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PsyB07 – Summer 2006
Midterm Exam
June 20, 2006
Name:
ID#:
Welcome to the midterm. It may take a few minutes to get everything and everyone set up, so in
the mean time please read (and think about) the following points.

Make sure you write your full name and ID# at the top of this page

The midterm is scheduled for three hours, but I expect it to take you about 100 minutes to
complete.

You should have ample room to write your answers in the spaces provided. Do not write
the answers on the back of the pages, use them for rough work.

Remember to show all your work. That means showing the equations in symbolic form,
substituting in the values, and then solving.

Round all final answers to two decimal places (unless the answer is a probability from a
table)

Make sure your writing is legible and in sentence form. Avoid using “point form”.

For some questions there are easy (i.e., fast) and hard (i.e., slow) ways to answer them.
Try to think about the most efficient way to answer each question before jumping in.

Please keep in mind that I would like nothing more than to catch someone cheating and to
make an example out of them. Please do not be that person – do not look at the work of
others and do not make it easy for others to look at your work.
GOOD LUCK!!!
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Multiple-Choice (each worth 2 marks)
Please make sure you write your answers to these questions in the booklet, as well as here.
1. If the null hypothesis is true, then (because we predetermine alpha) the probability of failing
to reject the null is:
a. .05
b. .5
c. based on the number of subjects
d. .95
e. unknown
2. When a distribution is ______ skewed, the ________ is necessarily greater than the _______.
a. negatively skewed; mode; median
b. negatively skewed; mean; median
c. positively skewed; mode; mean
d. positively skewed; median; mean
e. positively skewed; mode; median
3. Let’s say that on any given weekday you will go to school with a probability of 0.80, but will
stay in bed with a probability of 0.10. These probabilities provide:
a. an exhaustive set of mutually exclusive probabilities
b. an exhaustive set of independent probabilities
c. an exhaustive set of non-exclusive probabilities
d. a non-exhaustive set of independent probabilities
e. a non-exhaustive set of mutually exclusive probabilities
4. When doing hypothesis testing, we typically evaluate some value or mean with respect to its
probability of occurrence relative to:
a. the relevant sampling distribution assuming H0 is true
b. the relevant sampling distribution assuming H0 is false
c. the relevant sampling distribution assuming H1 is true
d. the relevant sampling distribution assuming H1 is false
e. the relevant sampling distribution assuming both H0 and H1 are false
5. Random sampling helps ensure ______ just as random assignment helps ensure ______
a. internal validity; external validity
b. external validity; internal validity
c. both help ensure internal validity
d. both help ensure external validity
e. I forgot the difference between internal and external validity (hint: don’t pick this)
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6. (4 marks) Identify the scale of measurement for each of the following variables:
a) Temperature (Celsius scale)
INTERVAL
b) Temperature (Fahrenheit scale)
INTERVAL
c) Smurf gender
NOMINAL
d) Rankings applicants to graduate schools
ORDINAL
7. (3 marks) Following are the moral development scores of a sample of Grade 1 children. Plot
the data as a frequency histogram.
7, 6, 2, 2, 3, 7, 5, 7, 4, 3, 8, 5, 7, 6, 7
Title
f
r
e
q
16
14
12
10
8
6
4
2
0
x
2
3
4
5
6
7
8
2
(2marks)
3
4 5 6
MDS
7
freq
2
2
1
2
2
5
1
8
(2 marks)
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8. (9 marks) For the above set of data calculate the (do the calculations on the back – fill in the
final answers below):
mean
5.27
range
6
median
6
variance
4.06
mode
7
SD
9. (6 marks) For a normally distributed population with μ=10 and σ=2:
a. the probability of selecting an x ≤ 15 =
2.01
.99
b. the probability of selecting a single score that falls between 11 and 16 =
c. what score cuts off the top 35.20% of the distribution?
.3072
10.76
10. (4 marks) A new researcher (let’s call him Lukmaan) wanted to change things around
campus a bit. He thought that being able to drink beer during class would make the usually
boring prof sound more interesting. He wanted to investigate this idea further before taking
it up with the powers that be, so he decided to run an experiment to see if there is any merit
to his idea. To do this, Lukmaan went to a local pub in order to recruit subjects for his
experiment. He talked 20 guys into participating. He then divided his subjects into two
groups, one group comprised of people under 25 years old, and the other of people over 25.
The first group was told to drink 5 bottles of beer during a 2 hour class, whereas the other
group was given 5 cups of coffee. Lukmaan then collected responses about how interesting
the prof sounded. Besides long line-ups to the bathroom after class, describe any problems
you might find with Lukmaan’s methodology, and reasons why they may be problematic.
Lack of random sampling (1mark) – no external validity (1mark)
Lack of random assignment (1mark) – no internal validity (1mark)
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11. (4 marks) Explain, step by step, the basic logic of hypothesis testing.
Propose null and alternate hypotheses
Obtain relevant sampling distribution (assuming null is correct)
Collect data and calculate the stats
Given the sampling distribution, determine the probability of observing the statistic by
chance
On the basis of the probability, make a decision
12. (4 marks) Let’s assume that a new sickness is around, commonly known as the Toronto Flu.
The suspicion is that it’s caused by watching the Toronto Maple Leafs (probably lose once
again). We know that the probability of any person in Toronto watching the Leafs is about
0.60. We also know that the probability of someone coming down with the Toronto Flu is
0.05. Well, if I were to tell you that the Toronto Flu has nothing to do with the Leafs, what
percentage of people who were sick would you expect to have watched the Leafs?
60%
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13. On a typical night Bianca watches taped episodes of American Idol for 48 minutes, with a
variance of 36 minutes. Based on this information, answer the following 3 questions:
a. (3 marks) on what percentage of nights do you expect Bianca to watch Idol within 1.2
standard deviations of the mean?
Essentially, this question is asking what percentage of the normal distribution lies between
a z of -1.2 and a z of +1.2. The “mean to z” when z=1.2 is 0.3849, thus about 76.98% of
the distribution lies within 1.2 SDs. Given this we can say that we’d expect Bianca to
watch Idol within 1.2 SDs of her mean 76.98% of the time.
b. (4 marks) on approximately 60% of nights, Bianca watches Idol more than 40 minutes
and less than how many minutes?
Given that 40 minutes is less than Bianca’s mean, the first thing we need to know is the
larger portion associated with that value. The z-score associated with 40 can be
computed as:
z
x   40  48

