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PSY 216 Assignment 12 Answers
1. Problem 1 from the text
Explain why the F-ratio is expected to be near 1.00 when the null hypothesis is true.
When H0 is true, the treatment had no systematic effect. In the formula for F:
F = (systematic treatment effects + random, unsystematic difference) / random, unsystematic
differences
The systematic treatment effects term will be 0 if the null hypothesis is true. Thus, when H0 is true, F
will equal
F = (0 + random, unsystematic differences) / random, unsystematic differences
Any value divided by itself equals 1.
2. Problem 3 from the text
Several factors influence the size of the F-ratio. For each of the following, indicate whether it would
influence the numerator or the denominator of the F-ratio, and indicate whether the size of the F-ratio
would increase or decrease.
a. Increase the differences between the sample means.
This affects the numerator of the F-ratio. As the sample means become more different, the
treatment has a larger and larger effect. The systematic effect of the treatment is in the
numerator of the F-ratio. The size of the F-ratio would increase.
b. Increase the size of the sample variance.
This affects both the numerator and denominator of the F-ratio. Sample variance is the
random, unsystematic change in the scores. Increasing sample variance will decrease the
size of the F-ratio
3. Problem 7 from the text
The following data summarize the results from an independent-measures study comparing three
treatment conditions: (Problem 7 from the text)
I II III n = 6 n = 6 n = 6 M = 1 M = 5 M = 6 N = 18 T = 6 T = 30 T = 36 G = 72 SS = 30 SS =35 SS = 40 ΣX2 = 477
a. Use an ANOVA with α = .05 to determine whether there are any significant differences
among the three treatment conditions.
Step 1: State the hypotheses and α
H0: μ1 = μ2 = μ3
H1: not H0 (or, at least one of the treatment means is different)
α = .05
Step 2: Locate the critical region
dfBetween Treatments = k (number of conditions) – 1 = 3 – 1 = 2
dfWithin Treatments = N – k = 18 – 3 = 15
Consult a table of critical F values with dfBetween Treatments = 2, dfWithin Treatments = 15, and α =
.05. The critical value of F = 3.682.
Step 3: Perform the ANOVA
𝑆𝑆𝑇𝑜𝑡𝑎𝑙 = ∑ 𝑋2 −𝐺2
𝑁= 477 −
722
18= 189
𝑆𝑆𝑊𝑖𝑡ℎ𝑖𝑛 = ∑ 𝑆𝑆𝐼𝑛𝑠𝑖𝑑𝑒 𝑒𝑎𝑐ℎ 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 = 30 + 35 + 40 = 105
𝑆𝑆𝐵𝑒𝑡𝑤𝑒𝑒𝑛 = ∑𝑇2
𝑛−
𝐺2
𝑁=
62
6+
302
6+
362
6−
722
18= 6 + 150 + 216 − 288 = 84
dfTotal = N – 1 = 18 – 1 = 17
dfBetween = 2 (from step 2)
dfWithin = 15 (from step 2)
MSBetween = SSBetween / dfBetween = 84 / 2 = 42
MSWithin = SSWithin / dfWithin = 105 / 15 = 7
F = MSBetween / MSWithin = 42 / 7 = 6
Source SS df MS F Between Treatments 84 2 42 6 Within Treatments 105 15 7 Total 189 17
Step 4: Make a decision
Because the calculated F is in the tail cut off by the critical F, we reject H0 and conclude that
the treatment likely had an effect.
b. Calculate η2 to measure the effect size for this study.
η2 = SSBetween Treatment / SSTotal = 84 / 189 = 0.444
Because η2 is larger than 0.25, this is a large effect.
c. Perform multiple comparisons if appropriate.
Because we rejected H0 and there were more than two conditions, it is appropriate to
perform multiple comparisons. In this case, I am calculating the Tukey’s Honestly
Significant Differences.
