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Before beginning the homework assignments, make sure you have reviewed this module's/lesson week's and presentations, as well as the practice data analysis. Finish all of your analyses in SPSS, then copy and paste your results and graphs into your homework document file. Any written questions (such as the text-based questions or the APA Participants section) should be answered in the appropriate area of the same file.
When we standardize scores, we transform individual scores from different normal distributions to a common normal distribution with known (a) ,mean____, (b) standard deviation____, and (c) percentiles____. Define the central limit theorem as described in Nolan and Heinzen (2016).
Answer: Refers to how a distribution of sample means is a more normal distribution than a distribution of scores, even when the population distribution is not normal.
3)
A z-score can be thought of as the number of _Standard Deviation_ that a score is from the mean.
4)
The formula for the standard error is the _standard deviation for the population_______ divided by the _square root of the sample__. (Use words, not symbols, to answer this question.)
5) Calculating z scores from raw scores: If a population has a mean of m=30 and a standard deviation of s = 3, calculate z scores for each of the following raw scores (X) from this population. Show work on the right hand side, put answers on the left in the space provided.
5a) X = 31; Z = Answer 0.3
Work: (31-30)/3=0.3
5b) X = 27; Z = Answer -1
Work: (27-30)/3= -1
5c) X = 37; Z = Answer 2.3
Work: (37-30)/3= 2.3
5d) X = 25.5; Z = Answer -1.5
Work: (25.5-30)/3= -1.5
6) Calculating raw scores from z scores: If a population has a mean of m=30 and a standard deviation of s = 3, calculate raw scores (X) for each of the following z scores from this population. Show work on the right hand side, put answers on the left in the space provided.
6a) Z = .57; X = Answer 31.71
Work: (0.57×3)+30=31.71
6b) Z = -2.03; X = Answer 23.91
Work: (-2.03×3)+30=23.91
6c) Z = -1.0; X = Answer 27
Work: (-1×3)+30=27
6d) Z = 1.68; X = Answer 35.04
Work: (1.68×3)+30=35.04
7) In a normal curve, what percentage of scores falls:
7a) Above the mean? Answer 50%
Work: where, s is the standard deviation.
7b) Between the mean and +1 standard deviation (SD) above the mean? Answer 34%
Work:
7c) In the left “tail” of the distribution, more than -2 SD from the mean. Answer (47.5%)
Work:
8) Compute the standard error (sm) for each of the following sample sizes, assuming a population mean m = 50 and a standard deviation of s = 4.5.
8a) 20 Answer 1.006
Work:
8b) 120 Answer 0.4108
Work:
8c) 1200 Answer 0.1299
Work:
9) Compute a z-statistic for each of the following sample means, assuming m = 85 and s = 9 (Remember to compute sM before computing the z statistic!)
9a) A sample of 46 scores has a mean of 89 Answer 3.0145
Work:
9b) A sample of 82 scores has a mean of 76 Answer -9.4340
Work:
9c) A sample of 100 scores has a mean of 86. Answer 1.1111
Work:
Part II: SPSS Analysis
Be sure you have viewed the SPSS tutorial presentation before proceeding.
Open “Module 5 Exercise File 1” (found in the course’s Assignment Instructions SPSS folder) in order to complete these exercises.
Reminder: Be sure to paste in the SPSS output and write out the answers in the spaces given.
Part II: Questions 1a-1d
Open Module 5 Exerise File 1. It contains the IQ scores of 38 participants.
Compute descriptive statistics of the raw IQ scores
Create a histogram of the raw IQ scores
Transform the raw scores to z-scores following the method in the SPSS tutorial and text
Be sure to save this transformed file, since you will be using it next week as well.
Identify the z-score that is closest to 0 and farthest from 0.
Evaluate whether the scores are normally distributed.
Support your answer.
1a)
Compute Descriptive statistics for the raw IQ scores and paste the table below.
Answer: SPSS: Descriptives table
Descriptive Statistics
Gender
N
Range
Minimum
Maximum
Sum
Mean
Std. Deviation
Men
High School GPA Scores
46
3.00
1.00
4.00
116.27
2.5276
.92214
College GPA Scores
46
1.77
1.17
2.94
103.65
2.2533
.39929
Valid N (listwise)
46
Women
High School GPA Scores
44
2.94
1.06
4.00
112.14
2.5486
.82349
College GPA Scores
44
1.40
2.05
3.45
120.19
2.7316
.36598
Valid N (listwise)
44
1b)
Create a histogram of the raw IQ scores and paste it below.
Answer: SPSS: Histogram
1c)
Using the descriptives method covered in the presentation and chapter, transform the IQ raw scores to z-scores, creating a new variable.
What is the z-score that is closest to 0 (on either side of the mean) in the data set?
What is the z-score that is the farthest from 0 (on either side of the mean) in the data set?
Answer: 2.48 and 2.54 when using the college GPA scores.
Answer: 1.17 and 3.45 when using the college GPA scores.
