100)*(34/99)*(33/98) = 39270/970200 = .04048
B. The probability that you will not get a Red marble in three draws.
The probability of not getting a red marble on the first draw is = number of not red
marbles divided by 100. The number of not red marbles are 75. When phrased in terms of not
red there is only one sequence that will yield 3 not reds in 3 draws, NR, NR, NR.
p(NR,NR,NR) = (75/100)*)(74/99)*(73/98) = 405150/970200 = .41759
The other way of doing this is to list all the sequences of only green and blue marbles in 3 draws.
GGG, GGB, GBG, BGG, GBB, BGB, BBG, BBB. Then calculate the probability of each
sequence and sum the probabilities.
B. The probability that out of 3 draws, you will only get Red or Green marbles.
The number of red and green marbles is 60, and red and green are mutually exclusive.
Therefore the probability of getting only red or green marbles in 3 draws is
(60/100)*(59/100)*(58/98) = 205320/970200 = .21163
C. The probability that in 4 draws, exactly 2 of the marbles are blue and 2 are green.
You need to construct all the sequences containing exactly two red and two green
marbles.
RRGG, RGRG, RGGR, GRRG, GRGR, GGRR. Or you could simply count them by
4!/(2!2!).
p(RRGG) = (25/100)*(24/99)*(35/98)*(34/97) = 714000/94109400 = .000758691
All sequences have the same probability. Therefore the probability of exactly 2 red and 2 green
is six times the probability of any one sequence = .00455215.
4. (25 points). A researcher wants to see if monkeys can remember the location of food, 24
hours after having seen the food hidden behind one of 10 doors. To test whether this is possible, Name____________________________ 7
the researcher, in the presence of a monkey, puts a cache of food behind a randomly picked door.
He then removes the monkey for 24 hours and then puts it back in the apparatus, and records the
first door that the monkey opens. He does this once to each of 20 monkeys. Of the 20 monkeys,
6 choose the door with the food behind it on the first try. What kind of experiment is this?
Why? State the null and research hypothesis in words and in statistical form. On the graph
below plot the probability histogram for this experiment assuming that the null hypothesis is
true. Using " = .05 determine the rejection region for the null hypothesis and indicate where it
falls on the graph. What would you conclude if 6 of the 20 monkeys choose the door with the
food behind it on the first try?
This is a binomial experiment because
1. There are two and only two mutually exclusive outcomes (picks the right door or not).
2. Under the null hypothesis the probability of choosing the correct door is constant for each
individual monkey.
3. The trials are independent. That is, what door individual X chooses will not influence
individual Y..
4. The statistic of interest is the number of monkeys out of 20 who choose the correct door.
The null hypothesis is that the monkey has no memory of where the food was cached.
The research hypothesis is that the monkey can remember or has some recollection of where the
food was cached.
Under the null hypothesis, the monkey will be choosing a door at random. Therefore the
probability of the monkey being correct is 1/10 or 0.1. If the monkey recollects anything about
where the food was cached, the probability of being correct should be greater than 0.1.
H 0: p = 0.1
H R p > 0.1
For " as close to but not exceeding .05, or for " as close .05 as possible, one-tail test, n = 20, the
rejection region is y > 4. The exact value of " is .0433.
Since the number of successes (6) falls in the rejection region, we reject the null hypothesis and
conclude that monkeys have some recollection of where the food was cached 24 hrs after the
event. Name____________________________ 8
0 1 2 3 4 5 6 7 8 11121141511718120
s
c0.25
c
S 0.2
o s
y0.15
t
l 0.1
b
a0.05
o
P f 0
0 1 2 3 4 5 6 7 8 11121141511718120
Successes Name____________________________ 9
5. (30 points total). Recently, a Senate Committee recommended that the growth, distribution
and possession of marijuana be legalized under the Controlled Substances Act. A number of
large corporations, sensing a potential for large profits, commissioned a number of studies on the
potential market for legalized marijuana. In particular, the advertising people in these
corporations wanted some information on the age distribution of potential purchasers so that they
could devise age-appropriate advertising strategies. Accordingly, they conducted a random
sample of 400 people. Of this random sample, 133 said that they already did or would start to
use marijuana on a regular basis if the product was legalized. The age distribution of these 133
respondents is given below.
A. (20 points). Plot a frequency histogram of these data using 13 class intervals. Be sure to
specify how you arrived at these class interval boundaries and show all of your work. The ages
of the respondents have been ordered for your convenience.
9,11,11,11,11,11,11,11,12,12,12,13,13,13,14,14,14,14,14,14,14,15
15,15,15,15,15,15,16,16,16,16,16,16,17,17,17,17,17,17,17,17,17,18
18,18,18,18,18,18,18,18,18,18,18,19,19,20,20,20,20,20,21,21,21,21
21,21,21,21,21,21,21,22,22,22,22,22,23,24,24,24,27,28,30,31,32,34
34,36,36,36,39,40,40,40,40,41,41,42,42,43,43,44,44,45,45,45,46,46
47,47,48,48,49,51,52,53,54,55,55,57,57,57,57,59,60,62,62,66,69,71
93
The measurement precision is to the first place to the left of the decimal point. There is some
question as to whether 93 should be treated as an outlier. No points are take off for not
identifying it as an outlier. If it is not identified as an outlier then
Range = 93 - 9 = 82
Lowest category boundary = 9 - half measurement precision = 9-.5 = 8.5
Category width = 82/13 carried to one decimal place beyond precision of measurement is
6.3, round up to 7.
