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```COURSE: JUST 3900
TIPS FOR APLIA
Chapters 7-9:
Test Review
Developed By:
John Lohman
Michael Mattocks
Aubrey Urwick
Chapter 7: Key Concepts



This chapter is on the distribution of sample means.
Sampling error is the natural discrepancy, or difference,
between a sample statistic and its corresponding
population parameter.
The distribution of sample means, or sampling
distribution, is the collection of sample means for all of
the possible random samples of a particular size (n) that
can be obtained from a population.
Chapter 7: Key Concepts


The mean of a sampling distribution, or expected value
of M, should always be equal to the population mean (µ).
The standard error of M (σM) is the standard deviation of
the distribution of sample means. It provides a measure
of how much distance is expected on average between a
sample mean (M) and the population mean (µ).

Standard Error of M: 𝜎𝑀 =
𝜎
𝑛
Chapter 7: Key Concepts

The central limit theorem states that for any population
with mean µ and standard deviation σ, the distribution of
sample means for sample size n will have a mean of µ
𝜎
and a standard deviation of 𝜎𝑀 = and will approach a
𝑛

normal distribution as n approaches infinity.
In other words, a sampling distribution has the following
characteristics:

A mean of µ. (Expected value of M)

A standard deviation of 𝜎𝑀 =

Is normal in shape if n is at least 30, or if the population
distribution is normal in shape.
𝜎
.
𝑛
(Standard error of M)
Chapter 7: Practice

Question 1: A population has a mean of µ = 100 and a
standard deviation of σ = 15.
a)
b)
c)
d)
For samples of size n = 5, what is the expected value and the
average difference between M and µ for the distribution of
sample means?
If the population distribution is not normal, describe the shape
of the distribution of sample means based on n = 5.
For samples of size n = 45, what is the expected value and the
average difference between M and µ for the distribution of
sample means?
If the population distribution is not normal, describe the shape
of the distribution of sample means based on n = 45.
Chapter 7: Practice

a)
Expected Value of M: µ = 100
Standard Error of M: 𝜎𝑀 =
b)
𝜎
𝑛
=
15
5
= 6.71
The distribution of sample means does not satisfy either
criterion to be normal. It would not be a normal distribution.
If the population distribution was normal, or if the sample size was at least 30,
the sampling distribution would be normal.
Chapter 7: Practice

c)
Expected Value of M: µ = 100
Standard Error of M: 𝜎𝑀 =
𝜎
𝑛
=
15
45
= 2.24
Notice how the expected value stays the same regardless of sample size, while
the standard error decreases as sample size increases.
d)
Because the sample size is greater than 30, the distribution of
sample means is a normal distribution.
Chapter 7: Key Concepts
Notice the relationship between sample size n and standard error 𝜎𝑀 .
As n increases, 𝜎𝑀 decreases.
Chapter 7: Key Concepts
Again, as sample size increases, standard error decreases.
Chapter 7: Key Concepts
Standard Deviation (σ)
1
5
10
15
20
Standard Error
1
𝜎𝑀 =
100
5
𝜎𝑀 =
100
10
𝜎𝑀 =
100
15
𝜎𝑀 =
100
20
𝜎𝑀 =
100
0.10
0.50
1.00
1.50
2.00
As standard deviation increases, so does standard error of M.
Chapter 7: Key Concepts
The relationship between sample size and standard error is an important concept.
Understanding this relationship can save you valuable time on the test.
Chapter 7: Practice

Question 2: If we have a sampling distribution with a
standard error of 𝜎𝑀 = 5 and sample size of n = 15, what
would the standard error be if we increased our sample
size to n = 30?
a)
b)
c)
d)
𝜎𝑀
𝜎𝑀
𝜎𝑀
𝜎𝑀
= 7.12
= 6.57
= 3.54
= 5.64
Chapter 7: Practice


c) 𝜎𝑀 = 3.54
𝜎
𝜎𝑀 =
𝑛
𝜎
5=
15
𝜎 = 19.36
𝜎𝑀 =
𝜎
19.36
=
= 3.54
𝑛
30
There are two ways to arrive at this answer.
1) Find the standard deviation, then re-calculate standard error for n = 30, or
2) Eliminate all of the other answer choices based on our rule: as sample size
increases standard error decreases. In this case, answer choice c) is the only
answer that is less than 5.
Chapter 7: Sampling
Distributions and Probability

Very similar to what we did in chapter 6, we can
calculate the probability of selecting a specific sample of
size n by calculating the z-score and looking up the
probability in the unit normal table.

