Week 2 Exercises
Deb Davis - Pg 1
Statistics for Those Who (Think They) Hate Statistics
Chapter 4 – Questions 1-5
C4-Q1-P72
Data set on web -- complete the following:
1a: Frequency Distribution & Histogram
2
6
7
8
11
12
14
15
16
21
22
25
26
27
29
31
33
34
36
38
41
42
43
44
45
47
49
51
53
54
55
56
57
59
1
1
1
1
1
2
1
2
2
2
2
1
1
1
2
1
1
1
1
1
1
2
2
2
2
2
1
1
1
3
1
3
2
1
0-5
6 - 10
1
3
11 - 15
6
16 - 20
21 - 25
2
5
26 - 30
4
31 - 35
3
36 - 40
2
41 - 45
9
46 - 50
3
51 - 55
6
56 - 60
6
Week 2 Exercises
Deb Davis - Pg 2
1b: Why the class interval you selected?
I selected intervals of 5 because it made for a reasonable size group.
1c: Is this distribution skewed? How would you know?
This distribution is skewed. This is visually obvious in the Cross Validation from
the lower limit of the first bin. It is also apparent from the above bar graph that the
distribution is not symmetrical, ergo, it is skewed.
Week 2 Exercises
C4-Q2-P72
Frequency distribution given:
Class
Frequency
90 - 100
12
80 - 89
14
70 - 79
20
60 - 69
24
50 - 59
28
40 - 49
29
30 - 39
21
20 - 29
15
10 - 19
17
0-9
12
Create a histogram:
Deb Davis - Pg 3
Week 2 Exercises
Deb Davis - Pg 4
C4-Q3-P73
Identify these distributions as negatively skewed, positively skewed, or not skewed
at all, and why.
3a. This talented group of athletes scored very high on the vertical jump takes.
This distribution is positively skewed as they are a talented group and apparently
all scored well.
3b. On this incredibly crummy test, everyone received the same score. This
distribution is not skewed at all as despite the “crumminess” of the test, all scores
were equal.
3c. On the most difficult spelling test of the year, the third graders wept as the
scores were delivered. It is impossible to tell if this is a skewed distribution
because the third grades may have wept for joy, for pity, or for either. There is no
quantifier to the distribution to indicate what the scores were.
C4-Q4-P73
For each of the following, indicate whether you would use a pie, line, or bar chart,
and why.
4a. Proportion of freshmen, sophomores, juniors, and seniors in a particular
university would easily lend itself to a pie chart. To visualize these groups by
pieces of a pie is very straightforward.
4b. Change in GPA over four semester would likely render best in a bar chart as
the visual change in grades could be color monitored to assign terms, and would
make tracking extremely visual.
4c. Number of applicant for four summer jobs would again render in a bar chart
for the same reasons.
4d. Reaction time to different stimuli would probably best render in a line chart as
the details could cloud and otherwise clear pie or bar chart.
4e. Number of scores in each of 10 categories could be rendered in any method,
but I would probably use a bar chart because of the clarity of image.
Week 2 Exercises
Deb Davis - Pg 5
C4-Q5-P73
Provide an example for each of the below and then draw the chart accordingly.
5a. A line graph is well geared for a large groups of numbers from which trends
may be gathered. For my area, I would use a line graph to chart the scores on
midterms taken over term of teaching.
For example, with a possible score of 200, the following scores were received
over the last two terms. Tplies
5b. A bar graph gives excellent comparatives, such as midterm paper grades to
final paper grades.
5c. A pie graph would give great information for group totals.
Week 2 Exercises
Deb Davis - Pg 6
Chapter 5 – Questions 1-8
C5-Q1-P93
Using the following data:
Number Correct Attitude
17
94
13
73
12
59
15
80
16
93
14
85
16
66
16
79
18
77
19
91
1a: Compute the Pearson product-moment correlation coefficient by hand and
show work.
