PSYC FPX 4700 Assessment 2 Central Tendency and Probability:

Complete the following problems within this Word document. Do not submit other files. Show your work for problem sets that require calculations. Ensure that your answer to each problem is clearly visible. You may want to highlight your answer or use a different type color to set it apart.

PSYC FPX 4700 Assessment 2 Central Tendency and Probability

Problem Set 2.1: Characteristics of the Mean 

Criterion: Describe a distribution.

Instructions: Read the following and answer the questions. 

Data: To study perception, a researcher selects a sample of participants (n = 12) and asks them to hold pairs of objects differing in weight, but not in size, one in each hand. The researcher asks participants to report when they notice a difference in the weight of the two objects. Below is a list of the difference in weight (in pounds) when participants first noticed a difference. Answer the following questions based on the data given in the table. 

Table 1

List of the Difference in Weight (in Pounds)

Difference in Weight
4	8
9	5
12	7
6	15
10	4
8	8

1. State the following values for this set of data:

1. Mean 8
2. Median 8
3. Mode(s) 8

1. What is the shape of this distribution? Hint: Use the values of the mean, median, and mode to infer the shape of this distribution. _______ 

As this is a distribution with a nominal variable, it will have a symmetrical shape.

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PSYC FPX 4700 Assessment 2 Central Tendency and Probability

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Problem Set 2.2.a: Interpret Means in a Chart

Criterion: Interpret means in a chart. 

Instructions: Read the information below and answer the questions. 

Data: General life satisfaction across culture. Gilman and colleagues (2008) measured general life satisfaction in 1,338 adolescents from two individualistic nations (Ireland, United States) and two collectivist nations (China, South Korea) using the Multidimensional Students’ Life Satisfaction Scale (MSLSS). Mean participant scores on the MSLSS are given in the following table.

Table 2

Mean Participant Scores on the MSLSS

Mean MSLSS Scores by Nation and Gender
Nation	Gender
	Men	Women
United States	4.39	4.61
Ireland	4.37	4.64
China	4.41	4.56
South Korea	3.92	3.78

1. Among which group was general life satisfaction lowest on average?   South Korean Women
2. Among which group was general life satisfaction highest on average?   Ireland Women

PSYC FPX 4700 Assessment 2 Central Tendency and Probability

Problem Set 2.2.b: Understanding Standard Deviations in a Chart

Criterion: Interpret standard deviations in a chart. 

Instructions: Read the following and answer the question based on the data in the chart. 

Data: Acceptable height preferences. Salska and colleagues (2008) studied height preferences among dating partners. In their first study, they reviewed Yahoo! Personals for heterosexual individuals living within 250 miles of Los Angeles, California, and recorded the acceptable range of heights for their dating partners. The following table lists some of the results. 

Table 3

Mean and Standard Deviation of Heights ( Shortest and Highest)

Preferences	Women		Men
	M	SD	M	SD
Shortest acceptable height, inches	68.9	2.6	60.6	3.7
Tallest acceptable height, inches	75.3	2.2	69.8	2.7

1. Overall, did men or women show greater variability in their responses? Explain.

The data collected shows that there was a greater degree of variability in the male sample. Specifically, when looking at the standard deviation values, it was found that men generally had more variation in their preferences from the mean values compared to women.

PSYC FPX 4700 Assessment 2 Central Tendency and Probability

Problem Set 2.3: Range, Variance, and Standard Deviation in Excel

Criterion: Calculate measures of variability in Excel from a group of raw scores.

Data: A sample of likes per post on Facebook: 45, 789, 16, 5, 486, 1, 87, 18, 48, 1

Instructions: Complete the following steps: 

1. Install the data analytics package in Excel.

• If you are unsure about how to do this, visit Load the Analysis Toolpak in Microsoft Excel, which has tutorials for both Windows and Mac. Used with permission from Microsoft.

1. Enter the data above into Excel using the variable name Data. In cell A1, type the word “Data.” Then, enter the data above in cells A2 to A11. 
2. In the Toolbar, click Data Analysis, Select Descriptive Statistics, then click Ok.
3. Next to input range type: $A$2:$A$11
4. Double check that summary statistics has a check next to it.
5. Click OK. A new sheet will appear to the right with your data. 
6. Copy and paste the descriptive statistics table below.
   • Highlight the range, mean, and standard deviation.  

Table 4

Statistics for likes per post on Facebook

Column1
Mean	149.6
Standard Error	84.81486
Median	31.5
Mode	1
Standard Deviation	268.2081
Sample Variance	71935.6
Kurtosis	3.346301
Skewness	2.035692
Range	788
Minimum	1
Maximum	789
Sum	1496
Count	10

PSYC FPX 4700 Assessment 2 Central Tendency and Probability

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Problem Set 2.4: Range, Variance, and Standard Deviation in JASP

Criterion: Calculate measures of variability in JASP.

Data: Use dataset likes.jasp. This dataset is a sample of likes per post on Facebook.   

Instructions: Complete the steps below.

1. Download likes.jasp. Double-click the icon to open the dataset in JASP. 
2. In the Toolbar, click Descriptives.
3. Select Likes and then click Arrow to send it over to the Variables box.
4. Check Transpose descriptives table. 
5. Select Statistics. In the menu that opens, check the boxes for Mean, Std. deviation, Variance, and Range. Deselect all other boxes. 
6. Copy and paste the descriptive statistics table below.

Table 5

Measures of Variability of Likes Per Post on Facebook

Descriptive Statistics
		Mean		Std. Deviation		Variance		Range
Likes		149.600		268.208		71935.600		788.000

1. Highlight the range, mean, and standard deviation.  
2. Answer: Does your mean equal the mean calculated in Problem Set 2.3? ___Yes_______

PSYC FPX 4700 Assessment 2 Central Tendency and Probability

Problem Set 2.5: Probability and Conditional Probability 

Criterion: Compute the probability. 