𝑧=
𝑀−𝜇
𝜎𝑀
Chapter 7: Practice

Question 3: What is the probability of obtaining a sample
mean greater than M = 60 for a random sample of n = 16
scores selected from a normal population with a mean of
µ = 65 and a standard deviation of σ = 20? What if we
changed our sample size to n = 5?
Chapter 7: Practice


Find 𝜎𝑀 .


𝜎
𝑛
=
20
16
=
20
4
= 5.00
Find the z-score.


𝜎𝑀 =
𝑧=
𝑀−𝜇
𝜎𝑀
=
60−65
5
=
−5
5
= −1.00
Look up z = -1.00 in the unit normal table. (Column B)

p(z > -1.00) = 0.8413 (or 84.13%)
Chapter 7: Practice


Find 𝜎𝑀 .


𝜎
𝑛
=
20
5
20
= 2.24 = 8.94
Find the z-score.


𝜎𝑀 =
𝑧=
𝑀−𝜇
𝜎𝑀
=
60−65
8.94
−5
= 8.94 = −0.56
Look up z = -0.56 in the unit normal table. (Column B)

p(z > -0.56) = 0.7123 (or 71.23%)
Chapter 8: Key Concepts



Chapter 8 covers hypothesis testing.
Hypothesis tests allow us to make generalizations from
samples to populations about whether our treatment has
an effect.
When there appears to be a treatment effect, one of two
things has happened.


The discrepancy between µ and M is the result of sampling error
or,
Chapter 8: Key Concepts

There are 4 steps to hypothesis testing:
1)
State the hypothesis.
1)
2)
We state both the null hypothesis H0, which states that there is no
change, and the alternative hypothesis H1, which states that there
is a change.
Set criteria for a decision.
1)
Next, we divide our distribution up into two sections:
1)
2)
2)
3)
Sample means close to the null hypothesis.
Sample means very different from the null hypothesis.
We use the alpha level, or level of significance, to mark the critical
region, which is composed of sample values that are very unlikely
to be obtained if the null is true.
These boundaries are generally set at α = 0.05, α = 0.01, or
α = 0.001
Chapter 8: Key Concepts
Critical region
for 2-tail test at
α = 0.05.
α Level
z-Score
.05
+/- 1.96
.01
+/- 2.58
.001
+/- 3.30
Chapter 8: Key Concepts
Critical
region
for 1-tail
test at
α = 0.05.
α Level
z-Score
.05
+/- 1.65
.01
+/- 2.33
.001
+/- 3.10
Chapter 8: Key Concepts

There are 4 steps to hypothesis testing:
3)
Compute the sample statistic.
1)
4)
𝑧=
𝑀−𝜇
𝜎𝑀
=
𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛 −ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑧𝑒𝑑 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑀 𝑎𝑛𝑑 𝜇
Make a decision.
1)
2)
3)
Does our sample statistic fall in the critical region?
If yes, reject the null. The treatment has an effect.
If no, fail to reject the null. The treatment does not have an effect.
Chapter 8: Practice

Question 1: What combination of factors will most likely
lead to rejecting the null hypothesis?
a)
b)
c)
d)
n = 30; α = 0.05
n = 5; α = 0.05
n = 30; α = 0.01
n = 5; α = 0.01
Chapter 8: Practice

a)
b)
c)
d)
n = 30; α = 0.05
n = 5; α = 0.05
n = 30; α = 0.01
n = 5; α = 0.01
A larger sample size leads to a smaller
Standard error, which in turn gives us
A larger z-score, increasing the likelihood
That it will fall in the critical region.
The larger alpha level increases the size of
the critical region, making it more likely that
our z-score will fall in this region.
Chapter 8: Practice

Question 2: On average, what would we expect our zscore to equal if the null hypothesis is true?
Chapter 8: Practice


z = 0, indicating that there was no change in µ and the treatment
Remember that our null hypothesis states that µ doesn’t change, and the µ for
the unit normal table (as with all z-distributions) is 0.
Chapter 8: Practice

