NZEOGU E
MIDTERM EXAMINATION (Ch. 17)
Spring 2014
Student’sname:_EmmanuelNzeogu
There are6 parts to this 100point exam you are to answer each part. Ifproblems need calculations and you use a calculator, please showexplicitly all the computational steps you followed to obtainyour results. [In case of a calculation mistake, you’ll get partialcredit. Don’t miss this opportunity]! For other problems just typeyour answers. [You can use reasonably unlimited space]. Do well!
Part 1(20POINTS, one for each question)
For eachquestion, please, highlight one answer.
1. Thechance of a randomly selected case having a score beyond ± 1.00standard deviation of the mean is
a. .6826
b. .5000
c. .3174
d. 1/2of the area of 1 standard deviation
2. Toobtain the area below a positive Z score or above a negative Z scoreyou would
a. subtract the value of the Z score from the mean
b.subtract the area in the "Area Beyond Z" column of the Zscore table from 50%
c. add the value of the Z score to the area beyond the mean
d. add the area between the Zscore and the mean to 50%
3. Thepurpose of inferential statistics is to acquire knowledge of the from the ___ by
means of the _ distribution
a. population, sample, sampling
b.sample, sampling, population
c.inductive, empirical, stratified
d.theoretical, empirical, theoretical
4. If aresearcher changes from the 90% confidence level to the 95% level,the confidence interval will
a. widen
b.decrease in width
c. notbe affected
d. widenonly if N is greater than 100
5. Apolice department reports that the number of reported rapes in theirjurisdiction increased by 100% over
thepast decade. This means that the number of reported rapes today is
a.double the number 10 years ago
b.actually about the same as 10 years ago
c. equalto the number 10 years ago plus 50%
d.triple the number 10 years ago
6. In apositively skewed distribution the mean is
a. equalin value to the median
b. greater in value than themedian
c. lessin value than the median
d.either a or b, depending on the value of the mode
7. Thevariable socioeconomic status ranges from upper class to lower classand is an example of the
a. nominal level of measurement
b. ordinal level of measurement
c.intervalratio level of measurement
d.ratio level of measurement
8. Choosethe nominal level variable below:
a.size of family unit
b. nursing home federal providernumber form Nursing Home Compare data set
c.speed of travel of a jet
d.your weight
9. Yourscore on the test is the same as the third quartile. You may concludethat
a. you scored higherthan 75% of the people who took the test
b. thedistribution of the scores is skewed
c. yourscore is `typical` since it is the same value as the median
d. youscored higher than 25% of the people who took the test
10. TheCentral Limit Theorem states that as sample size becomes large
a.the sampling distribution of sample means approaches normality
b.the sampling distribution of sample means becomes larger
c.the population distribution becomes normal
d. the sample distribution becomes normal
11. Aresearcher is preparing a report and wants to select a measure ofcentral tendency that shows the most
commonly occurring score in a particular distribution. Whichstatistic should she select?
a. mode
b.median
c. mean
d. anyof the above
12. In atwo person race, if the Republican presidential candidate isprojected to attract 46% 3% ofthe vote
andthe Democratic candidate is projected to attract 48%3%of the vote, then
a. theRepublican is the probable winner
b. theDemocrat is the probable winner
c.neither candidate will win a majority of votes in the electoralcollege
d. the race is “too close tocall”
13. Amongother things, the Z score table gives the area between a score andthe mean. For a Z score of 1.00
That area (in percentages) is
a.34.13%
b.34.13%
c.68.26%
d.68.26%
14. Whenintervalratio data are badly skewed, the appropriate measure ofcentral tendency is the
a. mode
b. median
c. mean
d. Firstquartile
15. Thearea between a negative Z score and a positive Z score can be foundby
a. subtracting the Z scores from each other
b.subtracting each Z score from the mean and adding the results
c. adding the Z scores and finding the area in the Z score table forthe summed Z scores
d.adding the areas between each Z score and the mean
16. A piechart shows the frequency distribution of
a.One nominal variable
b. Oneintervalratio variable
c. Twovariables
d.any number of variables
17. Ifthe scores on a variable are 11, 18, 14, 20, and 23, the median is
a.3
b. 18
c. 14
d. 19
18. Whichsample size will produce the confidence interval with the smallestwidth?
a. 100
b. 200
c. 500
d. 1000
19.Histograms and line charts or frequency polygons are generally usedwith data measured at the
a.nominal level
b.ordinal level
c. intervalratio level
d.any level
20. Ifdistribution A and distribution B have the same mean but A has alarger standard deviation, we may
Conclude that
a. themedians of both distributions would be the same
b.distribution B has a range of 0
c. A has a greater degreeof dispersion
d.distribution B displays more variety
Part 2 (10POINTS)
Readthe following article from The New York Times (on page 7) and answerthe following questions.