𝐻𝑆𝐷 = 𝑞√𝑀𝑆𝑊𝑖𝑡ℎ𝑖𝑛
𝑛= 3.675√
7
6= 3.969
The value of q comes from a table of critical values of the Studentized Range with α = .05,
three groups (number of conditions) and dfWithin = 15.
Any pair of means that are at least 3.969 different are reliably different by Tukey’s HSD
test. Conditions 1 (M = 1) and 2 (M = 5) are reliably different as are conditions 1 and 3 (M =
6). However, conditions 2 and 3 are not reliably different from each other.
d. Write a sentence demonstrating how a research report would present the results of the
hypothesis test and the measure of effect size.
The means and standard deviations of the three conditions can be found in Table 1. The
analysis of variance revealed an effect of the treatment, F(2, 15) = 6.00, p < .05, α = .05, η2 =
.444. According to Tukey’s HSD, conditions 1 and 2, and 1 and 3 are reliably different,
while conditions 2 and 3 are not reliably different from each other.
Table 1
Means and standard deviations of the conditions
Condition 1 Condition 2 Condition 3 M 1 5 6 s 2.45 2.65 2.68
4. Problem 15 from the text
The following summary table presents the results from an ANOVA comparing three treatment
conditions with n = 8 participants in each condition. Complete all missing values. (Hint: Start with the
df column.)
Source SS df MS F Between Treatments
__30 = 2 * 15__
__2 = 3 – 1__
15
__5 = 15 / 3__
Within Treatments
__63 = 93 – 30__
__21 = (8 – 1) * 3__
__3 = 63 / 21__
Total
93
__23 = 8 * 3 – 1__
There are different ways that you could solve this. Here is one possible way:
1. dfTotal = number of scores – 1 = (8 participants per condition X 3 conditions) – 1 = 24 – 1 = 23
2. dfBetween Treatments = number of conditions – 1 = 3 – 1 = 2
3. dfWithin Treatments = dfTotal – dfBetween Treatments = 23 – 2 = 21.
dfWithin Treatments = Σ(number of people in each condition - 1) = (8 – 1) + (8 – 1) + (8 – 1) = 21
(Sum across the conditions)
4. SSBetween Treatments = MSBetween Treatments X dfBetween Treatments (an algebraic rearrangement of MS = SS / df)
SSBetween Treatments = 15 X 2 = 30
5. SSWithin Treatments = SSTotal – SSBetween Treatments (an algebraic rearrangement of SSTotal = SSBetween Treatments
+ SSWithin Treatments)
SSWithin Treatments = 93 – 30 = 63
6. MSWithin Treatments = SSWithin Treatments / df Within Treatments = 63 / 21 = 3
7. F = MSBetween Treatments / MSWithin Treatments = 15 / 3 = 5
5. Problem 23 from the text
New research suggests that watching television, especially medical shows such as Grey’s Anatomy and
House can result in more concern about personal health (Ye, 2010). Surveys administered to college
students measure television viewing habits and health concerns such as fear of developing the
diseases and disorders seen on television. For the following data, students are classified into three
categories based on their television viewing patterns and health concerns are measured on a 10-point
scale with 0 indicating “none.”
