Exam 2 Review
(Chapter 6, 7, 8)
Professor: Dr. Max Owens
Teaching Assistant: Alex Goyette
Formulas to remember
• probability =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑐𝑙𝑎𝑠𝑠𝑖𝑓𝑖𝑒𝑑 𝐴
p(A) =
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
• To find z-scores related to the probability of scores in a
population :
𝑋−𝜇
𝑧=
𝜎
• To find z-scores related to the probability of sample means:
𝑧=
𝑀 − 𝜇
𝜎
standard error (𝜎𝑀 =
)
𝑛
Standard Error of M
standard error = 𝜎𝑀 =
𝜎
=
𝑛
𝛔2
𝐧
1. The standard deviation of the distribution of
sample means.
2. Provides a measure of how much distance is
expected on average between M and μ.
The Central Limit Theorem, shape and central
tendency.
• The Central limit theorem tells us what to
expect when we draw a sample.
• Shape = curve, normal or non-normal.
– The distribution of sample means is normal when
the population is normally distributed or the
sample size, n > 30.
• Central Tendency = 𝝁
– The expected value of the sample mean M is
equal to the population mean 𝜇.
• Variability = Standard Error of M.
4 Steps of Hypothesis Testing
1. State the hypothesis( H0= ?, HA = ?).
2. Setting the criteria for a decision( α= ?).
3. Collect data and compute sample statistics( z=?).
4. Make a decision.
Two-tailed vs One-tailed Hypothesis test
• A hypothesis test has only two possible outcomes.
Two-Tailed, Increase or Decrease
One-Tailed, Increase
One-Tailed, Decrease
Effect Size
1. Effect size: The absolute magnitude of a treatment
or experimental effect.
𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑀𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 𝜇 𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝐶𝑜ℎ𝑒𝑛 𝑠 𝑑 =
=
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝜎
′
Magnitude of d
d = 0.2
d = 0.5
d = 0.8
Evaluation of
Effect Size
Small
effect
Medium
effect
Large
effect
Statistical Power
• Power: Probability of correctly rejecting a false H0
– What proportion of samples would fall in the tails of the
distribution if a treatment works.
98%
Computing Power
• Five steps to compute power
1.
2.
3.
4.
Compute Standard Error (First Distribution).
Find our z-critical boundary (First Distribution).
Find the M of our critical boundary (First Distribution).
Calculate z scores for the treatment effect (Second
Distribution).
5. Find the proportion for the treatment z score = Power
(Second Distribution).
Factors Influencing Power
• Size of the Treatment Effect:  effect =  power.
– Both provide an indication of the strength or magnitude of a
treatment effect.
• Alpha () Level:   (e.g., from .05 to .01) =  power.
• One-tailed vs. Two-tailed Tests: 1-tailed test =  power.
• Sample Size:  sample size =  power
Review Quiz
10 questions highlighting central concepts of
each chapter.
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
A
• How much daily sugar intake corresponds to
the top 5% of the population
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
A
• How much daily sugar intake corresponds to the top
5% of the population?
p = 0.05, z = ?
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
A
• How much daily sugar intake corresponds to the top
5% of the population?
p = 0.05, z = 1.65
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
A
• How much daily sugar intake corresponds to the top
5% of the population?
p = 0.05, z= 1.65
X = 𝜇 + z𝜎
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
A
• How much daily sugar intake corresponds to the top
5% of the population?
p = 0.05, z= 1.65
X = 𝜇 + z𝜎
X = 22.7 + (1.65)(4.5)
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
A
• How much daily sugar intake corresponds to the top
5% of the population?
p = 0.05, z= 1.65
X = 𝜇 + z𝜎
X = 22.7 + (1.65)(4.5)
X = 30.125
Review Quiz
2. True or False
T/F
• The shape of a distribution of
sample means is always normal
T/F
• As sample size increases, the value
of the standard error decreases
Review Quiz
2. True or False
False
• The shape is normal only if the
population is normal or n > 30
Review Quiz
2. True or False
standard error = 𝜎𝑀 =
True
𝜎
𝑛
• Sample size is in the denominator
of the equation so as n grows
larger, standard error decreases
Review Quiz
3. What is Standard error?
A
• the average spread of a set of scores in a population
B
• the probability of concluding that there is an effect
when there is not.
