Transcript Chapter 1: Introduction to Statistics

COURSE: JUST 3900
INTRODUCTORY STATISTICS
FOR CRIMINAL JUSTICE
Final Review: Ch. 10, 12, 17
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Chapter 10: Independent
Measures t-test

Question 1: One sample from an independent-measures
study has n = 5 with SS = 48. The other sample has
n = 9 and SS = 32.
a)
b)
Compute the pooled variance for the sample.
Compute the estimated standard error for the mean difference.
Chapter 10: Independent
Measures t-test

Question 1 Answer:
a)
b)
𝑆𝑆 +𝑆𝑆
48+32
80
𝑠𝑝2 = 𝑑𝑓1+𝑑𝑓2 = 4+8 = 12 = 6.67
1
2
𝑠
𝑀1 −𝑀2
1.44
=
𝑠𝑝2
𝑛1
+
𝑠𝑝2
𝑛2
=
6.67
5
+
6.67
9
= 1.334 + 0.741 = 2.08 =
Chapter 10: Independent
Measures t-test

Question 2: What is the null hypothesis when using an
independent measures t-test?
Chapter 10: Independent
Measures t-test

Question 2 Answer:

𝜇1 − 𝜇2 = 0
Chapter 10: Independent
Measures t-test

Question 3: An independent measures t-test results in a
t-statistic of t = 3.53. Each sample consists of n = 8.
Compute the effect size using r2?
Chapter 10: Independent
Measures t-test

Question 3 Answer:
 𝑟2

=
𝑡2
𝑡 2 +𝑑𝑓
=
3.832
3.832 +7
=
This is a large effect.
14.67
14.67+7
=
14.67
21.67
= 0.677 = 67.7%
Chapter 10: Independent
Measures t-test

Question 4: A researcher is conducting an independent
measures t-test (two-tail) with two samples of n = 8.
M1 = 3 and M2 = 6. The estimated standard error is
s(M1 – M2) = 0.85. Is there a significant treatment effect?
(α = 0.05)
Chapter 10: Independent
Measures t-test

Question 4 Answer:

Critical t = ±2.131

𝑡=

-3.53 < -2.131, therefore we reject the null. There is a
significant effect.
𝑀1 −𝑀2 − 𝜇1 −𝜇2
𝑠 𝑀1−𝑀2
=
3−6 − 0
.85
=
−3
.85
= −3.53
Chapter 10: Independent
Measures t-test

Question 5: The boundaries for a confidence interval are
at t = ± 1.753. The sample sizes are n = 7 and n = 10.
How confident are we that the unknown mean difference
falls in this interval?
Chapter 10: Independent
Measures t-test

Question 5 Answer:

Find df.


Use the t distribution to find the alpha level where t = 1.753 and
df = 15 intersect.


α = 0.10
An alpha of 0.10 corresponds to 10% in the tails, leaving 90% in
the body.


df = 9 + 6 = 15
100 – 10 = 90
Therefore, we are 90% confident that our mean difference falls in
this interval.
Chapter 10: Independent
Measures t-test

Question 6: A researcher wants to conduct an
independent measures t-test, but first, he wants to make
sure the homogeneity assumption is not violated. Each
sample has n = 8 and the sum of squares for each are
SS = 16 and SS = 24. Use an F-Max test to see if the
assumption is violated (α = 0.05).
Chapter 10: Independent
Measures t-test

Question 6 Answer:


Find the variances.
16
7
= 2.29
𝑆𝑆
24
24
7
= 3.43

𝑠 2 = 𝑑𝑓 = 8−1 =
Find the critical F-Max statistic
F-Maxcrit = 4.99
Find F-Max


16
𝑠 2 = 𝑑𝑓 = 8−1 =


𝑆𝑆

𝐹 − 𝑀𝑎𝑥 =
2
𝑠𝑙𝑎𝑟𝑔𝑒𝑠𝑡
2
𝑠𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡
3.43
= 2.29 = 1.50
1.50 < 4.99, therefore the assumption of homogeneity is not
violated.
Chapter 12: ANOVA

Question 1: Which of the following F-ratios is most likely
to reject the null?
a)
b)
c)
d)
e)
F = 1.25
F = 1.00
F = 2.75
F = 4.50
None of the Above
Chapter 12: ANOVA

Question 1 Answer:


D) F = 4.50
F values closer to 1.00 are less likely to fall in the critical region
and thus, are less likely to reject the null.
Chapter 12: ANOVA

