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: 𝜎
𝑀
= 𝜎 𝑛
= 𝜎 2
= 𝑛 𝜎 2 𝑛
𝑀−𝜇 z-Score Formula: 𝑧 = 𝜎
𝑀
Cohen’s d : 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
= 𝜇 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
− 𝜇 𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 𝜎 estimated Cohen’s d : 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑀 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
− 𝜇 𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝜎
=

Logic of Hypothesis Testing
 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
 Question 1 Answers:
 For a two-tailed test:
 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:
 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.

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Alpha Level and the Critical
Region
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
 Question 2 Answer:
 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
 Question 3: Define Type 1 and Type II Error.

Possible Outcomes of a
Hypothesis Test
 Question 3 Answer:
 Type I error is rejecting a true null hypothesis – that is, saying that treatment has an effect when, in fact, it doesn’t.
 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.
 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
 Question 4 Answer:
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Step 1: State hypotheses
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H
0
: Texting has no effect on driving. (µ = 10.5)
H
1
: Texting has an effect on driving. (µ ≠ 10.5)
Step 2: Set Criteria for Decision ( α = 0.01) z = ± 2.58
Reject H
0
Reject H
0 z = - 2.58
z = 2.58

Two-Tailed Hypothesis Test
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Question 4 Answer:
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Step 3: Compute sample statistic 𝜎
𝑀
= 𝑧 = 𝜎 𝑛
=
𝑀−𝜇
= 𝜎
𝑀
4.8
16
=
4.8
4
= 1.20
15.9−10.5
=
1.20
5.4
1.20
= 4.50

Two-Tailed Hypothesis Test
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Question 4 Answer
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Step 4: Make a decision
For a Two-tailed Test:
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If -2.58 < z sample
If z sample
< 2.58, fail to reject H
0
≤ -2.58 or z sample
≥ 2.58, reject H
0 z sample
(4.50) > z critical
(2.58)
Thus, we reject the null and note that texting has a significant effect on driving.

Factors that Influence a
Hypothesis Test
 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
 Question 5 Answer:
 True. A larger sample produces a smaller standard error, which leads to a larger z-score.
For 𝑧 =
𝑀−𝜇
, where 𝜎 𝜎
𝑀
𝑀
= 𝜎 𝑛
, as sample size ( n ) increases, standard error ( 𝜎
𝑀
) decreases, which then increases z.
Consequently, as z increases so does the probability of rejecting the null hypothesis.

Factors that Influence a
Hypothesis Test
 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
 Question 6 answer:
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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.
𝜎
𝑀
= 𝜎
𝑀
= 𝜎
= 𝜎 𝑛 𝑛
=
10
=
25
20
25
=
10
= 2
5
20
= 4
5

One-tailed Hypothesis Test
 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
 Question 7 Answer:
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Step 1: State hypotheses
 H
0
: Endurance scores are not significantly higher with the sports drink. ( µ ≤ 50)
 H
1
: Endurance scores are significantly higher with the sports drink.
(µ > 50)
Step 2: Set Criteria for Decision ( α = 0.05) z = 1.65
Reject H
0 z = 1.65

One-tailed Hypothesis Test
 Question 7 Answer:
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Step 3: Compute sample statistic 𝜎
𝑀
= 𝑧 = 𝜎 𝑛
=
𝑀−𝜇
= 𝜎
𝑀
10
50
=
10
7.07
= 1.41
53−50
=
1.41
3
1.41
= 2.13

One-tailed Hypothesis Test
 Question 7 Answer:
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Step 4: Make a decision
For a One-tailed Test:
If z sample
If z sample
≤ 1.65, fail to reject H
0
> 1.65, reject H
0
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 z sample
(2.13) > z critical
(1.65)
Thus, we reject the null and note that the sports drink does raise endurance scores.

Effect Size and Cohen’s d
 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
 Question 8 Answer:
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 estimated Cohen’s d : 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 d =
47−40
8
=
7
8
= 0.875
=
This is a large effect.
𝑀 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
− 𝜇 𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 𝜎
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
 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
 Question 9 Answer:
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Step #1: Calculate standard error for sample
 𝜎
𝑀
= 𝜎 𝑛
=
20
25
=
20
= 4
5
Step #2: Locate Boundary of Critical Region 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.
 Step #3: Calculate the z-score
 𝑧 =
𝑀−𝜇
= 𝜎
𝑀
107.84 −110
=
4
−2.16
= −0.54
4

Computing Power
 Step #4: Interpret Power of the Hypothesis Test
 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
 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
 Question 10 Answer:
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Step #1: Calculate standard error for sample
 𝜎
𝑀
= 𝜎 𝑛
=
20
25
=
20
= 4
5
Step #2: Locate Boundary of Critical Region 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.
 Step #3: Calculate the z-score
 𝑧 =
𝑀−𝜇
= 𝜎
𝑀
87.84 −92
=
4
−4.16
= −1.04
4

Computing Power
 Question 10 Answer:
 Step #4: Interpret Power of the Hypothesis Test
 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.

Frequently Asked Questions
FAQs
 What is power?
 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.
 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.

Frequently Asked Questions
FAQs
 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?

Frequently Asked Questions
FAQs
 Step #1: Calculate standard error for sample
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In this step, we work from the population’s standard deviation ( σ) and the sample size (n)

Frequently Asked Questions
FAQs
 Step #2: Locate Boundary of Critical Region
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In this step, we find the exact boundary of the critical region
Pick a critical z-score based upon alpha ( α =.05
)

Frequently Asked Questions
FAQs
 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).

Frequently Asked Questions
FAQs
 Interpret Power of the Hypothesis Test
 Find probability associated with a z-score > - 2.04
 Look this probability up as the proportion in the body of the normal distribution (column B in your textbook)
 p = .9793
 Thus, with a sample of 25 people and an 8-point treatment effect, 97.93% of the time the hypothesis test will conclude that there is a significant effect.