 1.33

6
The larger portion of -1.33 is .9082
0.9082-.6=0.3082 (such that 60% of the distribution would lie between 40 and x). The
closest z-value with that smaller portion is 0.5. Now to convert that value to x we can do
the following:
x  z    0.5(6)  48  51
Thus, on 60% of the nights out, Bianca watches Idol between 40 and 51 minutes.
c. (6 marks) say that last night (while you were studying) Bianca watched the rerun of Kelly
winning the first Idol 1.5 hours. Formally test the hypothesis that last night’s behaviour was
not typical for Sarah.
H0: Watching idol for 1.5 hours is typical for Bianca
H1: Watching idol for 1.5 hours is not typical for Bianca
z
x   90  48

 7.00

6
From here they can answer this in one of two ways. They can either find the smaller
portion for z=7.00, which is 0.0001 or compare it to z(critical) of +-1.96 (it’s a two-tailed
test!). In either case they reject the null.
Given this, we can conclude that watching Idol for 1.5 hours is not typical for Bianca.
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14. (7 marks) The latest polls show that 25% of the population wants to be a contestant on the
Bachelor (or Bachlorette). I talked to 48 people in our class and found that on average 49%
(with a standard deviation of 18%) of them want to find true love on one of these shows.
Test the hypothesis that more stats students need reality TV to help them get a date than the
general population.
H0: Stats students don’t need (watch) reality TV more than general pop
H1: Stats students need (watch) reality TV more than gen. pop.
x 
s
n
49  25

18
48
 9.24
t
One tailed test, t(crit)=1.676, therefore reject null, and conclude that stats students need
reality TV more than gen. pop. to get a date.
15. (4 marks) Provide one example where it would be more important to avoid Type I error, and
one example where it would be more important to avoid Type II error. Explain your answers
in sufficient detail.
This is very open-ended, so it’s up to you how you mark it. 2marks for type I and 2 marks
for type II
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16. A drug company has recently created a new drug called MindHelp that they hope will
increase a person’s ability to think about things. Unfortunately, while they’re good at creating
drugs, they’re not so good at analyzing data. Given this, they’ve hired you as a consultant to
analyze the data and tell them what it means. Below is a description of their experiment, the
data, and the questions they want answered.
The experiment involved two groups of participants: a MindHelp group and a placebo group.
Both groups were first given a “digit-span” test that measures how many items a person can
“keep in mind” (the BEFORE variable). Specifically, subjects are first read 4 numbers, and then,
after an interval, they are asked to repeat those numbers. If they’re correct, then they’re given 5
numbers, then 6, and so on until we find the highest number of digits they can remember. After
this, the subjects were given either MindHelp or the placebo. Shortly after that, subjects were
given another digit-span task (the AFTER variable).
MINDHELP
BEFORE
6
5
7
7
8
6
5
6
7
5
x
s
6.2
1.03
PLACEBO
AFTER DIFFERENCE
7
-1
6
-1
6
1
8
-1
10
-2
6
0
6
-1
8
-2
9
-2
6
-1
7.2
1.48
-1
0.94
BEFORE
5
7
7
5
5
6
5
6
8
7
7
5
5
AFTER
6
9
8
4
6
7
6
8
9
8
8
7
6
DIFFERENCE
-1
-2
-1
1
-1
-1
-1
-2
-1
-1
-1
-2
-1
6
1.08
7.08
1.44
-1.08
0.76
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a. (5 marks) Focusing only on the MindHelp subjects, does it appear as though the drug had its
desired effect?
Ho: MindHelp did not increase the ability to think
H1: MindHelp did increase the ability to think
Matched-sample t-test comparing the before and after MindHelp conditions.
t=-3.36
one-tailed test
tcrit(9) = -1.833
Reject null
MindHelp did significantly improve the ability to think
b. (7 marks) What if you compare the drug’s effects across the MindHelp and placebo groups.
Do your conclusions with respect to the previous question change?
Ho: MindHelp did not increase the ability to think
H1: MindHelp did increase the ability to think
Independent-sample t-test comparing difference scores across MindHelp and placebo
Pooled variance
t= .226
one-tailed test
tcrit(21) = 1.721
Fail to reject null
MindHelp did not increase the ability to think