1d)
Based on the histogram from (1b) and your other answers above, would you describe the IQ data as being normally distributed? Why or why not? Support your answer with information from the chapter and presentations regarding normal distributions.
Answer: No, it is not entirely symmetric.
Justification: This is because there is the presence of a long tail to the left than o the right. However, we may conclude that the data presented in the histogram is moderately symmentric.
Part III: SPSS Data Entry and Analysis
Data provided below.
Stress Index Scores
101
60
10
27
89
60
16
184
34
17
78
141
11
104
76
65
87
19
126
98
Part III: Questions 1a-1d
The data in the column to the left represent Stress Index (SI) scores of a sample of 20 high-level business executives. In the general population, this Stress Index is normally distributed and has a mean of 60.2 and a standard deviation of 27.1. Enter this data into SPSS.
Generate descriptive statistics for this variable.
Generate a histogram for this variable.
In your data set, standardize the SI scores by transforming them into z-scores
SPSS will automatically name the new variable.
Which z-scores corresponds to a raw SI score of 60, 104, and 184?
Does the distribution reflect the distribution in the general population?
Support your answer.
1-a)
Generate descriptive statistics for this variable.
Answer: Descriptive Statistics Table
Descriptive Statistics
N
Range
Minimum
Maximum
Sum
Mean
Std. Deviation
Stress Index Score
20
174.00
10.00
184.00
1403.00
70.1500
47.90371
Valid N (listwise)
20
1-b)
Generate a histogram for this variable.
Answer: Histogram
1-c)
In your data set, standardize the SI scores by transforming them into z-scores under a new variable “ZSI.” Using your data set as a reference, what z-score corresponds to a raw SI score of 60?
To a raw SI score of 104? To a raw SI score of 184?
60
Answer -0.21188
104
Answer 0.70663
184
Answer 2.37664
1-d)
Based on what you have been told about SI scores in the beginning of the problem,
does this sample’s distribution seem to reflect the distribution of SI scores in the general population?
Why or why not?
Answer: The sample distribution does not reflect the scores in the general population.
Justification: This is because the data is not normally distributed as evident in the hstoram above. Besides, there are two peaks in the histogram one on the lower left side and the other in the middle of the graph. Moreover, the large value of the standard deviation reveals that the data deviates more away from the mean.
Part IV: Cumulative
Data provided below for respective questions.
Part IV: Question 1a-1d (Non-SPSS)
A counselor wants to find out whether adding a Bible study component to treatment leads to different scores on a measure of generalized anxiety as compared to treatment without a Bible study component.
He assigns clents to 1 of 2 groups:
Group 1 participates in treatment plus a Bible study component.
Group 2 participates in standard treatment without the Bible study component.
1-a)
What is the independent variable in this experiment?
The use of the Bible
1-b)
What is the dependent variable?
Generalized anxiety
1-c)
What is the likely null hypothesis for this experiment?
Adding a Bible to treatment has no effect on the measurement score of generalized anxiety patients
1-d)
What is the likely research hypothesis for this experiment?
Adding a Bible to treatment has an effect on the measurement score of generalized anxiety patients
Diagnosis
Narcissism
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
33
12
20
26
27
9
17
27
34
32
12
8
18
12
6
5
8
14
10
6
Part IV: Questions 2a & 2b (SPSS)
A forensic psychologist wants to examine the level of narcissistic personality traits in those who are diagnosed with antisocial personality disorder (ASPD) and those who do not qualify for ASPD within a local prison population. She administers a measure of narcissistic personality traits where higher scores indicate higher levels of narcissism and scores range from 0–35.
Create a new SPSS data file for these scores.
Name your variables in Variable View. Assign values to the “Diagnosis” variable as follows: 1 = ASPD; 2 = no ASPD
Be sure to assign the correct level of measurement to each variable in the “Measure” column. Then return to Data View to enter the data.
Compute descriptive statistics by diagnosis (that is, for each of the two groups in one table) using similar steps to those covered in Green and Salkind’s Lesson 21, Section 4 and in the Module 3 SPSS presentation (“High School GPA scores by Gender”).
Construct a boxplot to show the difference between the 2 groups
2-a)
Compute descriptive statistics by diagnosis (that is, for each of the two groups in one table) (2 pts)
Answer: SPSS Table- Descriptive Statistics for Score (level of narcissistic personality) grouped by Diagnosis (ASPD/No ASPD):
Descriptive Statistics
Diagnosis
N
Range
Minimum
Maximum
Sum
Mean
Std. Deviation
ASPD
Narcissism
10
25.00
9.00
34.00
237.00
23.7000
8.81980
Valid N (listwise)
10
no ASPD
Narcissism
10
13.00
5.00
18.00
99.00
9.9000
4.12176
Valid N (listwise)
10
2-b)
Construct a boxplot to show the difference between the scores of the 2 groups. (3 pts)
Answer: Boxplot
Submit Homework 5 by 11:59 p.m. (ET) on Monday of Module/Week 5. Remember to name file appropriately.
Done!
