### Chapter 1: Introduction to Statistics

```COURSE: JUST 3900
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Chapter 8:
Hypothesis Testing
Developed By:
Ethan Cooper (Lead Tutor)
John Lohman
Michael Mattocks
Aubrey Urwick
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)
Type I Error (p. 244)
Type II Error (p. 245)
Statistically Significant (p. 251)
Directional (one-tailed) Hypothesis Test (p. 256)
Effect Size (p. 262)
Power (p. 265)
Formulas
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Standard Error of M:  =
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−

z-Score Formula:  =

=
2

=
2

−
=

 estimated Cohen’s d:
=

−

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Cohen’s d:
Logic of Hypothesis Testing
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Question 1: The city school district is considering
increasing class size in the elementary schools.
However, some members of the school board are
concerned that larger classes may have a negative
effect on student learning. In words, what would the null
hypothesis say about the effect of class size on student
learning?
Logic of Hypothesis Testing
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For a two-tailed test:
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The null hypothesis would say that class size has no effect on
student learning.
The alternative hypothesis would say that class size does have an
effect on student learning.
For a one-tailed test:
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The null hypothesis would say that class size does not have a
negative effect on student learning.
The alternative hypothesis would say that class size has a negative
effect on student learning.
Alpha Level and the Critical
Region
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Question 2: If the alpha level is decreased from α = 0.01
to α = 0.001, then the boundaries for the critical region
move farther away from the center of the distribution.
(True or False?)
Alpha Level and the Critical
Region
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True. A smaller alpha level means that the boundaries for the
critical region move further away from the center of the
distribution.
Possible Outcomes of a
Hypothesis Test
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Question 3: Define Type 1 and Type II Error.
Possible Outcomes of a
Hypothesis Test
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Type I error is rejecting a true null hypothesis – that is, saying
that treatment has an effect when, in fact, it doesn’t.
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Type I error = false (+) = Alpha (α) = level of significance
Type II error is the failure to reject a null hypothesis. In terms of
a research study, a Type II error occurs when a study fails to
detect a treatment that really exists.
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Type II error = false (-) = beta error = (β)
A Type II error is likely to occur when a treatment effect is very small.
Two-Tailed Hypothesis Test
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Question 4: After years of teaching driver’s education, an
instructor knows that students hit an average of µ = 10.5
orange cones while driving the obstacle course in their
final exam. The distribution of run-over cones is
approximately normal with a standard deviation of
σ = 4.8. To test a theory about text messaging and
driving, the instructor recruits a sample of n = 16 student
drivers to attempt the obstacle course while sending a
text message. The individuals in this sample hit an
average of M = 15.9 cones. Do the data indicate that
texting has a significant effect on driving? Test with
α = 0.01.
Two-Tailed Hypothesis Test
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Step 1: State hypotheses
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H0: Texting has no effect on driving. (µ = 10.5)
H1: Texting has an effect on driving. (µ ≠ 10.5)
Step 2: Set Criteria for Decision (α = 0.01)
z = ± 2.58
Reject H0
Reject H0
z = - 2.58
z = 2.58
Two-Tailed Hypothesis Test
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Step 3: Compute sample statistic

=
4.8
16
−

=
15.9−10.5
1.20
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=
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=
=
4.8
4
= 1.20
=
5.4
1.20
= 4.50
Two-Tailed Hypothesis Test
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Step 4: Make a decision
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For a Two-tailed Test:
If -2.58 < zsample < 2.58, fail to reject H0
If zsample ≤ -2.58 or zsample ≥ 2.58, reject H0
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zsample (4.50) > zcritical (2.58)
Thus, we reject the null and note that texting has a significant
effect on driving.
Factors that Influence a
Hypothesis Test
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Question 5: If other factors are held constant, increasing
the size of a sample increases the likelihood of rejecting
the null hypothesis. (True or False?)
Factors that Influence a
Hypothesis Test
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True. A larger sample produces a smaller standard error, which
leads to a larger z-score.
For  =
−
,

where  =

,

as sample size (n) increases,
standard error ( ) decreases,
which then increases z.
Consequently, as z increases so
does the probability of rejecting the
null hypothesis.
Factors that Influence a
Hypothesis Test
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Question 6: If other factors remain constant, are you
more likely to reject the null hypothesis with a standard
deviation of σ = 2 or σ = 10?
Factors that Influence a
Hypothesis Test
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σ = 2. A smaller standard deviation produces a smaller standard
error, which leads to a larger z-score. Thus, increasing the
probability of rejecting the null hypothesis.

10
10
=
=
=2
5

25

20
20
=
=
=
=4
5

25
=
One-tailed Hypothesis Test
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Question 7: A researcher is testing the hypothesis that
consuming a sports drink during exercise improves
endurance. A sample of n = 50 male college students is
obtained and each student is given a series of three
endurance tasks and asked to consume 4 ounces of the
drink during each break between tasks. The overall
endurance score for this sample is M = 53. For the
general population of male college students, without any
sports drink, the scores average μ = 50 with a standard
deviation of σ = 10. Can the researcher conclude that
endurance scores with the sports drink are significantly
higher than score without the drink? (Use a one-tailed
test, α = 0.05)
One-tailed Hypothesis Test
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Step 1: State hypotheses
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H0: Endurance scores are not significantly higher with the sports
drink. (µ ≤ 50)
H1: Endurance scores are significantly higher with the sports drink.
