Decisionmaking methods and tools
DECISION MAKING METHODS & A ; TOOLS 
Problem 1
Workss:
Model 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
Cost 
220 
270 
280 
290 
310 
340 
340 
340 
370 
380 
400 
430 
450 
450 
Cost 
220 
270 
280 
290 
310 
340 
370 
380 
400 
430 
450 
Frequency 
1 
1 
1 
1 
1 
3 
1 
1 
1 
1 
2 
First quartile: 25th percentile ( 25 % x N ) = 25 % x 14 = 3.5 unit of ammunition up to 4
Second quartile: 50th percentile ( 50 % x N ) = 50 % x 14 = 7
Model 
Cost tenI 
Quartile 
Mean Deviation ( tenI– µ ) 
( tenI– µ )2 
( tenI– µ )3 
Omega tonss ( tenI– µ ) / ? 

1 
220 
127.857 
16347.412 
2090131.113 
1.893 

2 
270 
77.857 
6061.712 
471946.746 
1.153 

3 
280 
1^{st}quartile 
67.857 
4604.572 
312452.473 
1.005 

4 
290 
57.857 
3347.432 
193672.399 
0.857 

5 
310 
37.857 
1433.152 
54254.852 
0.561 

6 
340 
7.857 
61.732 
485.032 
0.116 

7 
340 
7.857 
61.732 
485.032 
0.116 

8 
340 
7.857 
61.732 
485.032 
0.116 

9 
370 
22.143 
490.312 
10856.988 
0.328 

10 
380 
3^{rd}quartile 
32.143 
1033.172 
33209.262 
0.476 

11 
400 
52.143 
2718.892 
141771.209 
0.772 

12 
430 
82.143 
6747.472 
554257.629 
1.216 

13 
450 
102.143 
10433.192 
1065677.576 
1.513 

14 
450 
102.143 
10433.192 
1065677.576 
1.513 

Sums 
4870 
0 
63835.708 
252462.439 

More workings: 

Nitrogen= 14 
?xi = 4870 == 347.857 
? ( eleven – µ ) 2 =63835.708 ?2== = 4559.693 ?= v ?2 = v4559.693 =67.525 
? ( eleven – µ ) 3 = 252462.439 

