COURSE: JUST 3900
INTRODUCTORY STATISTICS
FOR CRIMINAL JUSTICE
Chapter 10: Independent Measures
Peer Tutor Slides
Instructor:
Mr. Ethan W. Cooper, Lead Tutor
© 2013 - - PLEASE DO NOT CITE, QUOTE, OR REPRODUCE WITHOUT THE
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PLEASE EMAIL MR. COOPER AT THE FOLLWING: coopere07@students.ecu.edu
Key Terms: Don’t Forget
Notecards
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Hypothesis Test (p. 233)
Null Hypothesis (p. 236)
Alternative Hypothesis (p. 236)
Alpha Level (level of significance) (pp. 238 & 245)
Critical Region (p. 238)
Estimated Standard Error (p. 286)
t statistic (p. 286)
Degrees of Freedom (p. 287)
t distribution (p. 287)
Confidence Interval (p. 300)
Directional (one-tailed) Hypothesis Test (p. 304)
Independent-measures Research Design (p.318)
Formulas


Estimated Standard Error: 𝑠
Estimated Standard Error: 𝑠
𝑠𝑝2
=
𝑀1 −𝑀2
𝑀1 −𝑀2
𝑆𝑆1 +𝑆𝑆2
𝑑𝑓1 +𝑑𝑓2
=
=
=
𝑠12
𝑛1
𝑠22
+
𝑛2
𝑠𝑝2
𝑠𝑝2
𝑛1
+
𝑛2
n1 = n2
n1 ≠ n2
𝑑𝑓1 𝑠12 +𝑑𝑓2 𝑠22
𝑑𝑓1 +𝑑𝑓2

Pooled Variance:

t-Score Formula: 𝑡 =

Degrees of Freedom: 𝑑𝑓 = 𝑑𝑓1 + 𝑑𝑓2 = 𝑛1 + 𝑛2 − 2
𝑀1 −𝑀2 − 𝜇1 −𝜇2
𝑠 𝑀1 −𝑀2
More Formulas

Cohen’s d:
2
𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
=
𝑀1 −𝑀2
𝑠𝑝2
𝑡2
𝑡 2 +𝑑𝑓

𝑟 =

Confidence Interval:𝜇1 − 𝜇2 = 𝑀1 − 𝑀2 ± 𝑡𝑠

Hartley’s F-max Test:
𝑠 2 (𝑙𝑎𝑟𝑔𝑒𝑠𝑡)
𝑠 2 (𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡)
𝑀1 −𝑀2
Identifying the IndependentMeasures Design

Question 1: What is the defining characteristic of an
independent-measures research study?
Identifying the IndependentMeasures Design

Question 1 Answer:

An independent-measures study uses a separate group of
participants to represent each of the populations or treatment
conditions being compared.
Pooled Variance and Estimated
Standard Error

Question 2: One sample from an independent-measures
study has n = 4 with SS = 100. The other sample has
n = 8 and SS = 140.
a)
b)
Compute the pooled variance for the sample.
Compute the estimated standard error for the mean difference.
Pooled Variance and Estimated
Standard Error

Question 2 Answer:
a)
b)
𝑆𝑆 +𝑆𝑆
𝑠𝑝2 = 𝑑𝑓1+𝑑𝑓2 =
1
𝑠
𝑀1 −𝑀2
2
=
100+140
3+7
𝑠𝑝2
𝑛1
+
𝑠𝑝2
𝑛2
=
240
10
=
24
4
= 24
+
24
8
= 6+3= 9=3
t Test for Two Independent
Samples -- Two-tailed Example

Question 3: A researcher would like to determine
whether access to computers has an effect on grades for
high school students. One group of n = 16 students has
home room each day in a computer classroom in which
each student has a computer. A comparison group of
n = 16 students has home room in a traditional
classroom. At the end of the school year, the average
grade is recorded for each student The data are as
follows:
Computer
Traditional
M = 86
M = 82.5
SS = 1005
SS = 1155
t Test for Two Independent
Samples -- Two-tailed Example

Question 3:
a)
b)
c)
Is there a significant difference between the two groups? Use
a two-tailed test with α = 0.05.
Compute Cohen’s d to measure the size of the difference.
Compute the 90% confidence interval for the population mean
difference between a computer classroom and a regular
classroom.
t Test for Two Independent
Samples -- Two-tailed Example

