ANALYSIS OF VARIANCE

Question 10

A child psychologist treats four children who are afraid of snakes with a behavioral modification procedure called systematic desensitization. In this procedure, children were slowly introduced to a snake over four treatment sessions. Children rated how fearful they are of the snake before the first session (baseline) and following each treatment session. Higher ratings indicated greater fear. The hypothetical data are listed in the table.

	Sessions
Baseline	1	2	3	4
7	7	5	4	3
7	6	6	4	4
6	6	7	7	3
7	7	5	4	3

1. Complete the F-table. (Round your value for F to two decimal places)

 

Considering the given data, the calculations are done in the following table:

Sessions
Baseline	1	2	3	4	Sum	Sum of squares	Count
7	7	5	4	3	19	99	4	90.25
7	6	6	4	4	20	104	4	100
6	6	7	7	3	23	143	4	132.5
7	7	5	4	3	19	13	4	90.25
Sum	26	23	19	13
Sum of squares	170	135	97	43
Count	4	4	4	4
Sum2/count	262/4˭169	232/4˭132.5	192/4˭90.25	132/4˭42.5

 

For the given data, sum of all observations is G ˭ 81

There are 16 observations, i.e. N ˭ 16

Thus, raw sum of squares ˭ 445

SS_{T ˭} ∑^{X2- G2}/2 ˭ 445- 81²/2 ˭ 34.93

Therefore,

Sum of squares between groups (sessions) SS_G ( 26²/2  23²/2  19²/4 13²/4) – (81²/16)

˭ 433. 75 – 410.0625 ˭ 23.6875

Sum of Squares between persons (Baseline), SSP: ( 19²/4  20²/4  23²/4 19²/4) – (81²/16)

˭ 412.75 – 410.0625

˭ 2.6875

SS_E ˭ SS_T – SS_G – SS_P ˭ 34.93 – 23. 6875- 2.6875

˭ 8.555

There are four baselines and four sessions, therefore degrees of freedom of SSG and SSP are (4-1) ˭ 3

Degree of freedom for total ˭ N- 1 ˭ 16-1 ˭ 15

Degrees of freedom for error ˭ 15-3-3 ˭ 9

The mean sum of squares can be calculated by dividing the sum of squares by the corresponding degrees of freedom.

˭  ˭ 8.3065

˭  ˭ 0.9424

From this, the completed ANOVA table is as follows:

Sources of variation	SS	Df	MS	F
Between groups	23.69	3	7.90	8.306546
Between persons	2.69	3	0.90	0.942431
Error	8.56	9	0.95
Total	34.93	15

 

1. Compute a Bonferroni procedure and interpret the results. (Assume experiment wise alpha equal to 0.05.)

Ratings of fear significantly decreased from baseline to Session 3. Ratings of fear also significantly decreased from Session 1 to Session 4.

Ratings of fear significantly decreased from baseline to Session 3. Ratings of fear also significantly decreased from baseline to Session 4.

None of the pairwise comparisons are significant.

Ratings of fear significantly decreased from baseline to Session 4.

Ratings of fear significantly decreased from Session 1 to Session 4. Ratings of fear also significantly decreased from Session 2 to Session 4.

Question 11

A study investigated the effects of physical fatigue on the performance of professional tennis players. Researchers measured the number of unforced errors committed by a random sample of 12 professional tennis players during the first three sets of a match. They hypothesized that increased fatigue would be associated with a greater number of errors. The following is an F-table for this hypothetical study using the one-way within-subjects ANOVA.

Source of Variation	SS	df	MS	F
Between groups	36	2	13	Missing answer here
Between persons	Missing answer here	12	3
Within groups (error)	55	24	2.29
Total	Missing answer here	Missing answer here

 

The information given is related to study the effect of physical fatigue on the performance of professional tennis players from the given information numbers of players selected at random during the first three sets of a match is 12. Hence, the following is the output of F-table for this hypothetical study using the one-way with-in subjects Anova.

Degrees of freedom

Df_{between groups =k -1 =3-1 is 2}

Df_{between persons = h-1 which is 13-1 equals 12}

Df_{within = (k-1)(h-1) = 2 by 12 = 24,}

df_{total = (13by 3) equals (39-1) = 38}

Mean sum of squares

MS_{between groups} 18, MS _{between persons} 3.27, MS_{within groups}

            Thus, the ANOVA table is filled above

 

1. Make a decision to retain or reject the null hypothesis. (Assume alpha equal to 0.05.)

The F ratio is greater the 0.05 in all cases thus, we reject the null hypothesis.

