Statistics Management

Statistics Management

StatisticsManagement

StatisticsManagement

Question1(a)

Thediagram below is a tree diagram equally representing the proportionsor simply the probabilities of an Australian selected at random andhis/her opinion regarding the decision of Australian government tosend additional 500 Australian troops to Iraq.

0.73

LP

0.27

NLP

LPA

0.55

LP

0.73

0.27

0.45

DA

LPDAAA

NLP

0.73

LP

0.53

0.55

0.27

LPA

NLP

0.47

A

0.45

0.73

LP

LPD

0.27

NLP

KEY

DA&gtDisagree

A&gtAgree

LPA&gtLiberal Party Agree

LPD&gtLiberalParty Disagree

LP&gtLiberal Party Voter

NLP&gtNot Liberal Party Voter

Question1(b)

(0.53*0.55*0.73)+(0.53*0.45*0.73)+(0.47*0.55*0.73)+(0.47*0.45*0.73)=0.73

P=0.73

Fromthe tree diagram it is the probability of selecting a person atrandom you will find that 0.73 or 73% of the entire sample selectedrandomly is a liberal party voter (LP)

Question1 (c)

(0.47*0.55*0.27)=0.069795

P=0.069795

Fromthe tree diagram that represent the sample selected randomly it isevident that significantly at least 0.069795 or 6.9795% of therespondents werenotliberal party voters but significantly approved the additional 500Australian troops to be sent to Iraq.

Question1(d)

0.47*0.55*0.73=0.188705

P=0.188705

Significantlythe probability of a person selected at random is a liberal partyvoter and approves the decision to send more troops to Iraq is givenby p=0.188705or 18.8705% of the respondents selected at random were simply liberalparty voters and significantly approved the Australian governmentdecision to send additional 500 troops to Iraq

Question1 (e)

0.53*0.45*0.27

P=0.064395

Thesample presents that at least the proportion of or the probability ofa person not being a liberal party voter and disapproves thegovernment decision to send additional 500 Australian troops to Iraqas p=0.064395orsignificantly 6.4395% of persons selected at random is a non-liberalparty voter and disapproved the government decision to send moretroops to Iraq.

QUESTION2

t-Test: Two-Sample Assuming Equal Variances

&nbsp

IV

V

Mean

36

25.28571429

Variance

135.3333333

97.23809524

Observations

7

7

Pooled Variance

116.2857143

Hypothesized Mean Difference

0

d.f

12

t Stat

1.858805989

P(T&lt=t) one-tail

0.043871268

t Critical one-tail

1.782287548

P(T&lt=t) two-tail

0.087742536

t Critical two-tail

2.178812827

&nbsp

INTERPREATION

Thereis no significance difference in the means of the two samples. Thecritical value of the sample statistics is P(T&lt=t) on one tailtest is 0.043871268 which is below 0.05 suggesting that thedifference between the two means is relatively insignificant or at95% confidence level the differences between the two sample means isinsignificant, similarly the critical value of the sample statisticsis P(T&lt=t) on the two tail test is 0.087742536 is above 0.05suggesting that the differences between the two means at 95% C. I isinsignificant.

QUESTION3

Thefollowing data is selected to answer the following questions inregard to group number 7.

Group 7

&nbsp

&nbsp

&nbsp

&nbsp

&nbsp

&nbsp

&nbsp

Number of No-Shows

0

1

2

3

4

5

6

Number of days

20

37

23

15

4

0

1

100

FX

0

37

46

45

16

0

6

b) the following calculations shows the 99% confidence interval limit

Mean

21.42857

Standard Deviation

20.76742

Sample Size

7

Confidence Coefficient

0.99

Level of Significance

0.01

Margin of Error

20.21858

Upper Bound

41.64715

Lower Bound

1.209994

Max

37

Min

0

Range

37

c)INTERPRETATION

Theresults indicate that there is a 99% confidence that the mean numberof times with no show falls between 1.2 and 41.6

Thismeans that the minimum mean of time with no show is 1.2 and themaximum mean of times of no shows is 41.6

d)Another sample of 100 days would have produced a result that issimilar to the one produced. That is the mean of second sample wouldhave lied in between 1.2 and 41.6 showing that at 99% confident limitthe sample was fully representative and there was little or nobiasness.

e)1/5 of 20.74762= 4.149524

41.6=34.51+(2.58*4.149524/n)

N=100+2.2800

N=102.2800

Thereforefor the mean to remain the same with 0.99 confidence coefficient,then the sample size should increase to at least 102.2800.

QUESTION4

  1. The sampling method used in the survey is simple random sampling, that is any member within the population has equal chances of being selected and interviewed for the researchers survey data

  2. The advantage of simple random sampling is that every member in the population has equal chances of being selected i.e. it reduces the margin of error and biasness that would be caused during data collection.

Thedisadvantage of simple random sampling is that it may fall biased andimportant members of the population may be left out which may reducerelevance of the survey or consistently produce inaccurate resultsthat would be used to make future plans and policy arrangementtherefore producing unreliable survey and inaccurate futureinformation.

c)

Interpretationof output

Fromthe diagram it’s evident that respondents would relative orsignificantly prefer blue color of material of clothing relative tored color of materials.

d)

Thefollowing tables shows the color preferences for male respondents

Statistics

gender of respondent

color of material

N

Valid

105

105

Missing

0

0

gender of respondent

Frequency

Percent

Valid Percent

Cumulative Percent

Valid

male

105

100.0

100.0

100.0

color of material

Frequency

Percent

Valid Percent

Cumulative Percent

Valid

blue

31

29.5

29.5

29.5

white

27

25.7

25.7

55.2

black

15

14.3

14.3

69.5

purple

12

11.4

11.4

81.0

orange

12

11.4

11.4

92.4

red

8

7.6

7.6

100.0

Total

105

100.0

100.0

Interpretationof output

Fromthe table 29.5% of the male preferred blue which is the mostpreferred color while 7.6% of the male preferred red color which isthe least preferred

Thefollowing tables shows the color preferences for female respondents.