C
• The standard deviation of the distribution of sample
means
Review Quiz
3. What is Standard error?
A
• the average spread of a set of scores in a population
B
• the probability of concluding that there is an effect
when there is not.
C
• The standard deviation of the distribution of sample
means
What is Standard Error?
SE = the average
distance expected
between µ and M
Review Quiz
4. A normal population has μ = 60 with σ = 5;
the distribution of sample means for samples of
size n = 4 selected from this population would
have an expected value of _____
A •5
B • 60
C • 30
D • 15
Review Quiz
4. A normal population has μ = 60 with σ = 5;
the distribution of sample means for samples of
size n = 4 selected from this population would
have an expected value of _____
A •5
B • 60
C • 30
D • 15
Review Quiz
5. A normal distribution has a mean
of μ=58 and a standard deviation of σ=12.
Q
• What is the probability of selecting a sample
of n=9 scores with a mean less than M=52 ?
Review Quiz
5. A normal distribution has a mean
of μ=58 and a standard deviation of σ=12.
Q
• What is the probability of selecting a sample
of n=9 scores with a mean less than M=52 ?
𝜎𝑀 =
𝜎
𝑛
Review Quiz
5. A normal distribution has a mean
of μ=58 and a standard deviation of σ=12.
Q
• What is the probability of selecting a sample
of n=9 scores with a mean less than M=52 ?
𝜎𝑀 =
𝜎
𝑛
=
12
9
=
12
=
3
4
Review Quiz
5. A normal distribution has a mean
of μ=58 and a standard deviation of σ=12.
Q
• What is the probability of selecting a sample
of n=9 scores with a mean less than M=52 ?
𝜎𝑀 = 4
𝑀 − 𝜇
𝑧=
=
𝜎𝑀
Review Quiz
5. A normal distribution has a mean
of μ=58 and a standard deviation of σ=12.
Q
• What is the probability of selecting a sample
of n=9 scores with a mean less than M=52 ?
𝜎𝑀 = 4
𝑀 − 𝜇
52 − 58
𝑧=
=
= −1.5
𝜎𝑀
4
Review Quiz
5. A normal distribution has a mean
of μ=58 and a standard deviation of σ=12.
Q
• What is the probability of selecting a sample
of n=9 scores with a mean less than M=52 ?
𝜎𝑀 = 4
𝑀 − 𝜇
52 − 58
𝑧=
=
= −1.5
𝜎𝑀
4
𝑝 𝑧 < −1.5 = 0.0668
Review Quiz
6. A sample n = 9 had a cohen’s d of .5 after
treatment, with a population mean of µ = 80
and a standard deviation of σ = 12. What is the
mean for the treated sample?
A •M=6
B
• M = 82
C
• M = 86
D • Cannot answer without the sample error
Review Quiz
6. A sample n = 9 had a cohen’s d of .5 after
treatment, with a population mean of µ = 80
and a standard deviation of σ = 12. What is the
mean for the treated sample?
A •M=6
B
• M = 82
C
• M = 86
D • Cannot answer without the sample error
Review Quiz
6. A sample n = 9 had a cohen’s d of .5 after
treatment, with a population mean of µ = 80
and a standard deviation of σ = 12. What is the
mean for the treated sample?
Cohen’s d =
.5 =
𝜇𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 𝜇 𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝜎
𝜇𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 80
12
Review Quiz
6. A sample n = 9 had a cohen’s d of .5 after
treatment, with a population mean of µ = 80
and a standard deviation of σ = 12. What is the
mean for the treated sample?
.5 =
𝜇𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡−80
12
.5(12) = 𝜇𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 80
Review Quiz
6. A sample n = 9 had a cohen’s d of .5 after
treatment, with a population mean of µ = 80
and a standard deviation of σ = 12. What is the
mean for the treated sample?