Category boundaries and category frequencies are Name____________________________ 10
iLower Boundary Upper Boundary Fryquency
8.5 15.5 28
15.5 22.5 50
22.5 29.5 6
29.5 36.5 8
36.5 43.5 11
50.5 57.5 10
57.5 64.5 4
64.5 71.5 3
71.5 78.5 0
78.5 85.5 0
85.5 92.5 0
k 92.5 99.5 1 {
12 19 26 33 40 47 54 61 68 75 82 89 97
50
40
c
e30
q
e20
F
10
0
12 19 26 33 40 47 54 61 68 75 82 89 97
Age in Years Name____________________________ 11
If the highest measure is identified as the outliner, then
Range = 71-9 = 62
Lowest Boundary is 62/13 = 4.8, round up to 5.
iLower Boundary Upper Boundayy Frequency
8.5 13.5 14
18.5 23.5 24
23.5 28.5 5
28.5 33.5 3
33.5 38.5 5
43.5 48.5 11
48.5 53.5 4
53.5 58.5 7
58.5 63.5 4
63.5 68.5 1
k >890 73.5 1 {
11162126313641465156616671 > 90
40
30
y
c
n
u 20
q
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F
10
0
11162126313641465156616671 > 90
Age in Years Name____________________________ 12
B. (4 points). Describe the age distribution of marijuana users.
The frequency histogram shows two peaks, one for young users, one for middle-aged
users, with some positive skew at the higher ages.
C. (4 points). What measures of Central Tendency and Variability (if any) would be appropriate
for these data. Why?
Because there are two peaks or modes in this histogram, it would be appropriate to
specify the locations of the two modes as measures of central tendency. Because measures of
variability in a bimodal distribution do not indicate the degree of spread around either mode, you
would not normally use either the inter-quartile range or the sample standard deviation to
describe the variability in the data.
E. (2 points). What age groups should the advertisers target?
Advertisers should target two age ranges, one for young users in their late teens, and the
other being middle aged users.
6. (25 points) A recreational soccer league had 810 children aged 10-12 who wanted to play
soccer. From previous experience, league officials knew that the optimal number of teams in a
division was 8, and the optimal number of players per team was 20. Hence they decided that
they would create 5 divisions of 8 teams with approximately 20 players per team. They also
decided that they wanted the different leagues to have different levels of skill, with league A
being the league with the most skilled players, League B being those with the next level of skill,
League C being those with average skill, League D being those whose skills were one level
below average, and League E being those with the lowest level of skill. To assess skill, each
player was asked to go through a number of soccer drills with the player’s performance being
scored by a panel of expert coaches on a scale from 0 to 10. The frequency histogram of players’
skills is presented below, along with each of the categories used in constructing the histogram. Name____________________________ 13
1.20 2.52 3.84 5.16 6.48 7.80
100
y 80
n
u 60
q
r 40
F
20
0
1.20 2.52 3.84 5.16 6.48 7.80
Soccer skill score Name____________________________ 14
Category Boundaries, Frequency, Cum. Freq. and Cum. Percentage (Soccer Skill Scores)
iLower Boundary Upper Boundary Frequency Cumulative Freq. Cumulative Percenyage
1.035 1.365 12 12 1.48148
1.365 1.695 19 31 3.82716
1.695 2.025 54 85 10.4938
2.025 2.355 86 171 21.1111
2.355 2.685 116 287 35.4321
2.685 3.015 99 386 47.6543
3.015 3.345 116 502 61.9753
3.345 3.675 68 570 70.3704
3.675 4.005 65 635 78.3951
4.005 4.335 51 686 84.6914
4.335 4.665 39 725 89.5062
4.665 4.995 24 749 92.4691
4.995 5.325 21 770 95.0617
5.325 5.655 18 788 97.284
5.655 5.985 6 794 98.0247
5.985 6.315 5 799 98.642
6.315 6.645 1 800 98.7654
6.645 6.975 3 803 99.1358
6.975 7.305 2 805 99.3827
7.305 7.635 3 808 99.7531
7.635 7.965 0 808 99.7531
7.965 8.295 1 809 99.8765
k 8.295 8.625 1 810 100. {
A. (16 points). Find the skill boundaries between Leagues A, B, C, D, and E. Specify
how you arrived at these boundaries.
You need to find the skill boundaries that divide the sample into those below the 20
percentile value, those between the 20rcentile values, those between the 40 and 60
percentile values, those between the 60 and 80 percentile values, and those greater than the
80 percentile value.
The 20percentile value lies between the category boundaries, 2.025 and 2.355. The percentile
ranks corresponding to these two boundaries are 10.4938 and 21.1111. Hence {x1,y1) = (2.025,
10.4938) and (x2,y2) = (2.355, 21.1111). The score, x, that we are trying to locate is has the
coordinates (x, 20) on the cumulative relative frequency graph. Therefore,
21.1.111−104938 20−10.4983
=
23.5 2025 x −2.0235
Solving for x yields 2.32047. Hence kids with scores < 2.32047 go into League E. Name____________________________

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