Decision-making methods and tools Essay

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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

1stquartile

-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

3rdquartile

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

( hypertext transfer protocol: //onlinestatbook.com/2/estimation/mean.html )

=== 18.045

  • Omega tonss( as indicated in table above )

3.Outliers

To observe outliers, we can utilize the Z-scores, presuming that the information follows a normal distribution, and Z-scores that exceed ±3 in value are considered as outliers ( Seo, 2002 ) .

For this information set, the Z-scores of the values fall between: -1.893 and 1.513.

Therefore, there are no outliers because all the information values have a Z-score 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

Inter-quartile 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 z-table 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 z-table 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 z-table 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/z-table-left-of-curve/ )

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

( tenI– ? )

( tenI– ?I„ )2

( tenI– ? )

( tenI– ?I„ )2

( tenI– ? )

( tenI– ?I„ )2

( tenI– ? )

( tenI– ?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)

= ( tenI– ?I„ )2/ N

330 / 16 =

20.63

192.55/11 =

17.5

166.4/10 =

16.64

120.22/9 =

13.36

  1. Develop the hypotheses

Null hypotheses: Hydrogen0: All the agencies are equal: µA= µBacillus= µC= µCalciferol

Alternate hypotheses: HydrogenA: Not all the agencies are equal

  1. Specify degree of significance

? = 0.05

  1. Calculate the value of the trial statistic
  • Mean square due to interventions:

whereis 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 ) =whereis 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?2is the discrepancy

= () + () + ()

= 16-1 ( 20.63 ) + 11-1 ( 17.5 ) + 10-1 ( 16.64 ) + 9-1 ( 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 ) =whereis 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 F0.05 ; 3,42we use the F-distribution tabular array for ? = 0.05 ( see infusion of the F- distribution tabular array below bespeaking the F-value )

( Beginning:hypertext transfer protocol: //sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704_HypothesisTesting-ANOVA/ftable.pdf )

The trial statistic is the F value =0.453. Using the degree of significance ? = 0.05, the critical value for F0.05 ; 3,42=2.827

Decision:

Since the trial statistic ( F value =0.453) is much smaller than the critical value ( F0.05 ; 3,42=2.827) , wefail to reject the void hypothesis(Hydrogen0: All the agencies are equal) , and conclude that there is no important difference among the population means.

Problem 3

Part A

  1. 95 % Confidence Interval computation

On-line trial (N1= 17 )

Paper-pencil trial (N2= 23 )

for performance/scores achieved

for continuance

On-line trial: norm (1)

9.53

40.53

Paper-pencil trial: norm (2)

10.74

34.26

On-line trial: Std Deviation (s1)

3.710

7.559

Paper-pencil trial: Std Deviation (s2)

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: T-test 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 ) =n1+n2– 2

= 17 + 23 – 2 = 38

= 17 + 23 – 2 = 38

Look up critical value ofT0.025, DFfrom t-distribution tabular array

T0.025, 38= 2.0244

T0.025, 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: Omega-test for comparing two agencies

( To acquire the critical value, look up the country in the z-table 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

  1. Develop the hypotheses

Null hypotheses: Hydrogen0: There is no difference of mean tonss between online and paper-pencil trial: µA= µBacillus

Alternate hypotheses: HydrogenA: There is a difference of mean tonss between online and paper-pencil trial: µA? µBacillus

  1. Specify degree of significance

? = 0.05

  1. Calculate the value of the trial statistic

We are comparing agencies of two samples, therefore we use a two sample t-test.

A

Bacillus

Sample size (N )

17

23

Degrees of freedom (N-1)

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 t-score:

T === 1.0008

Entire grades of freedom is: 16 + 22 = 38

Get the critical value in the t-distribution critical values table ( see below ) , 0.05 significance, at 38 grades of freedom.

T0.05, 38= 2.0244, t-score = 1.0008& lt ;T0.05, 38

Decision:

The t-score 1.0008 is less than the critical value of T0.05, 38= 2.0244 from the tabular array, therefore wefail to reject the void hypothesis H0. This means that there is no important difference between the mean tonss of online and paper-pencil trial.

Hypothesis Testing – for continuance

  1. Develop the hypotheses

Null hypotheses: Hydrogen0: There is no difference in the continuance means between online and paper-pencil trial: µA= µBacillus

Alternate hypotheses: HydrogenA: There is a difference in the continuance means between online and paper-pencil trial: µA? µBacillus

  1. Specify degree of significance

? = 0.05

  1. Calculate the value of the trial statistic

We are comparing agencies of two samples, therefore we use a two sample t-test.

A

Bacillus

Sample size (N )

17

23

Degrees of freedom (N-1)

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

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