Psy 520 topic 2 exercises - Study guides, Class notes & Summaries
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PSY 520 TOPIC 3 EXERCISE, CHAPTER 7 AND 8 ANSWERS
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PSY 520 Topic 3 Exercise Chapter 7 and 8 
 
6.7 (a) Estimate whether the following pairs of scores for X and Y reflect a positive relationship, a negative relationship, or no relationship. Hint: Note any tendency for pairs of X and Y scores to occupy similar or dissimilar relative locations. 
 
(b) Construct a scatterplot for X and Y . Verify that the scatterplot does not describe a pronounced curvilinear trend. 
 
(c) Calculate r using the computation formula (6.1). 
 
6.10 On the basis of an ...
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PSY 520 TOPIC 3 EXERCISE, CHAPTER 7 AND 8 ANSWERS
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PSY 520 Topic 3 Exercise Chapter 7 and 8 6.7 (a) Estimate whether the following pairs of scores for X and Y reflect a positive relationship, a negative relationship, or no relationship. Hint: Note any tendency for pairs of X and Y scores to occupy similar or dissimilar relative locations. (b) Construct a scatterplot for X and Y . Verify that the scatterplot does not describe a pronounced curvilinear trend. (c) Calculate r using the computation formula (6.1). 6.10 On the basis of an extensive su...
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PSY-520 Topic 2 Exercises – Chapters 5 & 8 with Answers
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PSY-520 Topic 2 Exercises – Chapters 5 & 8 with Answers 
 
 
5.11 Scores on the Wechsler Adult Intelligence Scale (WAIS) approximate a normal curve with a mean of 100 and a standard deviation on 15. What proportion of the IQ scores are 
a. above Kristen’s 125? z= 125-100/15= 1.67 using the z table on page 536 = 0.0475 
 
b. below 82? z=82-100/15 = -1.2 
 
C’=0.1151 
 
c. Within 9 points of the mean? 100 +9=109 109-100/15=0.6 
100-9=91 
 
91-100/15=-0.6 
 
B+B’ 
 
.2257+.2257 
 
=0.4514...
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Exam (elaborations) PSY 520 Topic 2 Exercises 520.docx
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roportions Scores on the Wechsler Adult Intelligence Scale (WAIS) approximate 
a normal curve with a mean of 100 and a standard deviation of 15. What proportion of IQ scores 
are WAIS mean = 100; standards deviation = 15 
a. Above Kristen’s 125? 
µ = 100 ; = 15 ơ 
z= 
X−μ 
σ 
 = 
125−100 
15 = 25/15 => z = 1.67 looking at table A 
a 
, it gives us a 
value of C = 0.0475. 
b. Below 82 
z = 
82−100 
15 = 
−18 
15 = -1.2, so z value of -1.2 gives us a C value of 0.1151 
c. Within 9...
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PSY 520.
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PSY 520 Exercise.
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PSY 520 Topic 2 Exercises
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PSY 520 Topic 2 Exercises 
Chapter 5 
 
5.11 Scores on the Wechsler Adult Intelligence Scale (WAIS) approximate a normal curve with a mean of 100 and a standard deviation on 15. What proportion of the IQ scores are 
a. above Kristen’s 125? z= 125-100/15= 1.67 using the z table on page 536 = 0.0475 
 
b. below 82? z=82-100/15 = -1.2 
 
C’=0.1151 
 
c. Within 9 points of the mean? 100 +9=109 109-100/15=0.6 
100-9=91 
 
91-100/15=-0.6 
 
B+B’ 
 
.2257+.2257 
 
=0.4514 
 
d. More than 40 point...
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Exam (elaborations) PSY 520 ALL QUIZZES BUNDLED TOGETHER
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Exam (elaborations) PSY 520 ALL QUIZZES BUNDLED TOGETHER
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PSY 520 Topic 2 Exercises/GCU PSY 520
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Chapter 5 
 
5.11 Scores on the Wechsler Adult Intelligence Scale (WAIS) approximate a normal curve with a mean of 100 and a standard deviation on 15. What proportion of the IQ scores are 
a. above Kristen’s 125? z= 125-100/15= 1.67 using the z table on page 536 = 0.0475 
 
b. below 82? z=82-100/15 = -1.2 
 
C’=0.1151 
 
c. Within 9 points of the mean? 100 +9=109 109-100/15=0.6 
100-9=91 
 
91-100/15=-0.6 
 
B+B’ 
 
.2257+.2257 
 
=0.4514 
 
d. More than 40 points from the mean? 100+40=140...
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Grand Canyon University_PSY 520 Topic 6 Exercise:Chapter 16, 17, 18 COMPLETE SOLUTION
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PSY 520 Topic 6 Exercises

16.9 Given the aggression scores below for Outcome A of the sleep deprivation experiment, verify that, as suggested earlier, these mean differences shouldn’t be taken seriously by testing the null hypothesis at the .05 level of significance. Use the computation formulas for the various sums of squares and summarize results with an ANOVA table.

16.10 Another psychologist conducts a sleep deprivation experiment. For reasons beyond his control, unequal numbers of subj...
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PSY 520 COURSE BUNDLE
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PSY 520 COURSE BUNDLE
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