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2.14 Exercises – Solutions
1.
Contrast, by real examples, random and nonrandom samples.
A random sample of 50 trees in Oak Mountain State Park would be one taken such that
every possible sample of 50 trees has the same chance of being selected. A nonrandom
sample of 50 trees in Oak Mountain State Park would be one consisting of the first 50 trees
identified after entering the park.
2.
Describe a realistic situation in which a decision maker would prefer a stratified random
sample to a simple random sample. Why?
The population of undergraduate students at Samford University consists of 60% males and
40% females. A stratified random sample is preferable to a simple random sample since the
stratified sample would guarantee that males and females are represented in the sample in
the same proportion they occur in the population.
3.
What kinds of situations would lend themselves to the use of systematic samples instead of
simple random samples and why?
A systematic sample would be indicated when a printed list of the population is available (as
opposed to an electronic form of the list).
4.
Why should a decision maker be concerned with the size of the sample as well as its
randomness?
Size is related to the likelihood that a sample is representative; the larger the sample the
greater the chance that it reflects the make-up of the population.
5.
Describe a realistic situation in which it would be advantageous to use a grouped data
frequency distribution as opposed to a raw data frequency distribution.
A decision maker has a sample of 1500 observations consisting of the salaries of mid-level
managers in the IT field.
6.
What, if anything, does a histogram tell you that a frequency curve does not?
A histogram may communicate the number of observations within specific intervals of the
variable being displayed.
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7.
Provide a realistic situation which would likely result in:
a) a negatively skewed distribution
an easy finance exam
b) a positively skewed distribution
an extremely difficult finance exam
c) a unimodal symmetric distribution
a sample of the digits 0 through 9 taken from residential phone numbers
d) a bimodal distribution.
a capstone exam given to a random sample of freshman and senior business students
8.
What, if anything, does a relative frequency curve tell the decision maker that a frequency
curve does not?
the percentage of sample observation of a specific value
9.
Why does a cumulative relative frequency distribution go no higher than 1.0?
Because the greatest cumulative proportion (percentage) of observations is 1.0 (100%)
10. What kind of distribution would produce a cumulative frequency curve which is a
straight line?
a uniform distribution
11. In what kind of distributions would the mean median and mode be the same value?
unimodal symmetric
12. Give an example which shows the difference between:
a) X and μ
the mean age of a single classroom of Samford undergraduates versus the mean age of
the population of all current undergraduates at Samford
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b) S2 and σ2
the variance of the ages of a single classroom of Samford undergraduates versus the
variance of the ages in the population of all current undergraduates at Samford
13. Verify that ∑ (X - X ) = 0.
X
1
2
3
X=2
X−X
1 – 2 = -1
2–2=0
3–2=1
∑ ( X − X ) =0 √
14. What situation(s) would guarantee that the X value calculated with raw data would be the
same as X calculated with grouped data?
if the raw data were the same as the midpoints of the corresponding groups
15. When would it be desirable to utilize the median instead of the mean? Provide a real
example.
when a sample is skewed, such as a class of statistics students who generally do poorly on
the first exam but where a few “curve busters” score well
16. Why were absolute value signs used in the formula for mean deviation? What would be the
case if they were omitted?
Absolute value signs were used in the formula for MD to eliminate negative deviation scores
(i.e., X − X ); if the absolute value signs were omitted it would read, ∑ X − X - the sum of
all of the values of X, minus the mean
17. What, if anything, does S indicate that S2 does not. Why use S since S2 must be calculated
first?
S provides us with a measure of variability in the same units as the variable.
18. Why is n-1 used in equation (2.5) instead of n?
to make S2 an unbiased estimate of σ2
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19. When would one of the calculating formulas for S2 be more convenient than the
definitional formula, (2.5)?
When certain summary statistics are readily available.
20. What does the decision maker know for sure if (2.5), (2.7), and (2.8) produce different
answers for the same data set?
that a calculational error has been made
21. In transforming scores why is it usually advisable for the decision maker to consider the
standard deviation first when making the initial transformation?
Because changing the standard deviation first will impact both the mean and standard
deviation of the sample; changing the mean second insures that only the mean (and not the
standard deviation) is changed.
22. Why can the transformation into standard unit scores disregard the advice in question 21
above?
