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Basic Quantitative Methods in
the Social Sciences
(AKA Intro Stats)
02-250-01
Lecture 6 - Review
Change to Help Clinic Hours
• Help Clinic hours for next week are
changed as follows:
Tuesday June 10 - 12:00 - 5:00 PM (not
1:00-4:00)
Wednesday, June 11 - 11:00 AM - 4:00 PM
(not 12:30-3:30)
Thursday, June 12 - NO HELP CLINIC
HOURS
It’s Review Time!
• Here are some review questions that will
resemble exam questions
• Note: The answers are here – but don’t
look at them until you’ve completed the
questions!
Problem #1
• Suppose a researcher randomly selects 10 students
buying sandwiches from the CAW Centre cafeteria and
asks them to rate the sandwiches on a scale from 1
(bad taste) to 10 (great taste) (the researcher is
wondering how the entire university student
population rates the sandwiches but can’t afford to
interview the entire student body)
• He obtains the following ratings:
4, 6, 5, 8, 7, 3, 10, 2, 5, 5
• What are the mean, median, and mode of this data
set?
• What are the variance and standard deviation of this
data set?
Problem #1
• Data set (in order): 2, 3, 4, 5, 5, 5, 6, 7, 8, 10
• Mean: Add all numbers together and divide by n:
(2+3+4+5+5+5+6+7+8+10)/10 = 55/10 = 5.5
• Median: There are 10 scores, so find scores in the 5th
and 6th positions, add together, and find the average:
(5+5)/2 = 5
• Mode: Most frequently occurring score = 5
Problem 1
• Variance and standard deviation:
• Decide: Are we using a sample
or population formula? A sample
formula!
2
(

X
)
X 2 
n
s2 
n 1
3025
353 
10
s2 
9
353  302.5 2 50.5 s 2  5.6111
s 
s 
9
9
2
s  5.6111  2.3688
X
X2
2
4
3
9
4
16
5
25
5
25
5
25
6
36
7
49
8
64
10
100
ΣX=55
ΣX2=353
Problem #2
• Suppose the length of time spent studying for
a Stats exam is normally distributed with a
mean of 10 hours and a standard deviation of
2 hours. (N=200)
A. What proportion of students study for less
than 7.5 hours?
B. How many students study for between 11
and 14 hours?
Problem #2: A
• A:   10
 2
X  7 .5
z
X 

From Table E.10, find
the area in the
“smaller portion” for z
= -1.25 = .1056
Therefore, a
proportion of .1056
students study for less
than 7.5 hours for their
Stats exam
7.5  10  2.5


 1.25
2
2
z
-1.25 0
X
7.5 10
  10
Problem #2: B
• B:   2
11  10
14  10
z11 
 .50 z14 
 2.00
X 1  11
2
2
X 2  14
From Table E.10, find the
area in the “mean to z” for
z=.50 = .1915, and for
z=2.00 =.4772
Now: .4772-.1915=.2857
To find how many =
(.2857)(200)=57.14
Therefore, approx. 57
students study for
between 11 and 14 hours
for their Stats exam
z
0 .50 2.00
X
10 11
14
Problem #3
• The average University of Windsor student
eats 3000 calories a day with a standard
deviation of 400 calories. Professor X wants
to know whether students living in Residence
eat more than the average student. He takes
a sample of 36 students living in Residence
and find that their sample mean is x = 3175
calories. Test the hypothesis at the .05 level.
Example 3 cont.
• 1. State level of significance -  = 0.05 (what is
usually used)
• 2. State IV and DV
IV = living location (residence or not)
DV = calories
• 3. Null hypothesis:
H o : x    3000
 Students living in residence eat an equal amount of food
as does the average U of Windsor student.
• Alternative Hypothesis:
Students living in residence eat more than does the
average U of Windsor student.
Ha : x  
Example 3 continued
• 4. B/c this hypothesis is directional, we use a
one-tailed test
• 5. Find the rejection region:  = 0.05, so with
a one-tailed test we want a critical value that
represents a region of rejection that makes up
0.05 of the area of the tail. From Table E.10
we find that the critical value for z is equal to
1.64 (and since this is a one-tailed test, we
are interested in +1.64, and not -1.64).
Example 3 continued
• This means that zcrit = +1.64
• 6. Calculate your statistic
z 
x

n
3175  3000
z
400
36
175
z
 2.62
66.6667
Example 3 continued
• This means our zobs = +2.62
• 7. Compare zcrit to zobs
• Is zobs > zcrit??
Yes! 2.62 > 1.64
B/c our zobs lies beyond zcrit we say our z-value falls
into the region of rejection: the value of zobs is
greater than the value of zcrit so we choose to reject
the Ho
So: we reject the null hypothesis, students in
residence to in fact eat more than the average
University of Windsor student.
Example 4
Let’s say the average Canadian earns $40000 each year
with a standard deviation of $4300. Professor Y wants
to know if residents of Windsor earn less than the
average Canadian. She samples 49 Windsor residents
and finds that their mean yearly salary is $38050. Test
the hypothesis at the .01 level.
Example 4 cont.
• 1. State level of significance -  = 0.01
• 2. State IV and DV
IV = Living location (Windsor or not)
DV = yearly income
• 3. Null hypothesis:
H o : x    40000
 Residents of Windsor earn the same amount per
year as does the average Canadians
Ha : x  
Alternative Hypothesis:
Residents of Windsor earn less per year than
does the average Canadian.
Example 4 continued
• 4. B/c this hypothesis is directional, we use a
one-tailed test
• 5. Find the rejection region:  = 0.01, so with
a one-tailed test we want a critical value that
represents a region of rejection that makes up
0.01 of the area of the tail. From Table E.10
we find that the critical value for z is equal to
2.33 (and since this is a one-tailed test on the
left tail, we are interested in -2.33 and not
+2.33).
Example 4 continued
• This means that zcrit = -2.33
• 6. Calculate your statistic
z 
x

n
38050  40000
z
4300
49
 1950
 1950
 1950
z


 3.17
4300
4300 614.2857
7
49
Example 4 continued
• This means our zobs = -3.17
• 7. Compare zcrit to zobs
• Is zobs < zcrit??
Yes! -3.17 < -2.33
B/c our zobs lies beyond zcrit we say our z-value falls
into the region of rejection: the value of zobs is less
than the value of zcrit so we choose to reject the Ho
So: we reject the null hypothesis, residents of
Windsor earn significantly less per year than does
the average Canadian.
For Next Class
• Midterm #1
• Don’t forget: student ID card, pen,
pencil(s), eraser, calculator, your
textbook!