Question 2: Which of the following are possible
alternative hypotheses for ANOVA?
a)
b)
c)
d)
e)
H1: μ1 ≠ μ2 = μ3
H1: μ1 ≠ μ2 ≠ μ3
H1: μ1 = μ2 ≠ μ3
All of the above
None of the Above
Chapter 12: ANOVA

Question 2 Answer:


D) All of the above
The alternative hypothesis for ANOVA states that there is a
difference between our population means, but it does not identify
which means are different.
Chapter 12: ANOVA

Question 3: What is the advantage of using ANOVA over
an independent measures t-test?
Chapter 12: ANOVA

Question 3 Answer:

ANOVA can be used when comparing more than two
populations without increasing the risk of Type I error.
Chapter 12: ANOVA

Question 4: What accounts for between treatments
variance?
Chapter 12: ANOVA

Question 4 Answer:

Between treatments variance (MSbetween) is caused by both
treatment effects and random, unsystematic error.
Chapter 12: ANOVA

Question 5: Compute effect size (η2) for a data set with
SSbetween = 70 and SSwithin = 46.
Chapter 12: ANOVA

Question 5 Answer:

η2 =
70
70+46
=
70
116
= 0.603 = 60.3%
Chapter 12: ANOVA

Question 6: A researcher is using Tukey’s HSD to find
which treatments (k = 3 treatments) in his study had an
effect. MSwithin = 3.83 and n = 5. The mean difference
between treatments A and B is – 4. Is this a significant
mean difference? (α = 0.05)
Chapter 12: ANOVA

Question 6 Answer:

𝐻𝑆𝐷 = 𝑞


Find q.
 q = 3.77
Find HSD.


𝑀𝑆𝑤𝑖𝑡ℎ𝑖𝑛
𝑛
𝐻𝑆𝐷 = 𝑞
𝑀𝑆𝑤𝑖𝑡ℎ𝑖𝑛
𝑛
= 3.77
3.83
5
= 3.77 0.766 = 3.30
Compare HSD to 𝑀𝐴 − 𝑀𝐵 .
 4 > 3.30, therefore treatment A is significantly different from
treatment B.
Chapter 12: ANOVA

Question 7: Which of the following is not an assumption
required for the independent-measures ANOVA?
a)
b)
c)
d)
e)
The observations within each sample must be independent.
The samples must all be the same size.
The populations from which the samples are selected must be
normal.
The populations from which the samples are selected must
have equal variances (homogeneity of variance).
All of the above are required assumptions.
Chapter 12: ANOVA

Question 7 Answer:


B) The samples must all be the same size.
ANOVA requires the same assumption as independent
measures t tests:



The observations must be independent.
The populations variances must all be the same.
The populations must be normally distributed.
Chapter 12: ANOVA

Question 8:
Sources
SS
Between
20
Within
Total
200
n = 16 for each sample
k = 3 treatments
df
MS
F=
Chapter 12: ANOVA

Question 8 Answer:
Sources
SS
df
MS
Between
20
3–1=2
20/2 = 10
Within
200 – 20 =
180
47 – 2 = 45
180/45 = 4
Total
200
16*3 – 1 =
47
n = 16 for each sample
k = 3 treatments
F = 10/4 = 2.50
Chapter 17: Chi-Square

Question 1: In what situations would we use nonparametric tests
as substitutes for parametric tests?
Chapter 17: Chi-Square

Question 1 Answer:


The data do not meet the assumptions needed for a standard
parametric test.
The data consist of nominal or ordinal measurements, so that it
is impossible to compute standard descriptive statistics such as
the mean and standard deviation.
Chapter 17: Chi-Square

Question 2: To investigate the phenomenon of “hometeam advantage,” a researcher recorded the outcomes
from 64 college football games on one weekend in
October. Of the 64 games, 42 were won by home teams.
Does this result provided enough evidence that home
teams win significantly more than would be expected by
chance? Assume winning and losing are equally likely
events if there is no home-team advantage. Use α = 0.05
Chapter 17: Chi-Square

Question 2 Answer:
f0

Wins
Losses
42
22
Step 1: State the hypothesis


H0: There is no home-team advantage.
Wins
Losses
50%
50%
H1: There is a home-team advantage.
Chapter 17: Chi-Square

Question 2 Answer:

Step 2: Locate the critical region


Find df.
 df = C – 1 = 2 – 1 = 1
Use df and alpha level (α = 0.05) to find the critical Χ2 value for the
test.
 Χ2crit = 3.84
Chapter 17: Chi-Square