Sum of all correct (X) is 156
Sum of all Attitude (Y) is 797
Sum of each X-squared is 2476
Sum of each Y-squared is 64727
Sum of Products of X and Y is 12568
THEREFORE:
(10 x 12568) - (156 x 797)
=
-----------------------------------Sq rt of [(10 x 2476) - 1562][(10 x 64727) - 7972]
=======================================================
125680 - 124332
=
--------------------------------------------------------Sq rt of [24760-24336][647270-635209]
========================================================
1348
1348
1348
= --------------------------= --------------- =
------- =
0.59609
Sq rt of [424][12061]
sort 5113864
2261.39
1b. Construct a scatter plot for these 10 values by hand. Would you expect the
correlation to be direct or indirect?
Indirect correlation. Relationship is weak
Week 2 Exercises
Deb Davis - Pg 7
C5-Q2-P94
Use the below data for 2a and 2b.
Speed
21.6
23.4
26.5
25.5
20.8
19.5
20.9
18.7
29.8
28.7
Strength
135
213
243
167
120
134
209
176
156
177
Sum of all speed (X) is 235.4
Sum of all strength (Y) is 1730
Sum of each X-squared is 5677.74
Sum of each Y-squared is 313210
Sum of Products of X and Y is 41095.2
THEREFORE:
(10 x 41095.2) - (235.4 x 1730)
=
-----------------------------------Sq rt of [(10 x 5677.4) - 235.42][(10 x 313210) - 17302]
=======================================================
410952 - 407242
=
--------------------------------------------------------Sq rt of [56777.4-55413.16][3132100-2992900]
========================================================
1348
1348
1348
= --------------------------= --------------- =
------- =
0.26922
Sq rt of [1364.24][139200]
sqrt 189902208 1378.5
2b. A low correlation (.27) indicates that the contributing factors may not be a
huge influence.
Week 2 Exercises
Deb Davis - Pg 8
C5-Q3-P94
C5-Q3-P94
Budget + (X)
Acht (y)
7
3
5
7
2
1
5
4
4
38
x sq
11
14
13
26
8
3
6
12
11
104
y sq
49
9
25
49
4
1
25
16
16
194
121
196
169
676
64
9
36
144
121
1536
xy
77
42
65
182
16
3
30
48
44
507 SUMS
1a: Compute the Pearson product-moment correlation coefficient.
Sum of all Increase (X) is 38
Sum of all Acht (Y) is 104
Sum of each X-squared is 194
Sum of each Y-squared is 1536
Sum of Products of X and Y is 507
THEREFORE:
(10 x 507) - (38 x 104)
=
-----------------------------------Sq rt of [(10 x 194) - 382][(10 x 1536) - 1042]
=======================================================
5070 - 3952
=
--------------------------------------------------------Sq rt of [1940-1444][15360-10816]
========================================================
1118
1118
1118
= --------------------------= --------------- =
------- =
0.7447
Sq rt of [496][4544]
sqrt 2253824
1501.27
The correlation is slightly skewed indicating a relationship between increased
budget and increased scores.
Week 2 Exercises
C5-Q4-P95
Hours(x)
23
12
15
14
16
21
14
11
18
9
153
GPA (y)
x sq
3.95
529
3.9
144
4
225
3.76
196
3.97
256
3.89
441
3.66
196
3.91
121
3.8
324
3.89
81
38.73
2513
Deb Davis - Pg 9
y sq
xy
15.6025
90.85
15.21
46.8
16
60
14.1376
52.64
15.7609
63.52
15.1321
81.69
13.3956
51.24
15.2881
43.01
14.44
68.4
15.1321
35.01
150.099 593.16 SUMS
Sum of all hours (X) is 153
Sum of all GPA (Y) is 38.73
Sum of each X-squared is 2513
Sum of each Y-squared is 150.99
Sum of Products of X and Y is 593.16
THEREFORE:
(10 x 593.16) - (153 x 38.73)
=
-----------------------------------Sq rt of [(10 x 2513) - 1532][(10 x 150.099) - 38.732]
=======================================================
5931.6 - 5925.69
=
--------------------------------------------------------Sq rt of [25130-23409][1500.99-1500.0129]
========================================================
5.91
5.91
5.91
= --------------------------= --------------- =
------- =
0.1441
Sq rt of [1721][0.9771]
sqrt 1681.5891
41.007
A low correlation such as this would indicate a lack of relationship.