Third quartile: 75th percentile ( 75 % x N ) = 75 % x 14 = 10.5 unit of ammunition up to 11
1.
Mean ()=347.857
Median =340
Mode =340
First quartile =290
Third quartile =400
2.
 Discrepancy (?2) = 4559.693
?2= ? ( xi – µ ) 2 /Nitrogen= 63835.708 / 14 = 4559.693
 Standard divergence (? )= 67.525
? = v ?2 = v4559.693 = 67.525
 Range = 230
highest value – lowest value= 450 – 220 = 230
 Interquartile scope = 110
3rd quartile – foremost quartile= 400 – 290 = 110
 Coefficient of fluctuation = 19.412 %
(*100 ) % = 19.412 %
 Standard mistake of the average = 18.045
=== 18.045
 Omega tonss( as indicated in table above )
3.Outliers
To observe outliers, we can utilize the Zscores, presuming that the information follows a normal distribution, and Zscores that exceed ±3 in value are considered as outliers ( Seo, 2002 ) .
For this information set, the Zscores of the values fall between: 1.893 and 1.513.
Therefore, there are no outliers because all the information values have a Zscore greater than 3 or less than +3.
Another method we can utilize to observe outliers is by doing aBox Plotutilizing the following informations that we have:
Quartile 1 ( Q1 ) = 290
Quartile 2 ( Q2 ) = 340
Quartile 3 ( Q3 ) = 400
Interquartile scope ( IQR ) = 110
1.5 ten IQR = 165
Lower bound = Q1 – IQR = 290 – 165 = 125
Upper bound = Q3 + IQR = 400 + 165 = 565
From the above box secret plan, it is clear that there is no information in the outlier country ( shaded red ) , hence this information set has non outliers.
4.Lopsidedness
Skewness value =( Indiana University, no day of the month )
==<= 0.063
Lopsidedness= 0.063
A skewness value of 0.063 agencies that the informations are reasonably skewed left ( or instead, really somewhat skewed left ) .
5.Decisions
 The mean monetary value of digital cameras is 347.857
 Price scope of cameras: lowest monetary value is 220 and highest monetary value is 450
 75 % of cameras cost 400 or below
 Most common monetary value is 450
6.Assurance interval for the population mean cost of digital cameras
90 % Assurance Time interval:
Alpha ( ? ) = 1 – ( assurance interval / 100 )
= 1 – ( 90/100 )
= 1 – 0.9
? = 0.1
Critical chance ( p* ) = 1 –
= 1 – 0.05
p* = 0.95
To acquire the critical value, look up the country in the ztable for value of 0.95. The critical value is 1.645. ( see infusion of the Z tabular array below bespeaking the critical value )
Margin of mistake = critical value x standard mistake of the mean ( Stat Trek, no day of the month )
= 1.645 ten 18.045 = 29.684
So the border of mistake is ± 29.684, therefore:
The 90 % assurance interval is:347.857±29.684
or318.173 & lt ; = x & lt ; = 377.541
95 % Assurance Time interval:
Alpha ( ? ) = 1 – ( assurance interval / 100 )
= 1 – ( 95/100 )
= 1 – 0.95
? = 0.05
Critical chance ( p* ) = 1 –
= 1 – 0.025
p* = 0.975
To acquire the critical value, look up the country in the ztable for value of 0.975. The critical value is 1.96. ( see infusion of the Z tabular array below bespeaking the critical value )
Margin of mistake = critical value x standard mistake of the mean ( Stat Trek, no day of the month )
= 1.96 ten 18.045 = 35.368
So the border of mistake is ± 35.368, therefore:
The 95 % assurance interval is:347.857±35.368
or312.489 & lt ; = x & lt ; = 383.225
99 % Assurance Time interval:
Alpha ( ? ) = 1 – ( assurance interval / 100 )
= 1 – ( 99/100 )
= 1 – 0.99
? = 0.01
Critical chance ( p* ) = 1 –
= 1 – 0.005
p* = 0.995
To acquire the critical value, look up the country in the ztable for value of 0.995. The critical value is 2.578. ( see infusion of the Z tabular array below bespeaking the critical value )
Margin of mistake = critical value x standard mistake of the mean ( Stat Trek, no day of the month )
= 2.578 ten 18.045 = 46.52
So the border of mistake is ± 46.52, therefore:
The 99 % assurance interval is:347.857±46.52
or301.337 & lt ; = x & lt ; = 394.377
( Beginning:hypertext transfer protocol: //www.statisticshowto.com/tables/ztableleftofcurve/ )
7.Interpret significance of each interval
 The 90 % assurance interval is: 347.857 ±29.684. This means that we can be 90 % certain that the true mean of the population is contained in the 90 % assurance interval scope:318.173 & lt ; = x & lt ; = 377.541
 The 95 % assurance interval is: 347.857 ±35.368. This means that we can be 95 % certain that the true mean of the population is contained in the 95 % assurance interval scope:312.489 & lt ; = x & lt ; = 383.225
 The 99 % assurance interval is: 347.857 ±46.52. This means that we can be 99 % certain that the true mean of the population is contained in the 99 % assurance interval scope:301.337 & lt ; = x & lt ; = 394.377
8.Premises made:
 The information is usually distributed
 The monetary values of the digital cameras are right as given ( free from mistakes )
 The information given is independent ; the monetary value of one camera does non impact the monetary value of another
Problem 2
Trial for equality of the agencies for the 4 population sample informations for workss A, B, C, D.
Method to utilize: Analysis of Variance ( ANOVA ) .
ANOVA is a statistical trial that analyses discrepancy. It helps in doing comparing of two or more agencies which so allows us to do assorted anticipations about two or more sets of informations ( NCS Pearson, 2014 ) . Calculation of the mean and discrepancy for each population sample, as shown in table below:
A 
Deviation from Mean 
Bacillus 
Deviation from Mean 
C 
Deviation from Mean 
Calciferol 
Deviation from Mean 

Nitrogen 
( ten_{I}– ? ) 
( ten_{I}– ?I„ )^{2} 
( ten_{I}– ? ) 
( ten_{I}– ?I„ )^{2} 
( ten_{I}– ? ) 
( ten_{I}– ?I„ )^{2} 
( ten_{I}– ? ) 
( ten_{I}– ?I„ )^{2} 