Question 2 Answer:

Step 3: Calculate the Chi-Square Statistic



Identify proportions required to compute expected frequencies (fe).
 The null specifies the proportion for each cell. With a sample of
64 games, the expected frequencies for each category
(wins/losses) are equal in proportion (50%).
Calculate the expected frequencies with proportions from H0.
 fe = pn = (0.50) * (64) = 32 games in each category
Calculate the Chi-Square Statistic
Wins
Losses
f0
42
22
fe
32
32
Chapter 17: Chi-Square

Question 2 Answer:

Step 3: Calculate the Chi-Square Statistic


2
𝑋 =
=
102
32
+
𝑓0 −𝑓𝑒 2
𝑓𝑒
=
42−32 2
32
−10 2
32
=
100
100
+
32
32
+
=
22−32 2
32
200
32
= 6.25
Chapter 17: Chi-Square

Question 2 Answer:

Step 4: Make a Decision



If X2 ≤ 3.84, fail to reject H0
If X2 > 3.84, reject H0
6.25 > 3.84, thus, we reject H0, which means home-team advantage
does effect the outcome of college football games.
Chapter 17: Chi-Square

Question 3: A researcher obtains a sample of 200 high
school students. The students are given a description of
a psychological research study and asked whether they
would volunteer to participate. The researcher also
obtains an IQ score for each student and classifies the
students into high, medium, and low IQ groups. Do the
following data indicate a significant relationship between
IQ and volunteering? (α = 0.05)
Chapter 17: Chi-Square
High IQ
Medium IQ
Low IQ
Volunteer
43
73
34
150
Not Volunteer
7
27
16
50
50
100
50
Chapter 17: Chi-Square

Question 3 Answer:

Step 1: State the hypothesis


H0: There is no relationship between volunteering and IQ.
H1: There is a relationship between volunteering and IQ.
Chapter 17: Chi-Square

Question 3 Answer:

Step 2: Locate the critical region.


Find df.
 df = (R -1)*(C – 1) = (2 – 1)*(3 – 1) = (1)*(2) = 2
Use df (2) and alpha level (0.05) to find the critical X2 value.
 X2crit = 5.99
Chapter 17: Chi-Square

Question 3 Answer:

Step 3: Calculate the Chi-Square Statistic for Independence

Compute expected frequencies where 𝑓𝑒 =
Volunteer
Not Volunteer
𝑓𝑐 𝑓𝑟
𝑛
High IQ
150(50)
= 37.5
200
50(50)
= 12.5
200
Medium IQ
150(100)
= 75
200
50(100)
= 25
200
Low IQ
150(50)
= 37.5
200
50(50)
= 12.5
200
50
100
50
150
50
Chapter 17: Chi-Square

Question 3 Answer:

Step 3: Calculate the Chi-Square Statistic for Independence
High IQ
Medium IQ
Low IQ
Volunteer
(37.5)43
(75)73
(37.5)34
Not Volunteer
(12.5)7
(25)27
(12.5)16
𝑓0 −𝑓𝑒 2
43−37.5 2
73−75 2
34−37.5 2
=
= 37.5 + 75 + 37.5
𝑓𝑒
27−25 2
16−12.5 2
+ 12.5
25
30.25
4
12.25
30.25
4
12.25
 =
+
+
+
+
+
37.5
75
37.5
12.5
25
12.5
 𝑋2


+
= 0.81 + 0.05 + 0.33 + 2.42 + 0.16 + 0.98 = 4.75
4.75
7−12.5 2
12.5
+
Chapter 17: Chi-Square

Question 3 Answer:

Step 4: Make a Decision



If X2 ≤ 5.99, fail to reject H0
If X2 > 5.99, reject H0
4.75 > 5.99, thus, we fail to reject H0, which means there is not a
significant relationship between volunteering and IQ.
Chapter 17: Chi-Square

Question 4: A researcher completes a chi-square test for
independence and obtains X2 = 6.2 for a sample of
n = 40 participants.


If the frequency data formed a 2 X 2 matrix, what is the phicoefficient for the test?
If the frequency data formed a 3 X 3 matrix, what is Cramer’s V
for the test?
Chapter 17: Chi-Square

Question 4 Answer:


Φ=
𝑋2
𝑛
𝑉=
𝑋2
𝑛(𝑑𝑓)
=
=
6.2
40
= 0.155 = 0.394
6.2
40(2)
= 0.0775 = 0.278