Question 3a Answer:

Step 1: State hypotheses



H0: Treatment has no effect. (𝜇1 − 𝜇2 = 0)
H1: Treatment has an effect. (𝜇1 − 𝜇2 ≠ 0)
Step 2: Set Criteria for Decision (α = 0.05)
df = 16 + 16 – 2 = 30
Critical t = ± 2.042
If -2.042 ≤ tsample ≤ 2.042, fail to reject H0
If tsample < -2.042 or tsample > 2.042, reject H0
t Test for Two Independent
Samples -- Two-tailed Example
df = 30
t Distribution with
α = 0.05
Critical region
t = - 2.042
Critical region
t = + 2.042
t Test for Two Independent
Samples -- Two-tailed Example

Question 3a Answer:

Step 3: Compute sample statistic
a)
𝑠𝑝2 = 𝑑𝑓1+𝑑𝑓2 =
𝑆𝑆 +𝑆𝑆
1
b)
𝑠
c)
𝑡=
𝑀1 −𝑀2
2
=
1005+1155
15+15
𝑠𝑝2
𝑛1
+
𝑠𝑝2
𝑛2
𝑀1 −𝑀2 − 𝜇1 −𝜇2
𝑠 𝑀1−𝑀2
=
=
=
2160
30
72
16
+
= 72
72
16
= 4.5 + 4.5 = 9 = 3
86−82.5 − 0
3
=
3.5
3
= 1.17
Two-Tailed Hypothesis Test
Using the t Statistic
df = 30
t Distribution with
α = 0.05
Critical region
t = - 2.042
t = 1.17
Critical region
t = + 2.042
t Test for Two Independent
Samples -- Two-tailed Example

Question 3a Answer:

Step 4: Make a decision

For a Two-tailed Test:
If -2.042 ≤ tsample ≤ 2.042, fail to reject H0
If tsample < -2.042 or tsample > 2.042, reject H0


tsample (1.17) < tcritical (2.042)
Thus, we fail to reject the null and cannot conclude that access to
computers has an effect on grades.
t Test for Two Independent
Samples -- Two-tailed Example

Question 3b Answer:
𝑀1 −𝑀2
b)
This is a small, or small-to-medium, effect.
Magnitude of d
=
3.5
8.485
Cohen’s d: =
𝑠𝑝2
=
86−82.5
72
a)
= 0.412
Evaluation of Effect Size
d = 0.2
Small effect (mean difference around 0.2 standard deviations)
d = 0.5
Medium effect (mean difference around 0.5 standard deviations)
d = 0.8
Large effect (mean difference around 0.8 standard deviations)
t Test for Two Independent
Samples -- Two-tailed Example

Question 3c Answer:

df = 30
Critical t = ± 1.697
𝑠 𝑀1−𝑀2 = 3

𝜇1 − 𝜇2 = 𝑀1 − 𝑀2 ± 𝑡𝑠





Our α = 0.10 because our confidence interval
leaves 10% split between the 2-tails.
𝑀1 −𝑀2
86 − 82.5 + 1.697 3 = 3.5 + 5.091 = 8.591
86 − 82.5 − 1.697 3 = 3.5 − 5.091 = −1.591
Thus, the population mean difference is estimated to be between
– 1.591 and 8.591. The fact that zero is an acceptable value
(inside the interval) is consistent with the decision that there is
no significant difference between the two population means. (Fail
to reject the null)
t Test for Two Independent
Samples -- One-tailed Example

A researcher is using an independent-measures design
to evaluate the difference between two treatment
conditions with n = 8 in each treatment. The first
treatment produces M = 63 with a variance of s2 = 18,
and the second treatment has M = 58 with s2 = 14.
a)
b)
Use a one-tailed test with α = 0.05 to determine whether the
scores in the first treatment are significantly greater than scores
in the second.
Measure the effect size with r2.
t Test for Two Independent
Samples -- One-tailed Example

Question 4 Answer:

Step 1: State hypotheses



H0: No difference between treatments. (𝜇1 − 𝜇2 ≤ 0)
H1: Treatment 1 has a greater effect. (𝜇1 − 𝜇2 > 0)
Step 2: Set Criteria for Decision (α = 0.05) df = 8 + 8 – 2 = 14
Critical t = 1.761
If tsample ≤ 1.761, fail to reject H0
If tsample > 1.761, reject H0
t Test for Two Independent
Samples -- One-tailed Example
df = 14
t Distribution with
α = 0.05
Critical region
Because this is a
one-tailed test‚
there is only one
critical region.
t = + 1.761
t Test for Two Independent
Samples -- One-tailed Example