Because the target standard deviation of zero prevents the mean from changing in step 1.
23. Why can skew be negative and positive whereas kurtosis can only be positive?
It is a function of the mathematics; skewness involves a cube in the numerator (meaning that
a positive value cubed will remain positive and a negative value cubed will remain negative)
and kurtosis involves a numerator raised to the fourth power (meaning that a positive value
will remain positive and a negative value will become positive).
24. Calculate X , the median, mode, range, S and S2 using the following scores on an attitude
questionnaire (calculate both manually and with Minitab).
Attitude Score
0
1
2
3
4
5
X=
Frequency
3
5
2
1
0
2
0 + 0 + 0 + 1 + 1 + 1 + 1 + 1 + 2 + 2 + 3 + 5 + 5 22
=
= 1.7
13
13
5
S=
(0 − 1.7) 2 + (0 − 1.7) 2 + (0 − 1.7) 2 + (1 − 1.7) 2 + (1 − 1.7) 2 + L + (5 − 1.7) 2
= 1.7
13 − 1
S2 = (1.7)2 = 2.9
md = 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 3, 5, 5 = 1
mode = 1
range = 5 – 0 = 5
Minitab:
Descriptive Statistics: Attitude
Variable
Attitude
Total
Count
13
Mean
1.692
StDev
1.702
Variance
2.897
Median
1.000
Range
5.000
(Note: Minitab does not compute the mode.)
25. Calculate X using the following set of performance scores.
Performance
21 – 25
26 – 30
31 – 35
36 – 40
X=
Frequency
1
2
3
1
Midpoint
23
28
33
38
1(23) + 2(28) + 3(33) + 1(38) 23 + 56 + 99 + 38 216
=
=
= 30.9
7
7
7
26. In what way does the size of X affect the size of S?
Generally speaking, the sample mean has no affect on the size of the sample standard
deviation.
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27. Calculate, with Minitab, the mean, median, standard deviation and variance of the
pretest scores for the Section II groups in Appendix D.
Descriptive Statistics: Pretest
Variable
Pretest
Total
Count
20
Mean
14.050
StDev
3.103
Variance
9.629
Median
14.000
28. Using the Lecture Methods data set, using Mintab, create
a) a simple bar chart for ‘Grade.’
Chart of Grade
10
Count
8
6
4
2
0
A
B
C
Grade
D
F
b) a simple histogram for ‘Final.’ Manually draw a frequency polygon on the
histogram generated in b).
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c) a cumulative distribution curve for ‘Pretest.’
Empirical CDF of Pretest
Normal
Mean
StDev
N
100
Percent
80
60
40
20
0
0
5
10
15
20
Pretest
25
30
d) a raw data frequency distribution table for 'Grade.'
Tally for Discrete Variables: Grade
Grade
A
B
C
D
F
N=
Count
3
7
10
3
2
25
CumCnt
3
10
20
23
25
Percent
12.00
28.00
40.00
12.00
8.00
CumPct
12.00
40.00
80.00
92.00
100.00
e) a grouped data frequency distribution table for 'Final.'
Tally for Discrete Variables: FinalGrp
FinalGrp
35-39
40-44
45-49
50-54
55-59
60-64
65-69
N=
Count
3
5
5
4
4
2
2
25
CumCnt
3
8
13
17
21
23
25
Percent
12.00
20.00
20.00
16.00
16.00
8.00
8.00
CumPct
12.00
32.00
52.00
68.00
84.00
92.00
100.00
35
16.24
5.819
25
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29. Transform the following sample of observations (X) to a new set of observations with a
mean ( ) of 200 and a standard deviation (s) of 50.
X
2
1
1 x 21.7 →
6
5
X’
43.5
21.7
21.7 +134.8 →
130.4
108.7
X”
178.3
156.5
156.5
264.8
243.5
X = 3; s = 2.3
X' = 65.2; s ≅ 50
X" ≅ 200; s ≅ 50
30. Convert the sample of observations (in 29 above) into their corresponding z-scores.
X
2
1
1
6
5
X−X
s
2−3
2.3
1− 3
2.3
1− 3
2.3
6−3
2.3
5−3
2.3
z
-0.43
-0.87
-0.87
1.30
0.87
z = 0.00; s ≅ 1