Transcript Sooc2205 Midterm Review - University of Western Ontario

Soc2205 More Midterm
Review Questions
Additional Selected Problems and
Solutions For Midterm
**Let me know if there are any errors!**
Problem: Healey 1st #2.2 f - j,
2nd #2.2 f - j
• Note: This is a continuation of #2.2 in the
first review presentation.
Solutions to #2.2 f - j
•
•
•
•
•
f. 0.08
g. 25.47%
h. 0.68
i. 3.97 business to nursing majors
j. 0.06
Problem: Healey 1st #3.4,
2nd #3.4
• Use the data from this question to
calculate the appropriate measure of
central tendency – i.e. the mean, median
or mode - for #3.4 (ignore the instructions
on calculating dispersion in the 2nd Can.
Text)
Solutions to #3.4
• The mean income is $36,400.
• The modal marital status is "married".
Six respondents do not own BMWs, so
the modal category is "no".
• The mean number of years of
schooling is 4.6
• Note: median was not most appropriate
for any of the variables
Problems: Healey 1st Can. #3.8,
and 2nd Can. #3.8
• For both texts, compute
– the mean
– the median
– compare the two
Solutions: Healey 1st Can. #3.8,
and 2nd Can. #3.8
• 2000:
– Mean = 654.6, Median = 654
– Mean>Median. Very slight positive skew.
• 2005:
– Mean = 703, Median = 703
– Unskewed distribution.
Problem: Healey 1st #3.12,
2nd #3.12
• Compute the mean, median and
standard deviation for the pretest
and posttest data in these problems.
• For the standard deviation use the
computational (working) formula:
S
X
N
2
i
X
2
Solution: Healey 1st #3.12,
2nd #3.12
• Pretest: Mean = 9.33, Median = 10
• Pretest: s = 3.50
• Posttest: Mean = 12.93, Median = 12
• Posttest: s = 4.40
Problem: Healey 1st #5.2,
2nd #4.2
• Problem Information:
• X = 500
• S = 100
• *Remember to draw a “curve” to
find the % areas
Solution: Healey 1st #5.2,
2nd #4.2
Z score
1.50
-1.00
-1.25
.86
- .63
.26
1.21
- .02
.17
-1.02
% Area Above
6.68
84.13
89.44
19.49
73.57
39.74
11.31
50.80
43.25
84.61
% Area Below
93.32
15.87
10.56
80.51
26.43
60.26
88.69
49.20
56.75
15.39
Problem: Healey 1st #7.14,
2nd #6.14
• *Compare the widths of the intervals in
your answer before looking at the solution
on next slide
Solution: Healey 1st #7.14,
2nd #6.14
• Sample A (N = 100) : 0.40  0.10
• Sample B (N = 1000): 0.40  0.03
• Sample C (N = 10,000): 0.40  0.01
• *notice how as the sample size N gets
larger, the interval width decreases?
Larger samples produce a more efficient
estimate!
Problem: Healey 1st #7.8,
2nd #6.8
• Problem Information:
• This is the 90% CI
–  = .10, Z =  1.65
• N = 500
• Calculate Ps first = (50/500) = . 10
Solution: Healey 1st #7.8,
2nd #6.8
• CI = 0.10  0.04
• Expand this to:
• “I am 90% confident that the population
proportion of people who were victims of
violent crime is between .06 and .14 (6%
and 14%)
Problem: Healey 1st #7.18,
2nd #6.18
• Problem Information:
• This is the 99% CI
–  = .01,
Z =  2.58
• X = 3.1 litres/100km
• S = 3.7
• N = 120
Solution: Healey 1st #7.18,
2nd #6.18
• 99% CI = 3.1  0.87
• The 99% interval is between 2.23 and 3.97
litres/100 km.
• Manufacturer’s claim is 3.0 litres/100 km.
• This claim lies within the confidence
interval range so the manufacturer is
telling the truth!