.5 =
𝜇𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡−80
12
80 + 6 = 𝜇𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝜇𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 = 86
Review Quiz
7. A random sample(n=25) is selected from a normal
population with a mean of μ=20 and a standard
deviation of σ=10. After a treatment is administered to
the individuals in the sample, the sample mean is found
to be M=25 . Complete the following steps of the twotailed hypothesis test.
1
• What are the null and alternative hypotheses?
2
• Using a two-tailed test with α =0.05, what is the
critical value?
• What is the z-score associated with the mean for the
treatment?
• Pick one: Reject/fail to reject the null hypothesis.
There is/ is not a significant effect of treatment.
3
4
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the two-tailed hypothesis test.
1
• What are the null and alternative hypotheses?
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the two-tailed hypothesis test.
1
• What are the null and alternative hypotheses?
Null H0: There is no effect of treatment; μtreatment = 20
Alternative H1: There is an effect of treatment; μtreatment ≠ 20
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
2
• Using a two-tailed test with α =0.05, what is the
critical value?
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
2
• Using a two-tailed test with α =0.05, what is the
critical value?
P = 0.025 in each tail;
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
2
• Using a two-tailed test with α =0.05, what is the
critical value?
P = 0.025 in each tail; using Column C -> z = 1.96
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
3
• What is the z-score associated with the mean for the
treatment?
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
3
• What is the z-score associated with the mean for the
treatment?
𝜎𝑀 =
𝜎
𝑛
;𝑧=
𝑀−𝜇
𝜎𝑀
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
3
• What is the z-score associated with the mean for the
treatment?
𝜎𝑀 =
𝜎
𝑛
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
3
• What is the z-score associated with the mean for the
treatment?
𝜎𝑀 =
𝜎
𝑛
=
10
25
=
10
5
=2
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
3
• What is the z-score associated with the mean for the
treatment?
𝜎𝑀 = 2
𝑀 − 𝜇
𝑧=
𝜎𝑀
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
3
• What is the z-score associated with the mean for the
treatment?
𝜎𝑀 = 2
𝑧=
𝑀−𝜇
𝜎𝑀
=
25 −20
2
= 2.5
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
4
• Pick one: Reject/fail to reject the null hypothesis.
There is/ is not a significant effect of treatment.
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
4
• Pick one: Reject/fail to reject the null hypothesis.
There is/ is not a significant effect of treatment.
• Z critical < Z treatment score
Review Quiz
7. A random sample(n=25) is selected from a
normal population with a mean of μ=20 and a
standard deviation of σ=10. After a treatment is
administered to the individuals in the sample, the
sample mean is found to be M=25 . Complete the
following steps of the hypothesis test.
4
• Pick one: Reject/fail to reject the null hypothesis.
There is/ is not a significant effect of treatment.
• Z critical < Z treatment score
• 1.96 < 2.5
Review Quiz
8. A treatment is expected to lower cholesterol by
30 points for a population with µ = 240 (σ = 30). For
a sample of n = 9, what is the probability of a
successful treatment? (α = .05; two-tail test)
A
• 1.04
B
• .3508
C
• .8508
D
• .1492
Computing Power
• Five steps to compute power
1.
2.
3.
4.
Compute Standard Error (First Distribution).
Find our z-critical boundary (First Distribution).
Find the M of our critical boundary (First Distribution).
Calculate z scores for the treatment effect (Second
Distribution).
5. Find the proportion for the treatment z score = Power
(Second Distribution).
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05?