 

2. Estimate effect size using partial omega-squared: ω_P². (Round your answer to two decimal places)

0.00693

 

Question 12

Air traffic controllers perform the vital function of regulating the traffic of passenger planes. Frequently, air traffic controllers work long hours with little sleeps. Researchers wanted to test their ability to make basic decisions as they become increasingly sleep deprived. To test their abilities, a sample of 6 air traffic controllers is selected and given a decision-making skills test following 12-hour, 24-hour, and 48-hour sleep deprivation. Higher scores indicate better decision-making skills. The table lists the hypothetical results of this study.

Sleep Deprivation
12 Hours	24 Hours	48 Hours
22	18	16
19	21	21
33	23	22
26	20	13
22	15	15
21	21	15

1. Complete the F-table. (Round your answers to two decimal places)

H₀ : µ₁ = µ₂ =µ3

H_a : at least two µ_i differ

X1	X12	X2	X22	X3	X32
22	484	18	324	16	256
19	361	21	441	21	441
33	1089	23	529	22	484
26	676	20	400	13	169
22	484	15	225	15	225
21	441	21	441	15	225
∑x 1 143	∑x12 3535	∑x2118	∑x22 2360	∑X3102	∑X32 1800

 

            k  3, n

           

                         7695- 7320.5 374.5

           

                         7462.79 – 7320.50 142.29

                       

                                     374- 142. 29 231.71

                        _{Significance level is 0.05}

                               

                                 4.61

ANOVA

Source	SS	Df	MS	F
Among	142. 29	2	71.15	4.61
Within	231.71	15	Missing answer here
Total	374	17

(a) Complete the F-table. (Round your answers to two decimal places.)

Source of Variation	SS	df	MS	Fobt
Between groups	1	2	3	4
Between persons	5	6	7
Within groups (error)	8	9	10
Total	11	12

Parts are missing and rest are wrong I put in the above table what was correct!

Here P value > alpha 0.05 for Hours, we fail to reject null hypothesis

Thus we conclude that there is no significance difference among the means of 12hours, 24 hours and 48 hours

and P value > alpha 0.05 for Person, we fail to reject H₀.

Thus we conclude that there is no significance difference among the means of persons

1. b) Bonferroni correction simply divides the experimental-wise error rate by the number of orthogonal contrasts.

There is a significant difference in decision making for the 12-hour and 24-hour sleep deprivation conditions.

There is a significant difference in decision making for the 24-hour and 48-hour sleep deprivation conditions.

There is a significant difference in decision making for the 12-hour and 48-hour sleep deprivation conditions.

There are no significant differences between any of the groups.

 

 

Question 13

Some studies show that people who think they are intoxicated will show signs of intoxication, even if they did not consume alcohol. To test whether this is true, researchers had a group of five adults consume nonalcoholic drinks, which they were told contained alcohol. The participants completed a standard driving test before drinking and then after one nonalcoholic drink and after five nonalcoholic drinks. A standard driving test was conducted in a school parking lot where the participants had to maneuver through traffic cones. The number of cones knocked over during each test was recorded. The following table lists the data for this hypothetical study

 

 

Driving Test
Before Drinking	After One Drink	After Five Drinks
0	0	4
0	1	3
1	2	3
3	3	5
0	1	0

 

1. Complete the F-table. (Round your answers to two decimal places)

ANOVA
cone_knocked
	Sum of Squares	df	Mean Square	F	Sig.
Between Groups	12.93	2	6.47	2.99	.089
Missing Between Persons Within Groups	Missing answer   26.00	Missing answer	Missing answer     2.17
Total	38.93	14

 

 

1. Compute a Bonferroni procedure and interpret the results. (Assume experimentwise alpha equal to 0.05. Select all that apply.)

There were no significant differences between any of the groups.

Students knocked over significantly more cones after 1 nonalcoholic drink compared with the driving test prior to drinking.

Students knocked over significantly more cones after 5 nonalcoholic drinks compared with the driving test prior to drinking.

Students knocked over significantly more cones after 5 nonalcoholic drinks compared with the driving test after 1 nonalcoholic drink.

Question 14

Wilfley and colleagues (2008) tested whether the antiobesity drug sibutramine would be an effective treatment for people with binge eating disorder. They measured the frequency of binge eating every 2 weeks for 24 weeks during treatment. The following table lists a portion of the data similar to results reported by the authors for the frequency of binge eating over the first 8 weeks of the drug treatment.

Frequency of Binge Eating
Baseline	Week 2	Week 4	Week 6	Week 8
4	1	0	0	1
6	4	2	0	0
3	0	1	1	0
1	1	0	1	1
2	2	1	1	1
5	1	2	2	2

 

1. Complete the F-table. (Round your answers to two decimal places)

Source of variation	SS	Df	MS	F	p
Between groups	4	30.80	7.70000	6.31	0.002
Between persons	5	12.2667	2.45333	2.01	0.121
Error	20	24.4000	1.22000
Total	29	67.4667

 

Thus, S = 1.105 R-Sq = 63.83% R-Sq(adj) = 47.56%

Conclusion: Between groups: The calculated p-value of group effect is 0.002 and less than 0.05 level of significant. Hence, there is significant different of frequency of binge eating between the weeks.