Question 3: State the null and alternative hypotheses for
a one-tailed test with µ = 50. We expect our treatment to
have a positive effect.
a)
b)
c)
d)
H0: µ = 50
H1: µ ≠ 50
H0: µ = 0
H 1: µ ≠ 0
H0: µ > 50
H1: µ ≤ 50
H0: µ ≤ 50
H1: µ > 50
Chapter 8: Practice

a)
b)
c)
d)
H0: µ = 50
H1: µ ≠ 50
H 0: µ = 0
H 1: µ ≠ 0
H0: µ > 50
H1: µ ≤ 50
H0: µ ≤ 50
H1: µ > 50
I often find it less confusing to state the alternative
hypothesis first in a one-tailed test. We know that if
there’s an effect, µ will get larger, µ > 0. However, if µ
doesn’t get larger, it will either stay the same or get
smaller, µ ≤ 0.
Chapter 8: Key Concepts





Occasionally, hypothesis tests can lead to error.
There are two types of error in hypothesis testing: Type I
and Type II.
Type I error occurs when a researcher rejects a null
hypothesis that is actually true. In these cases, the
sample statistic falls in the critical region, not because of
a treatment effect, but as a result of sampling error.
Type I error is also known as a false positive.
The probability of a Type I error is equal to the alpha
level, or level of significance.
Chapter 8: Key Concepts


Type II error, on the other hand, occurs when a
researcher fails to reject a false null hypothesis.
This means that a treatment effect really exists, but the
hypothesis test fails to detect it. This is often the case
with very small treatment effects.
Chapter 8: Key Concepts
α
1-β
1-α
β
Chapter 8: Practice

Question 4: Which of the following will decrease the risk
of a type I error?
a)
b)
c)
d)
e)
f)
Increasing the sample size (n)
Decreasing the alpha from α = 0.05 to α = 0.01
Moving from a one- to a two-tailed test
All of the above
None of the above
Some of the above
Chapter 8: Practice

a)
b)
c)
d)
e)
f)
Increasing the sample size (n)
Decreasing the alpha from α = 0.05 to α = 0.01
Moving from a one- to a two-tailed test
All of the above
None of the above
Some of the above
Increasing n reduces 𝜎𝑀 , making it less likely that the effect is due to chance.
Decreasing alpha makes it more difficult to get into the critical region, reducing
the chance of getting a type I error. Here, we reduced the probability of a type I
error from 5% to 1%.
Just like when we decrease alpha, moving from a one- to a two-tailed test
reduces the probability of getting a type I error.
Chapter 8: Practice

Question 5: When are type II errors likely to occur?
Chapter 8: Practice


When the treatment effect is very small.
Remember: Type II errors occur when we fail to reject a false null hypothesis.
In other words, our treatment has an effect, but for some reason our z-score
did not fall in the critical region.
Chapter 8: Key Concepts


However, hypothesis testing doesn’t tell the whole story.
It tells us whether there is a significant treatment effect,
but not the size of the effect.
To find effect size, we calculate Cohen’s d.
𝑀𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝑀𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 −𝜇𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝜎

𝐶𝑜ℎ𝑒𝑛′ 𝑠 𝑑 =

Effect sizes are summarized in the following chart.
=
Cohen’s d is not influenced by sample size.
Chapter 8: Practice

Question 6: A researcher selects a sample from a
population with µ = 45 and σ = 8. A treatment is
administered to the sample and, after treatment, the
sample mean is found to be M = 47. What is the size of
the treatment effect?
Chapter 8: Practice

𝑀𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 −𝜇𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝜎

𝐶𝑜ℎ𝑒𝑛′ 𝑠 𝑑 =

This is a small effect.
=
47−45
8
2
8
= = 0.25
Chapter 8: Key Concepts


Power refers to the probability that a hypothesis test will
correctly reject a false null hypothesis. That is, power is
the likelihood that the test will identify a treatment effect
if one really exists.
Like hypothesis testing, power is a 4-step process:
1)
Calculate standard error.

2)
𝜎𝑀 =
𝜎
𝑛
Locate boundary of critical region.

M = µ + (Critical z-score * 𝜎𝑀 )
Chapter 8: Key Concepts
3)
Calculate z-score for the difference between the treated
sample mean for the critical region boundary and the
population mean with the treatment effect.