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 240 – 30 = 210
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210
1. Compute Standard Error.
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210
1. Compute Standard Error.
𝜎𝑀 =
𝜎
𝑛
=
30
9
=
30
3
= 10
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210
1. Compute Standard Error. 𝜎𝑀 = 10
2. Find our critical boundary, α = .05 = z =
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210
1. Compute Standard Error. 𝜎𝑀 = 10
2. Find our critical boundary, α = .05 = z = -1.96.
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210
1. Compute Standard Error. 𝜎𝑀 = 10
2. Find our critical boundary, α = .05 = z = -1.96.
3. Find the M of our critical boundary.
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210
1. Compute Standard Error. 𝜎𝑀 = 10
2. Find our critical boundary, α = .05 = z = -1.96.
3. Find the M of our critical boundary.
• 𝑧=
𝑀−𝜇
𝜎𝑀
= 𝑴 = 𝝁 + 𝒛 𝝈𝑴
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210
1. Compute Standard Error. 𝜎𝑀 = 10
2. Find our critical boundary, α = .05 = z = -1.96.
3. Find the M of our critical boundary.
• 𝑴 = 𝝁 + 𝒛 𝝈𝑴 = 𝟐𝟒𝟎 + (−𝟏. 𝟗𝟔) 𝟏𝟎
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210
1. Compute Standard Error. 𝜎𝑀 = 10
2. Find our critical boundary, α = .05 = z = -1.96.
3. Find the M of our critical boundary.
• 𝑴 = 𝝁 + 𝒛 𝝈𝑴 = 𝟐𝟐𝟎. 𝟒𝟎
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ = 30.
What is the power of a two tailed hypothesis test with n =
9, α = .05? µafter treatment = 210 𝝈𝑴 = 𝟏𝟎, M = 220.40.
4. Calculate z scores for the treatment effect.
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ = 30.
What is the power of a two tailed hypothesis test with n =
9, α = .05? µafter treatment = 210 𝝈𝑴 = 𝟏𝟎, M = 220.40.
4. Calculate z scores for the treatment effect.
𝑀 − 𝜇
𝑧=
𝝈 =
𝝈𝑴 =
𝒏
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ = 30.
What is the power of a two tailed hypothesis test with n =
9, α = .05? µafter treatment = 210 𝝈𝑴 = 𝟏𝟎, M = 220.40.
4. Calculate z scores for the treatment effect.
𝑀 − 𝜇
220.40 − 210 10.40
𝑧=
=
= 1.04
𝝈 =
30
10
𝝈𝑴 =
𝒏
9
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210 𝝈𝑴 = 𝟏𝟎, M =
220.40, z = 1.04
5. Find the proportion for the treatment z score
= Power.
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210 𝝈𝑴 = 𝟏𝟎, M =
220.40, z = 1.04
5. Find the proportion for the treatment z score
= Power. Which Column?
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210 𝝈𝑴 = 𝟏𝟎, M =
220.40, z = 1.04
5. Find the proportion for the treatment z score
= Power. Which Column? (B)
Computing Power
Treatment results in a 30 point reduction. µ = 240, σ =
30. What is the power of a two tailed hypothesis test
with n = 9, α = .05? µafter treatment = 210 𝝈𝑴 = 𝟏𝟎, M =
220.40, z = 1.04
5. Find the proportion for the treatment z score
= Power. Which Column? (B) = z = 1.04 = .8508
Review Quiz
8. A treatment is expected to lower cholesterol
by 30 points for a population with µ = 240 (σ =
30). For n = 9, at α = .05 (two-tailed), what is the
probability of a successful treatment?
A
• 1.04
B
• .3508
C
• .8508
D
• .1492
Review Quiz
9. Which of the following accurately describes
the critical region for a hypothesis test?
A
• Outcomes that have a low probability if the null
hypothesis is true
B
• Outcomes that have a high probability if the null
hypothesis is true
C
• Outcomes that have a low probability regardless of
whether the null hypothesis is true
D
• Outcomes that have a high probability regardless of
whether the null hypothesis is true
Review Quiz
9. Which of the following accurately describes
the critical region for a hypothesis test?
A
• Outcomes that have a low probability if the null
hypothesis is true
B
• Outcomes that have a high probability if the null
hypothesis is true
C
• Outcomes that have a low probability regardless of
whether the null hypothesis is true
D
• Outcomes that have a high probability regardless of
whether the null hypothesis is true
Review Quiz
9. Which of the following accurately describes
the critical region for a hypothesis test?