Accordingly, the plot is random.
Week 2 Exercises
Deb Davis - Pg 10
C5-Q5-P05 - A coefficient between two variables is 0.64. The Pearson correlation
is 8 [??????]; the relationship is quite strong, and the variance unaccounted is .36
(1-.64).
Chapter 6 - Questions 2-5
C6-Q2-P118
Provide an example of when you would want to establish test-retest and parallel
forms reliability.
C6-Q3-P118
You are developing an instrument that measures vocational preferences and you
need to administer the test several times during the year while students are
attending a vocational program. You need to assess the test-retest reliability of the
test and the data from two administrations (Ch6 data set 1) -- one fall and one
spring. Would you call this a reliable test? Why or why not?
C6-Q4-P118
How can a test be reliable and not valid, and not valid unless it is reliable?
C6-Q5-P118
When testing any experimental hypothesis, why is it important that the test you use
to measure the outcome be both reliable and valid?
Week 2 Exercises
Deb Davis - Pg 11
Chapter 7 - Questions 1-7 (Note: Teacher will provide the articles for #1)
C7-Q1-P113
Select five empirical research articles and detail the following information:
a-What is the null hypothesis?
b-What is the research hypothesis?
c-Create a null and research hypothesis for own area.
d-identify articles with no clear stated or implied hypothesis. Can a research
hypothesis be crafted?
C7-Q2-P113
Why does the scientific method work?
Steps:
Observe
Question
Hypothesize
Experiment
Accept or Reject
Change Hypothesis?
Experiment
Accept or Reject
Etc.
-- The scientific method generally works because of its circular perspective.
C7-Q3-P113
Why do good samples make for good tests of research hypotheses?
Good samples make for good tests of research hypotheses because good samples
are directed to incorporate specifics of a directed hypothesis (an educated guess).
C7-Q4-P113
For the following, create one null hypothesis, one directional research hypothesis,
and one nondirectional research hypothesis.
a-What are the effects of attention on out-of-seat classroom behavior?
-Diagnostically Severe ADHD students would have the same out-of-seat frequency
as those determined to be not ADHD-Severe.
-Diagnostically Severe ADHD students would have more out-of-seat frequency
than those determined to be not ADHD-Severe.
-Diagnostically Severe ADHD will differ in out-of-seat frequency than those
determined to be not ADHD-Severe.
Week 2 Exercises
Deb Davis - Pg 12
b-What is the relationship between the quality of a marriage and the quality of the
spouses relationships with their siblings?
-Those with a strong quality of marriage will always have a weak quality of sibling
relationships.
-Those with a strong quality of marriage will always have a strong quality of
sibling relationships.
-Those with a strong quality of marriage will have varying quality of sibling
relationships.
c-What’s the best way to treat an eating disorder?
- The best way to treat an eating disorder is always calories-in-calories-out.
- The best way to treat an eating disorder is never calories-in-calories-out.
- The best way to treat an eating disorder is completely dependent upon the cause
of the disorder, and even then, treatment may or may not be effective.
C7-Q5-P113
What do we mean when we say that the null hypothesis acts as a starting point?
To start at the null hypothesis allows for all possibilities. When there are a number
of unknowns, to start by eliminating as many variables as possible allows for
individual test methods.
C7-Q6-P113
Evaluate the hypotheses from C7-Q1 in terms of the five criteria discussed at the
end of the chapter.
