# Decision-making methods and tools

Flashcard maker : Lily Taylor
 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

=== 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

• 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:

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?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 ) =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 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

We use

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