What is the objective of Dr. Larry Kurdek’s research?
To determine howmarital quality changes over time and to explain why this happensbecause with time, a sense of reality sets in the union.

What is the population that he is studying?
522Americanmarried couplesover the first decade of their marriage.

What is the sample size?
522 couples

What is the unit of analysis?
A married couple

What is the dependent variable? (What level of measurement would you use for it? In other words, given what you know about measurement, how do you think the respondents were asked to reply to the questions?).
Marital quality (Yes or No)

What are the independent variables? (What level of measurement would you use for each variable? In other words, given what you know about measurement, how do you think the respondents were asked to reply to the questions?).

Divorce history,

The presence of children from the previous marriages and having children together are different variables in the model

Personality variables

Draw Kurdek’s theory (causes and effects) using boxes and arrows.
By ALISHA BERGER. New York Times. Published: October 5,1999

It is well known that nearly half of all marriages end in divorce. With that in mind, a psychology professor from Wright State University surveyed husbands and wives once a year over the first decade of their marriages to observe how marital quality changes over time. The researcher, Dr. Larry A. Kurdek, found that couples often began their unions with high levels of marital quality, but that it appeared to decrease twice: once rather steeply over the first four years and again after about seven. (The pattern of change was the same for both husbands and wives.) He also reported that couples with children experienced the steepest declines.
The research, in the September issue of the journal Development Psychology, began with a sample of 522 couples. Participants filled out an annual 32item questionnaire on various aspects of marital quality. Sample questions included these: “How satisfied are you with your marriage?“ “How affectionate is your partner?“ and “To what extent do you do things together?“ The husbands` and wives` responses were compared over time 93 couples participated for the entire decade.
“Most marriages start off with such high levels of quality that it can only change down,“ Dr. Kurdek said. “At the start of a relationship you can overlook the fact that he throws his socks around or that she leaves the refrigerator open. Over time, a sense of reality sets in. You`d started off making excuses for your partner. Then you don`t. It`s a natural evolution.
“The second dip is more difficult to explain,“ he said. “It may just be the result of being in something for a long time. You start reexamining. It might just be the natural curiosity — a sort of wondering about what else is out there.“
Dr. Kurdek also examined the factors that predict the rate of change. He looked at three major sets of causes: divorce history, the presence of children and personality variables. He found that couples who have children together, not children from previous marriages, experienced the steepest decline. “There is ample evidence to indicate that having kids changes the overall quality of marriage,“ Dr. Kurdek said. “For the most part these couples are dealing with young kids, and they require extensive levels of supervision. You`re spending less time together as a couple, may not have a lot of time and energy for sexual affection, and there`s a lot more to argue about.
“My own sense is that a lot of our emotional responses are based on expectation. If you can prepare for these declines, then chances are if you`re happy over all, the level of commitment can stay high.“ An expert on couples and mental health, Dr. Jerry I. Cooper, former chief of psychiatry at York Finch General Hospital near Toronto, called the study interesting but added: “People have to use common sense and judgment. In marriages, the damage is done from the beginning. Before you have children, you`re going to get a good preview of what your life is going to be like. If you`re not getting along now, later on you`re not going to get along.“
Part3. (20POINTS). The frequency distribution below displays the distribution of scoresassigned to a sample of employees as a part of the annual performancereview process. As you can see, employees’ performance scoresranged between 50 and 99.
Table1.
TEST GRADES 

Cumulative Frequency 
Cumulative Percentage 

Class Intervals 
Frequency 
Percentage 

50 – 59 
3 
12 
3 
4.1 
60 – 69 
4 
16 
7 
9.6 
70 – 79 
9 
36 
16 
21.9 
80 – 89 
6 
24 
22 
30.1 
90 – 99 
3 
12 
25 
34.2 
TOTALS 
25 
100 
73 
99.9 

Complete the table by filling in the columns for percentages, and cumulative percentages (in Table 1).
Totalfrequency: 3+4+9+6+3
=25
Percentage= divide the frequency by the total number ofresults and multiply by 100.
3/25*100
=12
4/25*100
=16
9/25*100
=36
6/25*100
=24
3/25*100
=12
Cum.Percentage
3/73*100
=4.1
7/73*100
=9.6
16/73*100
=21.9
22/73*100
=30.1
25/73*100
=34.2