Television Viewing Little or none Moderate Substantial 4 2 5 1 3 7 4 4 8 2
5 7 3 4 8 6 2 7 3 5
5 7 6 6 8 9 6 4 6 8
N = 30 G = 155 ΣX2 = 933
n1 = 10 T1 = 40 SS1 = 44
n2 = 10 T2 = 50 SS2 = 36
n3 = 10 T3 = 65 SS3 = 20.5
T1 = 4 + 2 + 5 + 1 + 3 + 7 + 4 + 4 + 8 + 2 = 40
SS1 = 42 + 22 + 52 + 12 + 32 + 72 + 42 + 42 + 82 + 22 - 402 / 10 = 204 – 160 = 44
T2 = 5 + 7 + 3 + 4 + 8 + 6 + 2 + 7 + 3 + 5 = 50
SS2 = 52 + 72 + 32 + 42 + 82 + 62 + 22 + 72 + 32 + 52 - 502 / 10 = 286 – 250 = 36
T3 = 5 + 7 + 6 + 6 + 8 + 9 + 6 + 4 + 6 + 8 = 65
SS3 = 52 + 72 + 62 + 62 + 82 + 92 + 62 + 42 + 62 + 82 – 652 / 10 = 443 – 422.5 = 20.5
N = n1 + n2 + n3 = 10 + 10 + 10 = 30
G = T1 + T2 + T3 = 40 + 50 + 65 = 155
ΣX2 = 42 + 22 + 52 + 12 + 32 + 72 + 42 + 42 + 82 + 22 + 52 + 72 + 32 + 42 + 82 + 62 + 22 + 72 + 32 + 52 + 52 +
72 + 62 + 62 + 82 + 92 + 62 + 42 + 62 + 82 = 933
a. Use an ANOVA with α = .05 to determine whether there are significant mean differences
among the three groups.
Step 1: State the hypotheses and α
H0: μ1 = μ2 = μ3
H1: not H0 (or, at least one of the treatment means is different)
α = .05
Step 2: Locate the critical region
dfBetween Treatments = k (number of conditions) – 1 = 3 – 1 = 2
dfWithin Treatments = N – k = 30 – 3 = 27
Consult a table of critical F values with dfBetween Treatments = 2, dfWithin Treatments = 27, and α =
.05. The critical value of F = 3.354.
Step 3: Perform the ANOVA
𝑆𝑆𝑇𝑜𝑡𝑎𝑙 = ∑ 𝑋2 −𝐺2
𝑁= 933 −
1552
30= 132.167
𝑆𝑆𝑊𝑖𝑡ℎ𝑖𝑛 = ∑ 𝑆𝑆𝐼𝑛𝑠𝑖𝑑𝑒 𝑒𝑎𝑐ℎ 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 = 44 + 36 + 20.5 = 100.5
𝑆𝑆𝐵𝑒𝑡𝑤𝑒𝑒𝑛 = ∑𝑇2
𝑛−
𝐺2
𝑁=
402
10+
502
10+
652
10−
1552
30= 160 + 250 + 422.5 − 800.833
= 31.667
dfTotal = N – 1 = 30 – 1 = 29
dfBetween = 2 (from step 2)
dfWithin = 27 (from step 2)
MSBetween = SSBetween / dfBetween = 31.667 / 2 = 15.833
MSWithin = SSWithin / dfWithin = 100.5 / 27 = 3.722
F = MSBetween / MSWithin = 15.833 / 3.722 = 4.254
Step 4: Make a decision
Because the calculated F is in the tail cut off by the critical F, we can reject H0 and conclude
that watching TV has an effect on health concerns.
b. Compute η2 to measure the size of the effect.
η2 = SSBetween Treatment / SSTotal = 31.667 / 132.167 = 0.240
Because η2 is between 0.09 and 0.25, this is a medium effect.
c. Use Tukey’s HSD test with α = .05 to determine which groups are significantly different.
𝐻𝑆𝐷 = 𝑞√𝑀𝑆𝑊𝑖𝑡ℎ𝑖𝑛
𝑛= 3.508√
3.722
10= 2.140
The value of q comes from a table of critical values of the Studentized Range with α = .05,
three groups (number of conditions) and dfWithin = 27.
Any pair of means that are at least 2.140 different are reliably different by Tukey’s HSD
test. Only conditions 1 (M = 4) and 3 (M = 6.5) are reliably different.
8. Enter the data from problem 21 into SPSS. Use SPSS to answer the following question: Do the data
indicate any significant mean differences among the three groups of birds? Give H0, H1 and α. Is this a
one-tailed or two-tailed test? Write a sentence or two in APA format that summarizes the results of
the analysis.