Review Quiz
10. A researcher uses a hypothesis test to
evaluate H0: µ = 80. Which combination of
factors is most likely to result in rejecting the
null hypothesis?
A
• σ = 5 and n = 25
B
• σ = 5 and n = 50
C
• σ = 10 and n = 25
D • σ = 10 and n = 50
Review Quiz
10. A researcher uses a hypothesis test to
evaluate H0: µ = 80. Which combination of
factors is most likely to result in rejecting the
null hypothesis?
A
• σ = 5 and n = 25
B
• σ = 5 and n = 50
C
• σ = 10 and n = 25
D • σ = 10 and n = 50
Review Quiz
10. A researcher uses a hypothesis test to
evaluate H0: µ = 80. Which combination of
factors is most likely to result in rejecting the
null hypothesis?
B
• σ = 5 and n = 50
𝜎𝑀 =
𝜎
𝑛
• We want a low SE( why?)
• Less variation between sample means
• More likely to detect a true effect
• Large n decreases SE (denominator)
• Small 𝜎 decreases SE (numerator)
Homework & Reading
Chapter 8 Homework is due FRIDAY, 10/6.
Practice Exam 2 is available.
Exam 2 will open today after class and close
SUNDAY, 10/8.
Extra Practice Answers
• Answers for the extra practice questions on
the following slide
Review Quiz
1. USFSP has four fraternities. If the selection of
“Best Fraternity” is completely random (it’s not, but
let’s assume it is), how likely is it that the same
fraternity would be selected 3 times in a row?
A
B
C
D
• 1.56%
• 15.6%
• 25%
• 4%
Review Quiz
2. USFSP has four fraternities. If the selection of
“Best Fraternity” is completely random (it’s not, but
let’s assume it is), how likely is it that the same
fraternity would be selected 3 times in a row?
probability = p(A)
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑐𝑙𝑎𝑠𝑠𝑖𝑓𝑖𝑒𝑑 𝐴
=
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
probability = p(A and A and A) = p(A) * p(A) * p(A)
Review Quiz
2. USFSP has four fraternities. If the selection of
“Best Fraternity” is completely random (it’s not, but
let’s assume it is), how likely is it that the same
fraternity would be selected 3 times in a row?
probability = p(A and A and A) = p(A) * p(A) * p(A)
= ¼ * ¼ * ¼ = 0.0156
Review Quiz
2. USFSP has four fraternities. If the selection of
“Best Fraternity” is completely random (it’s not, but
let’s assume it is), how likely is it that the same
fraternity would be selected 3 times in a row?
A
B
C
D
• 1.56%
• 15.6%
• 25%
• 4%
Review Quiz
2. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
B
• What percent of people consume less than 18
teaspoons of sugar a day?
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
B
• What percent of people consume less than 18
teaspoons of sugar a day?
𝑋−𝜇
𝑧=
𝜎
X= 18
µ=22.7
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
B
• What percent of people consume less than 18
teaspoons of sugar a day?
𝑧=
𝑋−𝜇 18−22.7
=
𝜎
4.5
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
B
• What percent of people consume less than 18
teaspoons of sugar a day?
𝑧=
𝑋−𝜇 18−22.7
=
=-1.04
𝜎
4.5
Review Quiz
1. According to a recent report, the average American
consumes 22.7 teaspoons of sugar each day (Cohen,
August 2013). Assuming that the distribution is
approximately normal with a standard deviation of σ=4.5,
find each of the following values.
B
• What percent of people consume less than 18
teaspoons of sugar a day?
𝑧 = -1.04
P( z < -1.04)= 0.1492
= 14.92%
X=-1.04
µ=0
Review Quiz
3. True or False.
Review Quiz
3. True or False.
T/F
• Increasing the sample size will also
increase the effect size
Review Quiz
3. True or False.
False
• Sample size does not affect
Cohen’s d
𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝜇𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 𝜇 𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝐶𝑜ℎ𝑒𝑛 𝑠 𝑑 =
=
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝜎
′