Based solely on the frequency distribution shown in Table 1, find the average performance score. (Feel free to add more columns to the table above in order to assist you in your computations).
Part 4. (20 points)
The Chicago police department was asked by the mayor’s office toestimate the cost of burglaries to the citizens of Chicago. Thepolice relied on a random sample of burglary files and found that theaverage dollar loss in a burglary was $678, with a standard deviationof $560. Assuming that the dollar loss was normally distributed,

What proportion of burglaries had dollar losses above $1,00
Zscore =Sample valuesample mean
Standard Deviation
Z score= 1000678
560
Z score0.575
Calculating the probability Pr(x>1000)= 0.2827
Area The Area that is 28.27% is above the $1,000

What is the probability that any one burglary had a dollar loss of above $400?
Z score =400678
560
Z score 0.496≈ 0.5
Pr(x>400)= 0.6915
Area 69.15% of the area is above $400

What proportion of burglaries had dollar losses below $500?
Z score =500678
560
Z score 0.318≈ 0.32
Pr.(x<500) = 0.3745
Area (%) 37.45% is below $500
Part 5(15points)
Thequestion that the respondents were asked in the General SocialSurvey, 2012 was:
How manyhours per week do you use email?
Usingstatistical information from the Table 2 below:
Table 2. Descriptive Statistics 

N 
Minimum 
Maximum 
Mean 
Std. Deviation 

WWW HOURS PER WEEK 
1018 
0 
184 
10.47 
14.523 
Valid N (listwise) 
1018 

Write a summary of the variable
Onaverage, respondents use 10.47 hours per week on email. The maximumnumber of hours that the respondents use per week is 184 hours. Inthe General survey 2012, it was found that a number of respondentsuse0 hours on email meaning that they don’t use email at all. Thestandard deviation of the hours spent per week is 14.523, meaning thedeviation around the mean (10.47) is big.

Construct 95% confidence interval (c.i.) for the population parameter (assuming that the variable is normally distributed)
s= 14.523 n= 1018 x =10.47
CL= 0.95
Α=1CL =10.95=0.05 hence σ/2=0.025
Z_{0.025}=1.96
EBM=1.96(14.523/√10181 = 0.8921
x EBM=10.470.8921= 9.5779
x+ EBM=10.47+0.8921=11.3621
Hence confidence interval is (9.58, 11.36)

Explain what this c.i. result means in plain English
TheBased on a random sample of 1018 respondents, it can be estimated atthe 95% confidence level that the 18 and over U.S. population averagetime per week using email is between 9.58 and 11.36 hours.
This can be explained as 95% of the confidence interval constructedi.e. (9.58, 11.36) contain a true sample mean of the hours used perweek on the internet.
Part 6 (15points)



1.Write a summary of the variable
The majority of 950respondents i.e 6 2.1% agreed that that even if the scientificresearch did not bring immediate benefits it should be supported.23.8 % strongly agreed for support of scientific research. 11.9%disagreed on its support while the least 2.2% strongly disagreed onthe support of the scientific research.
The bar chart below shows the percentage distribution of the validrespondent and their level of agreement
2. Construct one 95% confidence interval (c.i.) for 85.9% ofAmericans who say they (assuming that the variable is normallydistributed).
Ps= 0.859
Z_{0.025}=1.96
P_{u} = 0.5
N = 950

Explain what this c.i. result means in plain English.
At a 95% confidence level, between 82.72% and 89.08% of Americansagree or strongly agree with Federal government spending forscientific research
ExtraCredit (10 POINTS)
The United States has often adopted an isolationist foreign policydesigned to stay out of foreign entanglements (as George Washingtonadvised his fellow citizens more than two hundred years ago). In the2000 study, 1132 respondents from the US were asked whetherthey agreed that international agreements were a national priority. The data shows that 897 of the 1132 respondents eitherstrongly agreed or agreed that international agreementswere an important national priority. Estimate the proportion of alladult Americans who strongly agreed or agreed at the95% confidence level. Express the estimate in a summarysentence.
Ps= 0.79
Z_{0.025}=1.96
P_{u} = 0.5
n = 1132
c.i =79%8.82%
ci =70.18%87.82%
At 95% confidence interval between 70.18% and 87.82% respondentsagreed or strongly agreed on prioritizing the internationalagreements
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