PUB 504Summer2014Final Exam062114 Nzeogu Emmanuel
Part1 (25 POINTS)

For “tapphone“” variable produce the appropriate descriptive statistics SPSS table and graph. Use GSS 2006 Data.
Statistics 

SHOULD AUTHORITIES HAVE RIGHT TO TAP PHONE CONVERSATION 

N 
Valid 
1500 
Missing 
3010 

Mean 
2.47 

Median 
2.00 
SHOULD AUTHORITIES HAVE RIGHT TO TAP PHONE CONVERSATION 

Frequency 
Percent 
Valid Percent 
Cumulative Percent 

Valid 
Definitely should have right 
380 
8.4 
25.3 
25.3 
Probably should have right 
435 
9.6 
29.0 
54.3 

Probably should not have right 
280 
6.2 
18.7 
73.0 

Definitely should not have right 
405 
9.0 
27.0 
100.0 

Total 
1500 
33.3 
100.0 

Missing 
IAP 
2992 
66.3 

CANT CHOOSE 
18 
.4 

Total 
3010 
66.7 

Total 
4510 
100.0 

Write a summary /interpretation of the variable. Use information from the SPSS table.
Therespondents were asked if authorities should have right to tap phoneconversation. Of the 1500 respondents, majority 54.3% felt that theauthorities should have the right to tap phone.

Construct confidence intervals (c.i.) for the population proportions.
X=2.47 s= 1.139
N=1500
At95% CL Z=1.96
Hencethe confidence interval of the population is between 2.412 and 2.528

Interpret the results. Explain what these results mean in plain English.
Basedon a random sample of 1500 respondents, it can be estimated at the95% confidence level that the population agrees on authorities havingright to tap phone conversion.
Part2 (25 POINTS)

For “age” variable produce the appropriate descriptive statistics SPSS table and graph. Use GSS 2008 Data. The table below shows the descriptive statistics of the age.
Descriptive Statistics 

N 
Minimum 
Maximum 
Mean 
Std. Deviation 

AGE OF RESPONDENT 
2013 
18 
89 
47.71 
17.351 
Valid N (listwise) 
2013 

Using information from the SPSS table, write a summary /interpretation of the variable.
Theyoungest person to participate in GSS 2008 was aged 18 and the oldestwas aged 89. The average age of the respondent was 47.71 years. Thetable 2 above shows that all the respondents in 2008 GSS were adults.

Construct confidence intervals (c.i.) for a population mean.
_{Or}
Hencethe confidence interval of the population average age is between46.95 years and 48.47 years

Interpret the results. Explain what these results mean in plain English.
Basedon the Radom sample of 2013 respondents who participated in 2008 GSS,it can be estimated at the 95% confidence level that the populationaverage age lies between 46.95 years and 48.47 years.
Part3 (10 POINTS)
Variable“Costs per Vehicle”:
NoMaintenanceMaintained Cars
SampleMean= $625 Samplemean = $575
S_{1}= 150 S_{2}= 200
N_{1}= 75 N_{2}= 225
Step1.Assumptions and Test Requirements
Model:Independent random samples
Levelof measurement is intervalratio
Samplingdistribution is normal
Step2. State Null Hypothesis
H_{0}:µ_{1}= µ_{2}.
H_{1}:µ_{1}≠ µ_{2}
Step3. Sampling distribution = Z distribution
Alpha= 0.05, twotailed
Z(critical) = ± 1.96
Step4. Compute Test Statistics ( by hand)
_{}= _{}
_{} =21.96
Z(obtained) = _{}=_{}
_{}Z (obtained) =
Z(obtained) = 2.28
Z(obtained) = 2.28
Z(critical) = ± 1.96
Step5: Make Decision and Interpret Result
Fromthe hand calculations, the test statistic computed in step 4. Zobtained is greater than the z critical hence we conclude that thereis a significant difference between those vehicles that receivedroutine maintenance and those that did not. Given the direction ofthe difference, we can conclude that those vehicles that did notreceive routine maintenance had more costly repairs. The periodicmaintenance program should be continued.
Part4 (25 POINTS)
Comparethe number of hours watching TV between those with less than a highschool diploma and those with some graduate work.[Usethe SPSS output below. Do not run the test in SPSS. You do not havethe data that was used for this test].
TTest

What values should be put into two blanks (). [See the output above]
DF=95 and there are two variables
Hencethe total number of respondents is 95 + 2= 97
ForBlank 1 we have
9767=30
Meandifference is gotten by getting the difference between the 1st andthe 2nd mean in the group statistics table
=3.762.17
=1.59

State the null hypothesis (in words)
H0:µ1 = µ2. The null hypothesis state that the sample mean differenceof Less than High school student and that of Graduate student inaverage hours spent watching Tv is the same.

State the research (twotailed) hypothesis(in words)
Theresearch (two tail) hypothesis state that there is no statisticalsignificant difference in the sample means of hours per day thestudents spent in watching tv

State the research (onetailed) hypothesis(in words)
H_{0}:µ_{1}– µ_{2}>0the research (one tail) Hypothesis state that the difference in thepopulation meansin hours per day watching Tv is greater than zero

Based on this output at the 0.05 significance level do you reject or acceptthe null hypothesis? Why?
Basedon the output, our sig value is 0.017 this value is less than oursignificance level value 0.05 hence we reject the null hypothesis.This is because the area is in the critical region.