Hypotheses should:
Be stated in a declarative form
Posit a relationship between variables
Reflect a theory or a body of literature on which they are based
Be brief and to the point, and
Be testable!
C7-Q7-P113
Why does the null hypothesis presume no relationship between variables?
That defines “null” – having no relationship!
Week 2 Exercises
Deb Davis - Pg 13
C8-Q1-9
C8-Q1-P151
What are the characteristics of the normal curve? The three characteristics of a bell
curve are: 1) it is not skewed; 2) it is perfectly symmetrical about the mean; 3)
the tails are asymptotic (close to the axis but never quite reaches).
What human behavior is distributed normally? Generally, height and weight are
distributed normally in a population. In my classroom, grades turn from a reverse
bell to a bell through the course of the term.
C8-Q2-P151
Standard scores, such as z scores, allow us to make comparisons across different
samples. Why? A z score is the result of dividing the amount that a raw score
differs from the mean of the distribution by the standard deviation. So, scores
below the mean will have negative z scores, and scores above the mean will have
positive z scores. Positive z scores always fall to the right of the mean, and
negative always fall to the left. Remember that z scores across different
distributions are comparable.
C8-Q3-P151
Why is a z score a standard score, and why can standard scores be used to compare
scores from different distributions with one another? A z score is a standard score
because it is based on the degree of variability within its distribution.
C8-Q4-P151
Compute the z scores for the following raw scores where the X-bar is 50 and the
standard deviation is 5.
z = (rawscore – mean)/standarddeviation
a. 55
(55-50)/5 = 5/5 = 1
b. 50
(50-50)/5 = 0/5 = 0
c. 60
(60-50)/5 = 10/5 = 2
d. 57.5
(57.5 – 50)/5 = 7.5/5=1.5
e. 46
(46-50)/5 = -4/5 = -.8
Week 2 Exercises
Deb Davis - Pg 14
5. For the following set of scores, fill in the cells. The mean is 70 and the standard
deviation is 8.
z = (rawscore – mean)/standarddeviation
Raw Score
68.0
57.2
82.0
84.4
69.0
66.0
85.0
83.6
72.0
z score
(68-70)/8 = -2/8 = -.25
(x-70)/8 = -1.6
(82-70)/8 = 1.5
(x-70)/8 = 1.8
(69-70)/8 = -0.125
(x-70)/8=-0.5
(85.0-70)/8=1.875
(x-70)/8=1.7
(72.0-70)/8=0.25
6. Questions 6a through 6d are based on a distribution of scores with a mean of 75
and a standard deviation is 6.38.
z = (rawscore – mean)/standarddeviation
a. Wha is the probability of a score falling between a raw score of 70 and 80?
b. What is the probability of a score falling above a raw score of 80?
c. What is a probability of a score falling between a raw score of 81 and 81?
d. What is the probability of a score falling below a raw score of 63?
7. Jake needs to score in the top 10% in order to earn a physical fitness certificate.
The class mean is 78 and the standard deviation is 5.5. What raw score does he
need to get that valuable piece of paper?
(x-78)/5.5=.9
82.95 minimum required
8. So, why doesn’t it make sense to simply combine, for example, course grades
across different topics – just take and average and call it a day? Each raw score is
rated to different distributions which will make all the difference.
Week 2 Exercises
Deb Davis - Pg 15
9. Who is the better student, relative to his or her classmates? Here’s all the
information you ever needed to know . . . .
MATH
Class Mean
Class Standard Deviation
81
2
READING
Class Mean
Class Standard Deviation
87
10
z = (rawscore – mean)/standarddeviation
RAW
Mean
SD
z
math-n
math-t
85
87
81
81
2
2
2
3
rdg-n
rdg-t
88
81
87
87
10
10
0.1
-0.6
avg-n
avg-t
2
3
0.1
-0.6
2.1
2.4
1.05
1.2
Talya is the better student.