Part 1 (25 POINTS)

PUB 504-Summer-2014-Final Exam-06-21-14 Nzeogu Emmanuel

Part1 (25 POINTS)

    1. 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

    1. 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.

    1. 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

    1. 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)

  1. 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

  1. 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.

  1. 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

  1. 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

S1= 150 S2= 200

N1= 75 N2= 225

Step1.Assumptions and Test Requirements

Model:Independent random samples

Levelof measurement is interval-ratio

Samplingdistribution is normal

Step2. State Null Hypothesis

H01= µ2.

H11≠ µ2

Step3. Sampling distribution = Z distribution

Alpha= 0.05, two-tailed

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].

T-Test

  1. 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

97-67=30

Meandifference is gotten by getting the difference between the 1st andthe 2nd mean in the group statistics table

=3.76-2.17

=1.59

  1. 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.

  1. State the research (two-tailed) 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

  1. State the research (one-tailed) hypothesis(in words)

H01– µ2&gt0the research (one tail) Hypothesis state that the difference in thepopulation meansin hours per day watching Tv is greater than zero

  1. 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.

  1. Based on this output at the 0.05 significance level do you reject or accept the research (two-tailed) hypothesis? Why?

Sincewe rejected the null hypothesis, we accept the research hypothesis

  1. Based on this output at the 0.05 significance level do you reject or accept the research (one-tailed) 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

  1. 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

  1. State the null hypothesis

H0: The population means are the same.

µ1= µ2= µ3

  1. State the research hypothesis

H1:At least one of the population means is different.

  1. 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

  1. 158.333/2 =79.167

  2. 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

Fratio= 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:

  1. Government should reduce income differences

  2. Neutral ( Middle-of-the-road)

  3. 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.]

  1. What does the cross-tabulation show?

Thecross- tabulation below shows the relationship between reduction ofgovernment income differences and the political views

  1. State the null hypothesis (in words):

Thenull hypothesis state that there is no significant relationshipbetween political views and government reduction of incomedifferences

  1. 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

  1. Make a decision about the null hypothesis:

Inthe chi-square test, the value of chi-square (obtained) is 154.447.the df is 4 and the exact significance of the chi square is 0.000. At95% confidence level, the sig chi-square 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.

  1. 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%

Chi-Square Tests

Value

df

Asymp. Sig. (2-sided)

Pearson Chi-Square

154.447a

4

.000

Likelihood Ratio

152.268

4

.000

Linear-by-Linear 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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