1 
333 
6 
36 
328 
0.64 
0.4 
320 
6.4 
40.96 
333 
4.44 
19.75 

2 
332 
5 
25 
329 
1.64 
2.68 
321 
5.4 
29.16 
321 
7.56 
57.09 

3 
320 
7 
49 
323 
4.36 
19.04 
322 
4.4 
19.36 
329 
0.44 
0.2 

4 
326 
1 
1 
327 
0.36 
0.13 
325 
1.4 
1.96 
330 
1.44 
2.09 

5 
328 
1 
1 
330 
2.64 
6.95 
326 
0.4 
0.16 
328 
0.56 
0.31 

6 
325 
2 
4 
331 
3.64 
13.22 
328 
1.6 
2.56 
325 
3.56 
12.64 

7 
323 
4 
16 
332 
4.64 
21.5 
329 
2.6 
6.76 
333 
4.44 
19.75 

8 
321 
6 
36 
320 
7.36 
54.22 
330 
3.6 
12.96 
331 
2.44 
5.98 

9 
334 
7 
49 
324 
3.36 
11.31 
331 
4.6 
21.16 
327 
1.56 
2.42 

10 
322 
5 
25 
323 
4.36 
19.04 
332 
5.6 
31.36 

11 
320 
7 
49 
334 
6.64 
44.04 

12 
329 
2 
4 

13 
330 
3 
9 

14 
328 
1 
1 

15 
331 
4 
16 

16 
330 
3 
9 

TOTAL ( ? ) 
5232 
0 
330 
3601 
0 
192.55 
3264 
0 
166.4 
2957 
0 
120.22 

Sample Mean () = ? / N 
327 
327.36 
326.4 
328.56 

VARIANCE ( ?^{2}) = ( ten_{I}– ?I„ )^{2}/ N 
330 / 16 = 
20.63 
192.55/11 = 
17.5 
166.4/10 = 
16.64 
120.22/9 = 
13.36 
 Develop the hypotheses
Null hypotheses: Hydrogen_{0}: All the agencies are equal: µ_{A}= µ_{Bacillus}= µ_{C}= µ_{Calciferol}
Alternate hypotheses: Hydrogen_{A}: Not all the agencies are equal
 Specify degree of significance
? = 0.05
 Calculate the value of the trial statistic
 Mean square due to interventions:
where is the entire figure of populations
= = 327.33
SST ( Sum of Squares due to Treatment ) =
=
= 1.742 + 0.0099 + 8.649 + 13.616
SST = 24.0169
MST ( Mean Sum of Squares due to Treatment ) =where is the grade of freedom ( entire figure of populations – 1 )
MST = = = 8.0056
 Mean square due to mistake:
SSE ( Sum of Squares due to Error ) = whereis the entire figure of samples in a population and?^{2}is the discrepancy
= () + () + (
= 161 ( 20.63 ) + 111 ( 17.5 ) + 101 ( 16.64 ) + 91 ( 13.36 )
= 15 ( 20.63 ) + 10 ( 17.5 ) + 9 ( 16.64 ) + 8 ( 13.36 ) = 309.45 + 175 + 149.76 + 106.88
SSE = 741.09
MSE ( Mean Sum of Squares due to Error ) =where is the entire figure of observations ( N ) less entire figure of populations (== 17.645
F ( Anova coefficient ) === 0.453
ANOVA Table:
Beginning 
Sum of Squares 
Degrees of Freedom 
Mean Sum of Squares 
F 
Treatments 
24.0169 
3 
8.0056 
0.453 
Mistake 
741.09 
42 
17.645 

Entire 
765.1069 
45 
To acquire the critical value for F_{0.05 ; 3,42}we use the Fdistribution tabular array for ? = 0.05 ( see infusion of the F distribution tabular array below bespeaking the Fvalue )
( Beginning:hypertext transfer protocol: //sphweb.bumc.bu.edu/otlt/MPHModules/BS/BS704_HypothesisTestingANOVA/ftable.pdf )
The trial statistic is the F value =0.453. Using the degree of significance ? = 0.05, the critical value for F_{0.05 ; 3,42}=2.827
Decision:
Since the trial statistic ( F value =0.453) is much smaller than the critical value ( F_{0.05 ; 3,42}=2.827) , wefail to reject the void hypothesis(Hydrogen_{0}: All the agencies are equal) , and conclude that there is no important difference among the population means.
Problem 3
Part A
 95 % Confidence Interval computation
Online trial (N_{1}= 17 ) Paperpencil trial (N_{2}= 23 ) 
for performance/scores achieved 
for continuance 
Online trial: norm ( _{1}) 
9.53 
40.53 
Paperpencil trial: norm ( _{2}) 
10.74 
34.26 
Online trial: Std Deviation (s_{1}) 
3.710 
7.559 
Paperpencil trial: Std Deviation (s_{2}) 
3.840 
7.479 
Difference in average = 
= 9.53 – 10.74 = 1.21 
= 40.53 – 34.26 = 6.27 
Standard Error of = = = = 1.205 
= = = = = 2.407 

Alpha ( ? ) = 1 – ( assurance interval / 100 ) 
= 1 – ( 95/100 ) = 1 – 0.95 ? = 0.05 
= 1 – ( 95/100 ) = 1 – 0.95 ? = 0.05 
Critical chance ( p* ) = 1 = 1 – 0.025 = 0.975 
= 1 –= 1 – 0.025 = 0.975 