Instructions: Read the following and answer the questions.

Researchers are often interested in the likelihood of sampling outcomes. They may ask questions about the likelihood that a person with a particular characteristic will be selected to participate in a study. In this exercise, we will select a sample of one participant from the following hypothetical student population of new and returning students living on or off campus. The population is summarized in the following table.

Table 6

Number of Students Participating in the Study

1. What is the probability of selecting a new student participant? _50/100 = 0.5
2. What is the probability of selecting a returning student participant? 50/100 = 0.5
3. What is the probability of selecting a student who lives on campus? _______55%______

1. What is the probability of selecting a student who lives off campus? ____45%_________
2. What is the probability of selecting a new student, given that they live off campus? ______20/45= 0.44_________________
3. What is the probability of selecting a returning student, given that they live on campus? _________25/55= 0.45______________
4. What is the probability of selecting a new student, given that they live on campus? 30/55= 0.55____
   

The probability of selecting a new student, given that they live on campus, is 0.55 (30/55) since there are 30 new students who live on campus out of a total of 55 on-campus students.

1. What is the probability of selecting a returning student, given that they live off campus? 25/45= 0.56____
   

The probability of selecting a returning student, given that they live off campus, is 0.56 (25/45), since there are 25 returning students who live off campus out of a total of 45 off-campus students.

1. What is the probability of selecting a student who lives on campus, given that they are a new student? 30/50= 0.6________

The probability of selecting a student who lives on campus, given that they are a new student, is 0.6 (30/50), since there are 30 new students who live on campus out of a total of 50 new students.

________________

1. What is the probability of selecting a student who lives off campus, given that they are a new student? 20/50 = 0.4______

The probability of selecting a student who lives off campus, given that they are a new student, is 0.4 (20/50), since there are 20 new students who live off campus out of a total of 50 new students._____

1. What is the probability of selecting a student who lives on campus, given that they are a returning student? _______25/50 = 0.5______

The probability of selecting a student who lives on campus, given that they are a returning student, is 0.5 (25/50), since there are 25 returning students who live on campus out of a total of 50 returning students.

1. What is the probability of selecting a student who lives off campus, given that they are a returning student? P (lives off-campus | returning student) = 0.5 * 0.45 / 0.5 = 0.45_Therefore, the probability of selecting a student who lives off-campus, given that they are a returning student, is 0.45 (or 45%).____________

Problem Set 2.6: Determining Probability 

Criterion: Determine the probability.

Instructions: Read and answer the question below.

Probability of first marriage among women. A National Center for Health Statistics (NCHS) brief report by the Centers for Disease Control and Prevention (CDC) in 2009 identified that about 6% of women in the United States married for the first time by their 18th birthday, 50% married by their 25th birthday, and 74% married by their 30th birthday. 

Based on these data, what is the probability that in a family with two daughters, the first and second daughter will be married by each of the following ages?

1. 18 years of age:_______0.06 x 0.06 = 0.036
2. 25 years of age:__________ 0.5 x 0.5 = 0.25
3. 30 years of age:_________ 0.74 x 0.74 = 0.5476__________________

Problem Set 2.7: Understanding Normal Distribution

Criterion: Solve problems with information about normal distributions and probabilities.

Instructions: Read the following and answer the questions.

Judging the humorousness of “lawyer” jokes. Stillman et al. (2007) conducted a study where participants listened to a variety of jokes. To determine how funny the jokes were, the researchers asked a group of 86 undergraduates to rate the jokes on a scale from 1 (very unfunny) to 21 (very funny). Participants rated a “lawyer joke” as one of the funniest jokes, with a rating of 14.48 ± 4.38 (M ± SD). 

Assuming that these data are normally distributed,

1. What was the rating that marks the cutoff for the top 10% of participant ratings for this joke? ___________20.095________________
2. How many of the 86 undergraduates gave the joke a rating of at least 10? ___________73 Students (72.82)

Problem Set 2.8: Calculating z Scores in JASP

Criterion: Calculate z scores using JASP.

Data: Use the dataset ratings.jasp. This dataset is a record of how a sample of senior citizens rated the Internet on a 1–10 scale, with 1 being “really distrust it” and 10 being “completely trust it”:

Instructions: Complete the steps below.

1. Download ratings.jasp. Double-click the icon to open the dataset in JASP. 
2. Just below the Toolbar, click the + sign next to the column labeled Rating. 
3. Type “Z scores” in the box, then click Create Column. A formula box will appear below the Toolbar. 
4. Use the scrollbar on the right side of the formula box to scroll down and select zScores(y). 
5. On the left side of the box, click Rating, then drag it into the formula box to replace the word values. Click Compute column at the bottom of the formula box.  You will see that the z scores have been added to the file. Paste a screenshot of the dataset, showing the z scores, into the Word document.

Table 7

Internet Rating by Senior Citizen and their z Scores

Answer: Which number of ratings is closest to the z score of 0?

Looking at the given data, we can see that the z scores range from -2.122832333 to 1.081442886. The z score closest to 0 would be the one with the smallest absolute difference from 0, which means we need to find the z score that is closest to the midpoint between the minimum and maximum z scores.

The midpoint between -2.122832333 and 1.081442886 is approximately -0.520694723. The z score closest to this midpoint is -0.5206947231, which corresponds to a rating of 5. Therefore, the answer is that the number of ratings closest to the z score of 0 is 5.

PSYC FPX 4700 Assessment 2 Central Tendency and Probability

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