4.
𝑧=
𝑀−𝜇
𝜎𝑀
Interpret power of the hypothesis test.

Find probability associated with your z-score.
Chapter 8: Key Concepts
Chapter 8: Practice

Question 7: What is the power of a hypothesis test if the
probability of a type II error is β = 0.6046?
Chapter 8: Practice


Power: 1 − β = 1 − 0.6046 = 0.3954 or 39.54%
Chapter 8: Practice

Question 8: What 3 factors increase power?
Chapter 8: Practice

1)
2)
3)
Sample size is increased.
Alpha is increased (e.g., from .01 to .05).
You go from a 2- to a 1-tail test.
Chapter 9: Practice

Question 1: Which of the following is a fundamental
difference between the t statistic and a z-score?
a)
b)
c)
d)
The t statistic uses the sample mean in place of the population
mean.
The t statistic uses the sample variance in place of the
population variance.
The t statistic computes the standard error by dividing the
standard deviation by n-1 instead of dividing by n.
All of the above are differences between z and t.
Chapter 9: Practice

a)
b)
c)
d)
The t statistic uses the sample mean in place of the population
mean.
The t statistic uses the sample variance in place of the
population variance.
The t statistic computes the standard error by dividing the
standard deviation by n-1 instead of dividing by n.
All of the above are differences between z and t.
Without the population standard deviation or variance, we cannot calculate
a z-score.
Chapter 9: Practice

Question 2: A sample of n = 25 is selected from a
population with a mean of µ = 50. A treatment is
administered to the individuals in the sample and, after
treatment, the sample has a mean of M = 56 and a
variance of s = 5.

If all other factors are held constant and the sample size is
increased to n = 25, is the sample sufficient to conclude that the
treatment has a significant effect? (Use a two-tailed test with
α = 0.05)
Chapter 9: Practice


Step 1: State hypotheses


H0: Treatment has no effect. (µ = 50)
H1: Treatment has an effect. (µ ≠ 50)
Chapter 9: Practice

Step 2: Set Criteria for Decision (α = 0.05)
t Critical: ± 2.064
Chapter 9: Practice
df = 24
t Distribution with
α = 0.05
Critical region
t = - 2.064
Critical region
t = + 2.064
Chapter 9: Practice

a)
b)
Step 3: Compute sample statistic
𝑠𝑀 =
𝑡=
𝑠
𝑛
𝑀−𝜇
𝑠𝑀
=
5
25
5
5
=
56−50
1
= = 1.00
=
6
1
= 6.00
Chapter 9: Practice
df = 24
t Distribution with
α = 0.05
Critical region
t = - 2.064
Critical region
t = + 2.064
t = 6.00
Chapter 9: Practice

Step 4: Make a decision

For a Two-tailed Test:
If -2.064 < tsample < 2.064, fail to reject H0
If tsample ≤ -2.064 or tsample ≥ 2.064, reject H0


tsample (6.00) > tcritical (2.064)
Thus, we reject the null and conclude that the treatment has an
effect.
Chapter 9: Practice
Question 3: A sample of n = 16 is selected from a
population. A treatment is administered to the sample
and, after treatment, the sample is found to be M = 86
with a standard deviation of s = 8. A confidence interval
is constructed and the interval spans μlower = 81.738 and
μupper = 90.262. How confident are we that our µ falls
within this interval?
Chapter 9: Practice

1)
Find the corresponding t-statistics for μupper and μlower.
𝑠
𝑛
=
8
16
8

𝑠𝑀 =
= 4 = 2.00

𝑡=
𝑀−𝜇
𝑠𝑀
=
81.738−86
2
=
−4.262
2

𝑡=
𝑀−𝜇
𝑠𝑀
=
90.262−86
2
=
4.262
2
= −2.131
= 2.131
Chapter 9: Practice

df = 15
Middle ?%
of t distribution
?% in the lower tail
t = - 2.131
?% in the upper tail
t = + 2.131
Chapter 9: Practice

2)
Use the t-distribution to find the alpha level (percentage
between both tails)
100% - 5% = 95%
Chapter 9: Practice

df = 15
Middle 95%
of t distribution
2.5% in the lower tail
t = - 2.131
2.5% in the upper tail
t = + 2.131
Chapter 9: Practice