Between the Persons: The calculated p-value is 0.121 and larger than 0.05 level of significant. Hence, there is no significant different of frequency of binge eating between the persons.

1. Use the Bonferroni procedure to make the post hoc test. In which week do we first see significant differences compared to baseline?

Week 2 is the first week where significant differences from baseline are evident.

Week 4 is the first week where significant differences from baseline are evident.

Week 6 is the first week where significant differences from baseline are evident.

Week 8 is the first week where significant differences from baseline are evident.

None of the weeks are significantly different from the baseline.

Question 15

A psychotherapist asks a sample of violent and nonviolent criminals from impoverished, middle-class, and upper-class backgrounds to indicate how accountable they feel they are for their crimes. Identify each factor and the levels of each factor in this example. (Select all that apply.)

Accountability (two levels: low accountability, high accountability)

Type of crime (three levels: infraction, misdemeanor, felony)

Type of crime (two levels: violent, nonviolent)

Accountability (three levels: negative accountability, neutral accountability, positive accountability)

Social background (three levels: impoverished, middle-class, upper-class)

Social background (four levels: impoverished, middle-class, psychotherapists, upper-class)

Question 16

Seasonal affective disorder (SAD) is a type of depression during seasons with less daylight (e.g., winter months). One therapy for SAD is phototherapy, which is increased exposure to light used to improve mood. A researcher tests this therapy by exposing a sample of SAD patients to different intensities of light (low, medium, high) in a light box, either in the morning or at night (these are the times thought to be most effective for light therapy). All participants rated their mood following this therapy on a scale from 1 (poor mood) to 9 (improved mood). The hypothetical results are given in the following table

		Light Intensity
		Low	Medium	High
Time of Day	Morning	5	5	7
		6	6	8
		4	4	6
		7	7	9
		5	9	5
		6	8	8
	Night	4	6	9
		8	8	7
		6	7	6
		7	5	8
		4	9	7
		3	8	6

1. Complete the F-table and make a decision to retain or reject the null hypothesis for each hypothesis test. (Round your answers to two decimal places. Assume experimentwise alpha equal to 0.05)

 

Two factor Anova
			Intensity
	Means
		Low	Medium	High
Time of the day	Morning	5.5	6.5	7.2	6.4
	Night	5.5	7.2	7.2	6.6
		5.5	6.8	7.2	6.5
	6	Replications per cell

 

 

 

 

 

 

ANOVA

Source	SS	Df	MS	F	P-Value
Time of the day	0.44	1	0.44	0.19	.6634
Intensity	18.67	2	9.33	4.06	.0276
Interactions	0.89	2	0.44	0.19	.8253
Error	69.00	30	2.30
Total	89.00	35

 

 

Post-Hoc Analysis: Tukey simultaneous comparison t-values (d.f. = 30)

		Low	Medium	High
		5.5	6.8	7.2
Low	5.5
Medium	6.8	2.15
High	7.2	2.69	0.54
critical values for experimentwise error rate:

 

Tukey test shows that at 0.05 level of significance, low intensity is significantly different from high intensity. Not all other pairs are significant

 

(b) Compute Tukey’s HSD to analyze the significant main effect.

The critical value is 21for each pairwise comparison

 

 

Summarize the results for this test using APA format

State the decision for the main effect of the time of day.

Calculated F=0.19, P=0.6634 which is > 0.05 level.

Retain the null hypothesis.

State the decision for the main effect of intensity.

Calculated F=4.06, P=0.0276 which is < 0.05 level.

Reject the null hypothesis.

State the decision for the interaction effect.

Calculated F=0.19, P=0.8253 which is > 0.05 level.

Retain the null hypothesis.

 

 

Question 17

To test the relationship between gender and ratings of a promiscuous partner, a group of men and women was given a vignette describing a person of the opposite sex who was in a dating relationship with one, two, or three partners. Participants rated how positively they felt about the individual described in the vignette, with higher ratings indicating feelings that are more positive.

Source of Variation	SS	df	MS	F
Gender	15
Promiscuity
Gender × Promiscuity	144
Error	570	114
Total	819

1. Complete the F-table and make a decision to retain or reject the null hypothesis for each hypothesis test. (Assume experiment wise alpha equal to 0.05.)

 

Source of variation	SS	Df	MS	F	F at 5% level	Decision
Gender	15	1	15	3	3.9243	Retain H0
Promiscuity	90	2	45	15	3.0758	Reject H0
G× P	144	2	72	14.4	3.0758	Reject H0
Error	570	114	5
Total	819	119

 

 

.

 

 

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