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