Learning Activity Statistics Exercises Template
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Liberty University *
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Health Science
Date
Dec 6, 2023
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Learning Activity: Statistics Exercises
Briauna A. Dobbs
School of Education, Liberty University
EDLC 606
L
EARNING
A
CTIVITY
: S
TATISTICS
E
XERCISES
S
TUDENT
T
EMPLATE
Type your answers directly in the document in the spaces provided. Please consider
highlighting,
starring*, or
changing
the font color of answers for ease of instructor grading.
You MUST show your work to be eligible for partial credit
.
1.
(20 Pts,
1 pt each
). Calculate the mean, median, mode, standard deviation, and range for
the following sets of measurements (fill out the table):
a.
20, 18, 17, 17, 19
b.
15, 10, 7, 6, 4
c.
28, 28, 28, 28, 28
d.
10, 10, 7, 6, 4, 79
x
i
¿
¿
s
=
√
1
n
−
1
∑
¿
DISTRIB
MEAN
MEDIAN
MODE
SD
RANGE
a.
20+18+17+17+1
9 =
91/5 = 18.2
18
17
6.8/4 =
√
1.7
=
1.3
20-17 = 3
b.
15+10+ 7+6+ 4 =
42/5 = 8.4
7
none
73.2/4 =
√
18.3
=
4.27
15-4 = 11
c.
28
28
28
0
0
d.
10+10+ 7+
6+4+79 = 116/6 =
19.33
7+10 =
17/2 = 8.5
10
4299.33/5 =
√
859.87
=
29.32
79-4 = 75
2.
(20 Pts,
5 pts each
) Answer the following questions.
a.
Why is the SD in (d) so large compared to the SD in (b)?
Because the data in set D deviates more from the mean than the data in set B
b.
Why is the mean so much higher in (d) than in (b)?
The maximum value in set D data is 79, which is larger than the maximum value
of 15 in set B data
c.
Why is the median relatively unaffected?
Page 1 of 7
EDLC 606
The median is not affected by outliers because it only looks at the middle value of
the dataset. Once the dataset has been sorted, the median is the value in the
middle, regardless of the size of the other values.
d.
Which measure of central tendency best represents the set of scores in (d)? Why?
Median best measures the central tendency the set of scores in set (d) because the
value of mean is influences by outliers in the dataset.
3.
(
4 pts
) Determine the semi-interquartile range for the following set of scores.
92
95
89
65
99
100
85
67
72
99
85
100
Lower half: Median = (72 + 85)/2 = 78.5
Upper half: Median = (99 + 99)/2 = 99
Semi-interquartile range = (Upper median - Lower median)/2 =( 99 - 78.5 )/2= 10.25
4.
(24 pts,
2 pts each
) Fill in the blanks on the table with the appropriate raw scores,
z
-
scores, T-scores, and approximate
percentile ranks. You may refer to the distribution
curve below.
Note:
the Mean = 50, SD = 5.
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EDLC 606
RAW
z
T
Percentile
40
-2
30
2.3%
62.5
2.5
75
99.4%
42.5
-1.5
35
6.7%
55
1
60
84.13
5.
(6 pts, 3 pts each) The following are the means and standard deviations of some well-
known standardized tests, referred to as Test A, Test B, and Test C. All three yield normal
distributions.
Test
Mean
Standard Deviation
Test A
300
75
Test B
250
4
Test C
40
12
a. (
3 pts
) A score of 275 on Test A corresponds to what score on Test B?
-0.33x4 = -1.3 = 250 - 1.3 =
248.7
____
b. (
3 pts
) A score of 400 on Test A corresponds to what score on Test C?
1.33 x 12 = 15.96+40 =
sc
__
6.
(12 pts, 2 pts each) The Graduate Record Exam (GRE) has a combined verbal and
quantitative mean of 1000 and a standard deviation of 200. Scores range from 200 to 1600
and are approximately normally distributed. For each of the following problems, indicate the
percentage or score called for by the problem and select the appropriate distribution curve
(from below) that relates to the problem.
Page 3 of 7
EDLC 606
a.
(
2 pts
) What percentage of the persons who take the test score below 600?
2.28%
b.
(
2 pts
) Type the curve best representing your answer:
E
c.
(
2 pts
) What percentage of the persons who take the test score below 1200?
84.13%
d.
(
2 pts
) Type the curve best representing your answer:
C
e.
(
2 pts
) Above what score do the top 2.27% of the test-takers score?
1400
f.
(
2 pts
) Type the curve best representing your answer:
B
7.
(14 pts, varied) Refer to the following data and scatterplots to respond to questions 7a-e.
Individua
l
Years of School
Body Mass
Index
A
21
18
B
18
20
C
17
33
D
17
29
E
14
31
F
11
32
G
22
19
H
23
21
I
16
33
J
22
36
K
17
30
L
15
28
M
17
20
N
12
28
O
14
33
P
13
29
Page 4 of 7
10
12
14
16
18
20
22
24
0
5
10
15
20
25
30
35
40
Figure A
Years of School
Body Mass Index
EDLC 606
10
12
14
16
18
20
22
24
0
5
10
15
20
25
30
35
40
Figure B
Years of School
Body Mass Index
Figure A represents a scatterplot constructed from the data; Figure B represents a regression line
drawn through the scatterplot that “fits” the data points reasonably well; Figure C represents an
ellipse drawn around the data points.
a.
(
2 pts
.) What is the overall direction of the correlation?
Negative
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EDLC 606
b.
(
2 pts
.) Estimate the strength of the correlation coefficient:
Moderate Strength
Consider Figure D (below).
10
12
14
16
18
20
22
24
0
5
10
15
20
25
30
35
40
Figure D
Years of School
Body Mass Index
c.
(
2 pts
.) Using only the data points associated with the years of school above 16;
what effect does this have on the direction
and
strength
of the correlation
coefficient?
Looking at the data point associated with years above 16, it shows a weak negative trend
of BMI to years in school. The points are all scattered after year 16 versus years
before 16.
d.
(
4 pts
.) Explain why this is the case.
Due to the points being scattered away from the trend line it shows a weak correlation
between BMI and years in school.
e.
(
4 pts
.) Identify how likely it is that a causal relationship has been indicated.
It is not likely that a causal relationship has been identified by the data points. As students
age they may join more sports, but they may also care more about appearance and health versus
school. Other factors such as activity level, genetics, and diet would need to be taken into
account.
Page 6 of 7