One possible explanation for why some birds migrate and others maintain year round residency in a
single location is intelligence. Specifically, birds with small brains, relative to their body size, are
simply not smart enough to find food during the winter and must migrate to warmer climates where
food is easily available (Sol, Lefebvre, & Rodriguez-Teijeiro, 2005). Birds with bigger brains, on the
other hand, are more creative and can find food even when the weather turns harsh. Following are
hypothetical data similar to the actual research results. The numbers represent relative brain size for
the individual birds in each sample:
Non-Migrating
Short-Distance Migrants
Long-Distance Migrants
18 13 19 12 16 12
6 11 7 9 8
13
4 9 5 6 5 7
Step 1:
H0: μNon-migratingd = μShort-distance migrant = μLong-distance migrant
H1: not H0
α = .05
Step 2:
Skip – use p value from SPSS output
Two Tailed – ANOVA is always two tailed
a. Step 3:
In SPSS create two variables – one for the IV and one for the DV. – In variable view (Ctrl-T, click on
the Variable View tab in the lower left corner, or click on View | Variables), enter migratory in row
one and brainsize in row two of the Name column.
Click the cell at the intersection of the “migratory” row and “Value” column.
A button with “…” should appear in the cell. Click on it.
Enter 1 in the value box and Non-Migrating in the label box. Click Add
Enter 2 in the value box and Short-Distance Migrants in the label box. Click Add
Enter 3 in the value box and Long-Distance Migrants in the label box. Click Add
Click OK
Switch to Data View (Ctrl-T, click on the Data View tab in the lower left corner, or click on View |
Data)
Enter the data from the table into SPSS. For each “non-migrating” bird, enter a 1 in the
“migratory” column and its brain size in the “brainsize” column. For each “short-distance
migrants” bird, enter a 2 in the “migratory” column and its brain size in the “brainsize” column.
For each “long-distance migrants” bird, enter a 3 in the “migratory” column and its brain size in
the “brainsize” column.
Click on Analyze | General Linear Model | Univariate…
Move “brainsize” into the Dependent Variables box
Move “migratory” into the Fixed Factor(s) box (the IV box)
Click the “Post Hoc…” button
Move “migratory” into the “Post Hoc Tests for:” box.
Click in the check box to the left of “Tukeys”
Click Continue
Click on “Options…”
Move “migratory” into the “Display Means for:” box
Click in the check box to the left of “Descriptive statistics” and “Estimates of effect size”
Click on “Continue”
Click on “OK”
Descriptive Statistics
Dependent Variable:BrainSize
MigrantType Mean Std. Deviation N
Non-Migrating 15.0000 3.09839 6
Short-Distance Migrants 9.0000 2.60768 6
Long-Distance Migrants 6.0000 1.78885 6
Total 10.0000 4.53743 18
Tests of Between-Subjects Effects
Dependent Variable:BrainSize
Source
Type III Sum of
Squares df Mean Square F Sig.
Partial Eta
Squared
Corrected Model 252.000a 2 126.000 19.286 .000 .720
Intercept 1800.000 1 1800.000 275.510 .000 .948
MigrantType 252.000 2 126.000 19.286 .000 .720
Error 98.000 15 6.533
Total 2150.000 18
Corrected Total 350.000 17
a. R Squared = .720 (Adjusted R Squared = .683)
Multiple Comparisons
BrainSize
Tukey HSD
(I) MigrantType (J) MigrantType
Mean
Difference (I-J) Std. Error Sig.
95% Confidence Interval
Lower Bound Upper Bound
Non-Migrating Short-Distance Migrants 6.0000* 1.47573 .003 2.1668 9.8332
Long-Distance Migrants 9.0000* 1.47573 .000 5.1668 12.8332
Short-Distance Migrants Non-Migrating -6.0000* 1.47573 .003 -9.8332 -2.1668
Long-Distance Migrants 3.0000 1.47573 .138 -.8332 6.8332
Long-Distance Migrants Non-Migrating -9.0000* 1.47573 .000 -12.8332 -5.1668
Short-Distance Migrants -3.0000 1.47573 .138 -6.8332 .8332
Based on observed means.
The error term is Mean Square(Error) = 6.533.
