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>Disagree
A>Agree
LPA>Liberal Party Agree
LPD>LiberalParty Disagree
LP>Liberal Party Voter
NLP>Not 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 nonliberalparty voter and disapproved the government decision to send moretroops to Iraq.
QUESTION2
tTest: TwoSample Assuming Equal Variances 

  
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<=t) onetail 
0.043871268 

t Critical onetail 
1.782287548 

P(T<=t) twotail 
0.087742536 

t Critical twotail 
2.178812827 
  
INTERPREATION
Thereis no significance difference in the means of the two samples. Thecritical value of the sample statistics is P(T<=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<=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 
  
  
  
  
  
  
  

Number of NoShows 
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

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

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
Statistics^{a} 

Gender of respondent 
Color of material 

N 
Valid 
93 
93 
Missing 
0 
0 

a. Gender of respondent = female 
FrequencyTable
Color of material^{a} 

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 
.000^{a} 
.000 
white 
16 
1.00 
.000^{a} 
.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<0.25 

Mean 
5.277777778 

Standard Deviation 
0.044426166 

z 
1.362402419 

P(z<1.362) 
0.008933572 
Interpretation:
Thenull hypothesis is rejected since p (0.009) <0.05
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 pvalue (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 Goodenough 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: SpringerVerlag
Neyman,Jerzy, and Egon S. Pearson 1936a. “Contributions to the Theory ofTesting Statistical Hypothesis” StatisticalResearch Memorandum1, 137.
Oakes, M. 1968,StatisticalInferences: A Commentary for the Social and Behavioral Sciences.New York: John Wiley & Sons
ECON309 Lecture 7B: Hypothesis Testing
Dixon,W. J., & Massey, F. J. (1969). Introductionto statistical analysis (Vol.344). New York: McGrawHill.
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)