(µ > 50)
Step 2: Set Criteria for Decision (α = 0.05)
z = 1.65
Reject H0
z = 1.65
One-tailed Hypothesis Test
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Step 3: Compute sample statistic

=
10
50
−

=
53−50
1.41
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=
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=
=
10
7.07
=
= 1.41
3
1.41
= 2.13
One-tailed Hypothesis Test
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Step 4: Make a decision
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For a One-tailed Test:
If zsample ≤ 1.65, fail to reject H0
If zsample > 1.65, reject H0
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zsample (2.13) > zcritical (1.65)
Thus, we reject the null and note that the sports drink does raise
endurance scores.
Effect Size and Cohen’s d
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Question 8: A researcher selects a sample from a
population with µ = 40 and σ = 8. A treatment is
administered to the sample and, after treatment, the
sample mean is found to be M = 47. Compute Cohen’s d
to measure the size of the treatment effect.
Effect Size and Cohen’s d
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estimated Cohen’s d:
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d=
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This is a large effect.
47−40
8

=
−

7
8
= = 0.875
Remember: These are thresholds. Any effect less than d = 0.2 is a trivial effect and
should be treated as having no effect. Any effect between d = 0.2 and d = 0.5 is a small
effect. And between d = 0.5 and d = 0.8 is a medium effect.
Computing Power
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Question 9: A researcher is evaluating the influence of a
treatment using a sample selected from a normally
distributed population with a mean of µ = 100 and a
standard deviation of σ = 20. The researcher expects a
10-point treatment effect and plans to use a two-tailed
hypothesis test with α = 0.05. Compute the power of the
test if the researcher uses a sample of n = 25 individuals.
Computing Power
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Step #1: Calculate standard error for sample
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=

=
20
25
=
20
5
=4
Step #2: Locate Boundary of Critical Region
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z = 1.96, for α = 0.05
1.96 * 4 = 7.84 points
Thus, the critical boundary corresponds to M = 100 + 7.84 = 107.84.
Any sample mean greater than 107.84 falls in the critical region.
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Step #3: Calculate the z-score
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=
−

=
107.84 −110
4
=
−2.16
4
= −0.54
Computing Power
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Step #4: Interpret Power of the Hypothesis Test
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Find probability associated with a z-score > - 0.54
Look this probability up as the proportion in the body of the normal
distribution (column B in your textbook)
p(z > -0.54) = 0.7054
Thus, with a sample of 25 people and a 10-point treatment effect,
70.54% of the time the hypothesis test will conclude that there is a
significant effect.
Computing Power
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Question 10: A researcher is evaluating the influence of
a treatment using a sample selected from a normally
distributed population with a mean of µ = 80 and a
standard deviation of σ = 20. The researcher expects a
12-point treatment effect and plans to use a two-tailed
hypothesis test with α = 0.05. Compute the power of the
test if the researcher uses a sample of n = 25 individuals.
Computing Power
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Step #1: Calculate standard error for sample
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=

=
20
25
=
20
5
=4
Step #2: Locate Boundary of Critical Region
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z = 1.96, for α = 0.05
1.96 * 4 = 7.84 points
Thus, the critical boundary corresponds to M = 80 + 7.84 = 87.84.
Any sample mean greater than 87.84 falls in the critical region.
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Step #3: Calculate the z-score
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=
−

=
87.84 −92
4
=
−4.16
4
= −1.04
Computing Power
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Step #4: Interpret Power of the Hypothesis Test
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Find probability associated with a z-score > - 1.04
Look this probability up as the proportion in the body of the normal
distribution (column B in your textbook)
p(z > -1.04) = 0.8508
Thus, with a sample of 25 people and a 12-point treatment effect,
85.08% of the time the hypothesis test will conclude that there is a
significant effect.
FAQs
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What is power?
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Power is the probability that a hypothesis test will reject the null
hypothesis, if there is a treatment effect.
β is the probability of a type II error (false negative). Therefore, power is 1 – β.
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There are 4 steps involved in finding power.
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Step #1: Calculate the standard error.
Step #2: Locate the boundary of the critical region.
Step #3: Calculate the z-score.
Step #4: Find the probability.
Using the example from the lecture notes, let’s go through each
step.
FAQs
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The previous slide was based upon a study from your
book with μ = 80, σ = 10, and a sample (n=25) that is
drawn with an 8-point treatment effect (M=88). What is
the power of the related statistical test for detecting the
difference between the population and sample mean?
FAQs
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Step #1: Calculate standard error for sample
 In this step, we work from the population’s standard
deviation (σ) and the sample size (n)
FAQs
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Step #2: Locate Boundary of Critical Region
 In this step, we find the exact boundary of the critical
region
 Pick a critical z-score based upon alpha (α =.05)
FAQs
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Step #3: Calculate the z-score for the difference between
the treated sample mean (M=83.92) for the critical region
boundary and the population mean with an 8-point
treatment effect (μ = 88).