Statistics

gender of respondent

color of material

N

Valid

190

190

Missing

0

0

gender of respondent

Frequency

Percent

Valid Percent

Cumulative Percent

Valid

female

190

100.0

100.0

100.0

color of material

Frequency

Percent

Valid Percent

Cumulative Percent

Valid

blue

60

31.6

31.6

31.6

white

32

16.8

16.8

48.4

black

26

13.7

13.7

62.1

purple

28

14.7

14.7

76.8

orange

26

13.7

13.7

90.5

red

18

9.5

9.5

100.0

Total

190

100.0

100.0

Interpretationof output

Fromthe table 31.6% of the female preferred blue which is the mostpreferred color while 9.5% of the female preferred red color which isthe least preferred

E(i)

Genderof respondent = female

Statisticsa

Gender of respondent

Color of material

N

Valid

93

93

Missing

0

0

a. Gender of respondent = female

FrequencyTable

Color of materiala

Frequency

Percent

Valid Percent

Cumulative Percent

Valid

blue

30

32.3

32.3

32.3

white

19

20.4

20.4

52.7

black

12

12.9

12.9

65.6

purple

14

15.1

15.1

80.6

orange

13

14.0

14.0

94.6

red

5

5.4

5.4

100.0

Total

93

100.0

100.0

a. Gender of respondent = female

Interpretingthe output

Thefirst present the valid cases for this test which is 93. The secondtable presents frequencies, statistics for the six colors of thematerials.

Thepercentages of cases in each category for the sample as a whole isdisplayed. This includes any values that may be labeled missingvalues which in this case is none. For example 32.3% of the samplecases are coded blue, which implies that amongst the femalerespondents interviewed 32.3% of the preferred color blue of thematerial of clothing.

Thecumulative percentages excluding the missing values is found in thelast column of numbers. Since there are no missing values in thiscategory then the cumulative frequencies of percentages add up or sumup to 100%.

Thethird chart shows a bar graph comparing the materials preferencesamong female respondents in terms of percentages

E(ii)

E(iii)

Group Statistics

gender of respondent

color of material

N

Mean

Std. Deviation

Std. Error Mean

gendr=2 (FILTER)

female

blue

29

1.00

.000a

.000

white

16

1.00

.000a

.000

t cannot be computed because the standard deviations of both groups are 0.

Themeans of white and blue materials are 1.00, therefore aresignificantly equal. The difference between the means is 0.00therefore implying that at 95% confidence level for the differencesis 0.00.Since this interval does not include 0.00 the differences inmean is statistically insignificant at two tailed test of 5% level.

E(IV)

Female Sample

95

Female with white

18

Null Hypothesis

W In White≥0.25

claim

Alternative Hypothesis

W In White&lt0.25

Mean

5.277777778

Standard Deviation

0.044426166

z

-1.362402419

P(z&lt-1.362)

0.008933572

Interpretation:

Thenull hypothesis is rejected since p (0.009) &lt0.05

E(V)

Importanceof hypothesis testing

Hypotheses

Ahypothesis is a claim about a parameter that you’re interested in. The simplest hypotheses are about the parameters of a singlevariable, such as the mean of a population.

(ECON309 Lecture 7B: Hypothesis Testing)

Ahypothesis helps the researcher to prove or disapprove a claim thatis believed to be true or false and therefore making appropriatedecision in regard to the outcome of the research. Should a claimfrom the field has an invalid result, the conclusion, made anddecision recommended to the person interested in the research wouldbe unreliable for future planning.

Sincethe p-value (0.09) from the field data is greater that thehypothetical value of 0.05, the researcher opted to reject or failedto accept the null hypothesis which stated that less than 25% ofwomen did not prefer white clothes but from the sample it’s clearthat over 25% preferred white clothes.

Thereforethe manager will make a decision of stocking more materials/ clothesof white color and fail to appropriately plan for the future.

REFERENCES

Sterlin,Ronald C, and Dniel K. Lapsley. ((1993). “Rational Appraisal ofPsychological Research and the Good-enough Principle.” In G. Keren’and C Lewis, eds.Ahandbook for data analysis in the Behaviorial SciencesMetholodogical issues. Hilldaleds, NJ: Lawrence Erlibaum Associates

StiglerStephen M. 1986. The Historyof Statistics The measurements of Uncertainity before 1900,Cambridge,MA: Harvard University Press.

TannerMartins,1996. Toolsfior Statistical inferencing: Methods for the exploration ofPosterior Distributions and Livelihood Functions. ThirdEdition. New York: Springer-Verlag

Neyman,Jerzy, and Egon S. Pearson 1936a. “Contributions to the Theory ofTesting Statistical Hypothesis” StatisticalResearch Memorandum1, 1-37.

Oakes, M. 1968,StatisticalInferences: A Commentary for the Social and Behavioral Sciences.New York: John Wiley &amp Sons

ECON309 Lecture 7B: Hypothesis Testing

Dixon,W. J., &amp Massey, F. J. (1969).&nbspIntroductionto statistical analysis&nbsp(Vol.344). New York: McGraw-Hill.

Schmist,Frank L. 1996 “Statistical Signifance Testing and CumulativeKnowledge in Psychology. Implications for the Training OfResearchers” PsychologicalMethods.

Learmer,Edward E. 1983. “Lets Takes the Con Out of Econometrics”American Economic Review 73, No 1(march)