Question 4 Answer:

Step 3: Compute sample statistic
a)
𝑠
b)
𝑡=
𝑀1 −𝑀2
=
𝑠12
𝑛1
+
𝑠22
𝑛2
𝑀1 −𝑀2 − 𝜇1 −𝜇2
𝑠 𝑀1−𝑀2
=
=
18
8
+
14
8
63−58 − 0
2
= 2.25 + 1.75 = 4 = 2
5
2
= = 2.50
t Test for Two Independent
Samples -- One-tailed Example
df = 14
t Distribution with
α = 0.05
Critical region
Because this is a
one-tailed test‚
there is only one
critical region.
t = + 1.761
t = 2.50
t Test for Two Independent
Samples -- One-tailed Example

Question 4 Answer:

Step 4: Make a decision

For a One-tailed Test:
If tsample ≤ 1.761, fail to reject H0
If tsample > 1.761, reject H0


tsample (2.50) > tcritical (1.761)
Thus, we reject the null and conclude that the treatment 1 has a
significantly greater effect.
t Test for Two Independent
Samples -- One-tailed Example

Question 4 Answer:
𝑡2
𝑡 2 +𝑑𝑓
2.502
2.502 +14
b)
𝑟2
c)
This is a large effect.
=
=
=
6.25
6.25+14
=
6.25
20.25
= 0.309 = 30.9%
Percent of Variance Explained as Measured by r2
Evaluation of Effect Size
r2 = 0.01 (0.01*100 = 1%)
Small effect
r2 = 0.09 (0.09*100 = 9%)
Medium effect
r2 = 0.25 (0.25*100 = 25%)
Large effect
Assumptions Underlying the
Independent-Measures t Test

Question 5: What three assumptions must be satisfied
before you use the independent-measures t formula for
hypothesis testing?
Assumptions Underlying the
Independent-Measures t Test

Question 5 Answer:
1)
2)
3)
The observations within each sample must be independent.
The two populations from which the samples are selected must
be normal.
The two populations from which the samples are selected must
have equal variances (homogeneity of variance).
Testing for Homogeneity of
Variance

Question 6: Suppose that two independent samples
each have n = 10 with sample variances of 12.34 and
9.15. Do these samples violate the homogeneity of
variance assumption? (Use Hartley’s F-Max Test with
α = 0.05)
Testing for Homogeneity of
Variance

Question 6 Answer:
𝑠 2 (𝑙𝑎𝑟𝑔𝑒𝑠𝑡)
𝑠 2 (𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡)

F-max =

df = 10 – 1 = 9
k=2

Critical F-max = 4.03
=
12.34
9.15
= 1.35
Critical levels for α = 0.05 are in regular type;
critical levels for α = 0.01 are in bold type.
Testing for Homogeneity of
Variance

Question 6 Answer:

Because the obtained F-max value (1.35) is smaller than the
critical value (4.03), we can conclude that the data do not
provide evidence that the homogeneity of variance assumption
has been violated.
Frequently Asked Questions
FAQs

What’s the difference between 𝑠𝑝2 =
𝑆𝑆1
𝑑𝑓1


+
𝑆𝑆2
?
𝑑𝑓2
𝑆𝑆1 +𝑆𝑆2
𝑑𝑓1 +𝑑𝑓2
and 𝑠𝑝2 =
There seems to be some confusion as to how to work the
formula for pooled variance. In the first example (the correct
one), the numerator is the sum of SS1 and SS2. The denominator
is the sum of df1 and df2.
In the second example (the incorrect method), we’re dividing,
then adding. This yields a completely different, and ultimately
wrong, result.
SS = 10; df = 15
SS = 5; df = 20
𝑠𝑝2
𝑆𝑆1 + 𝑆𝑆2
10 + 5
15
=
=
=
= 0.429
𝑑𝑓1 + 𝑑𝑓2 15 + 20 35
𝑠𝑝2
𝑆𝑆1 𝑆𝑆2 10 5
=
+
=
+
= 0.67 + 0.25 = 0.92
𝑑𝑓1 𝑑𝑓2 15 20