*. The mean difference is significant at the .05 level.
F(2, 15) = 19.29, p = .000, MSerror = 6.53, η2 =.72
Step 4:
Reject H0 because p = .000 which is less than α = .05
Multiple Comparisons:
Step 1:
H0: μNon = μShort
H1: μNon ≠ μShort
H0: μNon = μLong
H1: μNon ≠ μLong
H0: μShort = μLong
H1: μShort ≠ μLong
Step 2:
Skip – use p value from output
Step 3:
Skip
Step 4:
Non vs. short –reject H0 (p = .003)
Non vs. long – reject H0 (p = .000)
Short vs. long – fail to reject H0 (p = .138)
Summary:
The mean and standard deviations of brain size for the three groups of birds (non-migrating, short-
distance migrants and long-distance migrants) are shown in Table 1. The analysis of variance
revealed a significant effect of the type of bird on brain size, F(2, 15) = 19.286, p = .000, MSerror = 6.53,
η2 =.72. Tukey multiple comparisons revealed that non-migrating birds had a larger brain than both
short-distance migrants (p =.003) and long-distance migrants (p = .000). There was insufficient
evidence to suggest that the brain size of short- and long-distance migrants were reliably different (p
= .138).
Table 1
Brain Size of Different Types of Birds
Group M SD
Non-Migrating 15.00 3.10
Short-Distance Migrant 9.00 2.61
Long-Distance Migrant 6.00 1.79
Hand calculations:
Non-Migrating Short-Distance Migrants Long-Distance Migrants
18 13 19 12 16 12
6 11 7 9 8 13
4 9 5 6 5 7
ΣX = 90 ΣX2 = 1398 M = 15 SSNon = 1398 – 902 / 6 = 48
ΣX = 54 ΣX2 = 520 M = 9 SSShort = 520 – 542 / 6 = 34
ΣX = 36 ΣX2 = 232 M = 6 SSLong = 232 – 362 / 6 = 16
ΣX = 90 + 54 + 36 = 180 ΣX2=1398+520+232=2150 M = 10 SSTotal = 2150 – 1802 / 6 = 350
SSBetween-Treatments = n · SSM
SSM = ΣM2 – (ΣM)2 / k
ΣM = 15 + 9 + 6 = 30
ΣM2 = 152 + 92 + 62 = 342
SSM = 342 – 302 / 3 = 42
SSBetween-Treatments = 6 · 42 = 252
SSWithin-Treatments = ΣSSfor each group = 48 + 34 + 16 = 98
SSTotal = 350
Check: SSTotal = SSBetween-Treatments + SSWithin-Treatments = 252 + 98 + 350
dfBetween-Treatments = k – 1 = 3 – 1 = 2
dfWithin-Treatments = Σ(n – 1) = (6 – 1) + (6 – 1) + (6 – 1) = 15
dfTotal = N - 1 = 18 – 1 = 17
Check: dfTotal = dfBetween-Treatments + dfWithin-Treatments = 2 + 15 = 17
MSBetween-Treatments = SSBetween-Treatments / dfBetween-Treatments = 252 / 2 = 126
MSWithin-Treatments = SSWithin-Treatments / dfWithin-Treatments = 98 / 15 = 6.53
F = MSBetween-Treatments / MSWithin-Treatments = 126 / 6.53 = 19.29
η2 = SSBetween-Treatments / SSTotal = 252 / 350 = .72
HSD = q ∙ √MSWithin−Treatment
n= 3.675 ∙ √
6.53
6= 3.83
| MNon-Migrating – MShort-Distance Migrants | > HSD → Reject H0: μNon-Migrating = μShort-Distance Migrants
| MNon-Migrating – MLong-Distance Migrants | > HSD → Reject H0: μNon-Migrating = μLong-Distance Migrants
| MShort-Distance Migrants - MLong-Distance Migrants | < HSD → Fail to reject H0: μShort-Distance Migrants = μLong-Distance-Migrants