Download Exam 1 Practice Questions - KEY

SAMPLE QUESTIONS
EXAMINATION NO. 1 – KEY
1.
Using the following frequency distribution,
Class Interval
69-73
64-68
59-63
54-58
49-53
44-48
39-43
34-38
29-33
24-28
Exact Limits
68.5-73.5
63.5-68.5
58.5-63.5
53.5-58.5
48.5-53.5
43.5-48.5
38.5-43.5
33.5-38.5
28.5-33.5
23.5-28.5
f
1
9
14
41
34
35
21
16
7
2
and the following formula: Percentile point (PX) = X L +
a.
cf
180
179
170
156
115
81
46
25
9
2
c%
100.00
99.44
94.44
86.67
63.89
45.00
25.56
13.89
5.00
1.11
i
(cum f P − cum f L )
fi
Determine P15 = 38.98
(Where cum fP = .15 × 180 = 27)
P15 = 38.5 +
b.
5
(27 − 25) = 38.5 + .238095 × 2 = 38.5 + .47619 = 38.97619
21
Determine P80 = 57.04
(Where cum fP = .80 × 180 = 144)
P80 = 53.5 +
5
(144 − 115) = 53.5 + .121951 × 29 = 53.5 + 3.536579 = 57.036579
41
Using the above frequency distribution, and without using any formula calculation:
c.
A reasonable value for P25 would be
a.
23.46
b.
25.00
c.
43.26
d.
64.23
2.
Using the frequency distribution from question 1,
cum f L +
and the following formula: Percentile rank (PRX) =
a.
fi
(X − XL)
i
× 100
N
Determine PR40
25 +
PR40 =
=
21
(40 − 38.5)
25 + 4.2 × 1.5
25 + 6.3
5
× 100 =
× 100 =
× 100
180
180
180
31.3
× 100 = .173889 × 100 = 17.388889
180
= 17.39
b.
Determine PR67
170 +
PR67 =
=
9
(67 − 63.5)
170 + 1.8 × 3.5
170 + 3.6
5
× 100 =
× 100 =
× 100
180
180
180
173.6
× 100 = .979444 × 100 = 97.944444
180
= 97.94
Using the above frequency distribution, and without using any formula calculation:
c.
.
A reasonable value for PR25 would be
a.
.3333
b.
13.33
c.
25.00
d.
33.33
SAMPLE QUESTIONS – EXAM 1 – KEY
PAGE 2
3.
Using the following sample of scores:
a.
12
10
12
11
6
15
14
17
9
12
13
8
7
15
14
15
18
19
14
10
14
14
16
8
9
Determine the mean using Formula: X =
ΣX i 312
=
= 12.48
N
25
ΣXi = 6 + 7 + 8 + 8 + 9 + 9 + 10 + 10 + 11 + 12 + 12 + 12 + 13 + 14 + 14 + 14 +
14 + 14 + 15 + 15 + 15 + 16 + 17 + 18 + 19 = 312
N=
25
Add up all the scores (ΣXi) and divide by the total number of scores (N)
b.
Determine the median using the guidelines outlined in the text and class = 13
6 7 8 8 9 9 10 10 11 12 12 12 13 14 14 14 14 14 15 15 15 16 17 18 19
6 7 8 8 9 9 10 10 11 12 12 12 ) 13 ( 14 14 14 14 14 15 15 15 16 17 18 19
12 scores
12 scores
Arrange the scores in order and find the middle point – since there are an odd
number of scores – the median will be the middle score.
c.
Determine the mode using the guidelines outlined in the text and class
The most frequently occurring score (there are 5) in the distribution = 14
d.
Determine the range using – (Highest Score – Lowest Score) + 1
Highest Score = 19 and Lowest Score = 6, therefore 19 – 6 + 1 = 14
SAMPLE QUESTIONS – EXAM 1 – KEY
PAGE 3
e.
Determine the variance using the Raw Score (or observed score) Formula:
s2 =
SS
=
N −1
N=
25
( ΣX i ) 2
N
N −1
ΣX i2 −
ΣXi = 6 + 7 + 8 + 8 + 9 + 9 + 10 + 10 + 11 + 12 + 12 + 12 + 13 + 14 + 14 + 14 +
14 + 14 + 15 + 15 + 15 + 16 + 17 + 18 + 19 = 312
ΣX i2 = 62 + 72 + 82 + 82 + 92 + 92 + 102 + 102 + 112 + 122 + 122 + 122 + 132 + 142
+ 142 + 142 + 142 + 142 + 152 + 152 + 152 + 162 + 172 + 182 + 192
= 36 + 49 + 64+ 64 + 81+ 81 + 100 + 100 + 121 + 144 + 144 + 144 + 169
+ 196 + 196 + 196 + 196 + 196 + 225 + 225 + 225 + 256 + 289 + 324 +
361 = 4182
s2 =
f.
(312) 2
97344
4182 −
25 =
25 = 4182 − 3893.76 = 288.24 = 12.01
25 − 1
24
24
24
4182 −
Determine the standard deviation using the Raw Score Formula:
s = s2 =
SS
=
( N − 1)
(ΣX i ) 2
N
( N − 1)
ΣX i2 −
s = 12.01 = 3.46554469 = 3.47
SAMPLE QUESTIONS – EXAM 1 – KEY
PAGE 4
4.
Given the following values for the mean, median, and mode – indicate whether the
distribution would be negatively skewed, normal (symmetrical), or positively skewed.
You may want to sketch out the distribution to assist in your answer.
a.
Mean = 43
Median = 43 Mode = 43
Symmetric = Normally distributed
f
X
b.
Mean = 46
Median = 43 Mode = 40
Skewed to the right = Positively skewed
f
X
c.
Mean = 40
Median = 43 Mode = 46
Skewed to the left = Negatively skewed
f
X
SAMPLE QUESTIONS – EXAM 1 – KEY
PAGE 5
5.
John took three standardized tests with the following results:
Test
Score
X
s
z
Chemistry
85
82
10
.30
Mathematics
80
75
8
.63
English
80
90
12
-.83
a.
b.
Compute a z score for each of John’s test scores.
X−X
s
Using the Formula:
z=
For Chemistry:
zC =
X C − X C 85 − 82 3
=
=
= .30
sC
10
10
For Mathematics:
zM =
X M − X M 80 − 75 5
=
= = .625
sM
8
8
For English:
zE =
X E − X E 80 − 90 − 10
=
=
= −.833333333
sE
12
12
What appears to be John’s strongest subject area?
Mathematics (z = .63) is John’s strongest subject.
c.
What appears to be John’s weakest subject area?
English (z = -.83) is John’s weakest subject.
SAMPLE QUESTIONS – EXAM 1 – KEY
PAGE 6