METHOD 1: Ttest for comparing two agencies 

utilizing POOLED method since the comparing of the largest to smallest sample standard divergence is ? 2 ( Eberly College of Science, no day of the month ) 
comparing of largest to smallest sample SD 3.84 / 3.71 = 1.035 
comparing of largest to smallest sample SD 7.559 / 7.479 = 1.01 
Degree of freedom ( DF ) =n_{1}+n_{2}– 2 
= 17 + 23 – 2 = 38 
= 17 + 23 – 2 = 38 
Look up critical value ofT_{0.025, DF}from tdistribution tabular array 
T_{0.025, 38}= 2.0244 
T_{0.025, 38}= 2.0244 
Margin of mistake ( ME ) = critical value x standard mistake of the mean 
= 2.0244 ten 1.205 = 2.439 
= 2.0244 ten 2.407 = 4.873 
95 % Confidence Interval is: 
1.21 ± 2.439 or 3.649 to 1.229 
6.27 ± 4.873 or 1.397 to 11.143 
METHOD 2: Omegatest for comparing two agencies 

( To acquire the critical value, look up the country in the ztable for value of p* ) ( see infusion of the Z tabular array below bespeaking the critical value ) 
The critical value for p*=0.975 is 1.96. 
The critical value for p*=0.975 is 1.96. 
Margin of mistake ( ME ) = critical value x standard mistake of the mean ( Stat Trek, no day of the month ) 
= 1.96 tens 1.205 = 2.362 
= 1.96 tens 2.407 = 4.718 
95 % Confidence Interval is: 
1.21 ± 2.362 or 3.572 & lt ; 1.21 & lt ; 1.152 
6.27 ± 4.718 or 1.552 & lt ; 6.27 & lt ; 10.988 
Hypothesis Testing – for tonss achieved
 Develop the hypotheses
Null hypotheses: Hydrogen_{0}: There is no difference of mean tonss between online and paperpencil trial: µ_{A}= µ_{Bacillus}
Alternate hypotheses: Hydrogen_{A}: There is a difference of mean tonss between online and paperpencil trial: µ_{A}? µ_{Bacillus}
 Specify degree of significance
? = 0.05
 Calculate the value of the trial statistic
We are comparing agencies of two samples, therefore we use a two sample ttest.
A 
Bacillus 

Sample size (N ) 
17 
23 
Degrees of freedom (N1) 
16 
22 
Sample Mean () 
9.53 
10.74 
Std Deviation ( SD ) 
3.710 
3.840 
We so calculate the pooled criterion divergence ( ) by ciphering the leaden norms of the standard divergences squared ( discrepancies ) . We are utilizing weighted norms because one sample is larger than the other and therefore will hold more grades of freedom.
This is calculated as follows: ( Simon Fraser University, no day of the month )
====3.786
We use= 3.786 to cipher the standard mistake ( SE ) as follows:
Selenium===1.209
To acquire the tscore:
T === 1.0008
Entire grades of freedom is: 16 + 22 = 38
Get the critical value in the tdistribution critical values table ( see below ) , 0.05 significance, at 38 grades of freedom.
T_{0.05, 38}= 2.0244, tscore = 1.0008& lt ;T_{0.05, 38}
Decision:
The tscore 1.0008 is less than the critical value of T_{0.05, 38}= 2.0244 from the tabular array, therefore wefail to reject the void hypothesis H_{0}. This means that there is no important difference between the mean tonss of online and paperpencil trial.
Hypothesis Testing – for continuance
 Develop the hypotheses
Null hypotheses: Hydrogen_{0}: There is no difference in the continuance means between online and paperpencil trial: µ_{A}= µ_{Bacillus}
Alternate hypotheses: Hydrogen_{A}: There is a difference in the continuance means between online and paperpencil trial: µ_{A}? µ_{Bacillus}
 Specify degree of significance
? = 0.05
 Calculate the value of the trial statistic
We are comparing agencies of two samples, therefore we use a two sample ttest.
A 
Bacillus 

Sample size (N ) 
17 
23 
Degrees of freedom (N1) 
16 
22 
Sample Mean () 
40.53 
34.26 
Std Deviation ( SD ) 
7.559 
7.479 
We so calculate the pooled criterion divergence by ciphering the leaden norms of the standard divergences squared ( discrepancies ) . We are utilizing weighted norms because one sample is larger than the other and therefore will hold more grades of freedom.
This is calculated as follows: ( Simon Fraser University, no day of the month )
====7.513
We use