Based on this output at the 0.05 significance level do you reject or accept the research (twotailed) hypothesis? Why?
Sincewe rejected the null hypothesis, we accept the research hypothesis

Based on this output at the 0.05 significance level do you reject or accept the research (onetailed) hypothesis? Why?
Sig( one tail) =0.017/2 = 0.0085Whichis still less than our significance indicator 0.05 hence we rejectthe null hypothesissincethe area is not in the critical area

Summarize your conclusions.
Inconclusion, sig( 2tail) which is 0.017 which is less than theindicator value of 0.05 hence we reject the null hypothesis and conclude that there is statistically significant difference in hoursper day watching tv between the students. We conclude that, onaverage LT high school and graduates have significantly different tvwatching habit.
Part5 (15 POINTS)
Threedifferent sections of the same Interstate highway, with roughly equaltraffic volumes have been patrolled by the State Police at differentlevels of intensity for the past six months. The posted speed limitis 55 and the speeds of random samples of motorists have beenregistered for each of the threesections.Is there any statistically significant difference in the speeds? [Usethe SPSS output below. Do not run the test in SPSS. You do not havethe data that was used for this test]
Oneway

State the null hypothesis
H_{0}: The population means are the same.
µ_{1}= µ_{2}= µ_{3}

State the research hypothesis
H_{1}:At least one of the population means is different.

Calculate the F obtained using the intermediate results from ANOVA Table. (Hint: SPSS uses the same formula for calculation of F obtained as you have in the book).
Meansquare in the ANOVA table

158.333/2 =79.167

557/21 = 2.524
F(ratio) is gotten by dividing the mean square of speed between thegroup and that of speed within Groups
Meansquare within =SSW/dfw = 557.000/21 = 26.52
=
Meansquare between = SSB/dfb = 158.333/2 = 79.17
F_{ratio}= MSB/MSW = 79.167/26.524 = 2.985
4.Interpret the results
TheANOVA output shows that the F ratio (31.366) and the exactprobability (“sig”) of getting the null hypothesis are not true.The sig is 0.72, which is higher than our usual level of 0.05. Wefail to reject the null hypothesis and conclude that there is nostatistically difference in speed.
Extracredit (10 POINTS)
In2010 General Social Survey the respondents were asked the followingquestion: Should government reduce income differences? The possibleanswers:

Government should reduce income differences

Neutral ( Middleoftheroad)

Government should not take any actions to reduce income differences
Isthe opinion about government redistribution policy (variable “Shouldgovernment reduce income differences?”) dependent on politicalviews? [Use the tables below to answer the question.]

What does the crosstabulation show?
Thecross tabulation below shows the relationship between reduction ofgovernment income differences and the political views

State the null hypothesis (in words):
Thenull hypothesis state that there is no significant relationshipbetween political views and government reduction of incomedifferences

State the research hypothesis (in words):
Thealternative or research hypothesis state that there is astatistically significant relationship between the governmentreduction in income differences and the political views

Make a decision about the null hypothesis:
Inthe chisquare test, the value of chisquare (obtained) is 154.447.the df is 4 and the exact significance of the chi square is 0.000. At95% confidence level, the sig chisquare is less than 0.05 hencereject the null hypothesis. We hence conclude that there is astatistically significant relationship between the Government incomedifference reduction and the political views.

Interpret the results of the test:
Inspectingthe column percent in the case processing table below show thatmajority 59.2% of those who agree on reduction are liberal 2.8% aremoderate and 25.3% are conservative. This proves that there isrelationship between the government income difference reduction andthe political views.
Crosstabs
Case Processing Summary 

Cases 

Valid 
Missing 
Total 

N 
Percent 
N 
Percent 
N 
Percent 

Should Government Reduce Income Differences * Political Views 
1377 
67.4% 
667 
32.6% 
2044 
100.0% 
Should Government Reduce Income Differences * Political Views Crosstabulation 

Political Views 
Total 

LIBERAL 
MODERATE 
CONSERVATIVE 

Should Government Reduce Income Differences 
GOVT REDUCE DIFF 
Count 
234 
212 
123 
569 
% within Political Views 
59.2% 
42.8% 
25.3% 
41.3% 

NEUTRAL 
Count 
57 
122 
70 
249 

% within Political Views 
14.4% 
24.6% 
14.4% 
18.1% 

NO GOVT ACTION 
Count 
104 
161 
294 
559 

% within Political Views 
26.3% 
32.5% 
60.4% 
40.6% 

Total 
Count 
395 
495 
487 
1377 

% within Political Views 
100.0% 
100.0% 
100.0% 
100.0% 
ChiSquare Tests 

Value 
df 
Asymp. Sig. (2sided) 

Pearson ChiSquare 
154.447^{a} 
4 
.000 
Likelihood Ratio 
152.268 
4 
.000 
LinearbyLinear Association 
126.755 
1 
.000 
N of Valid Cases 
1377 

a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 71.43. 
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