Stewart Flecknoe-Brown MARK: /20

STUDENT ID NO. 30385385 I certify that this assignment is entirely my own work and I have not copied, discussed or shown my solutions to anyone else. I understand that the penalty for doing so is a mark of zero in the assignment. SIGNATURE __________________________

MGSM960

INFORMATION AND DECISION ANALYSIS

North Ryde Campus TERM 1 2017

INDIVIDUAL ASSIGNMENT

Due at the lecture on Wednesday 22 February

Show all working

Lecturer: Dr Vik Kortian

Email: vik.kortian@pi2.com.au

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3 A manufacturer of outboard motors receives a shipment of shear pins to be used in the assembly of its motors. A random sample of 30 pins is selected and tested to determine the amount of pressure required to cause the pin to break. The test results are summarised on the table below.

Table 1: Lab results of shearpins (kNm-2)

Original Data Set Sorted by Break Point Value Sample No.

Break Point (kN/m2) Sample No.

Break Point (kN/m2)

Sorted Sample No.

1 76.05 29 49.60 1 2 77.70 9 49.77 2 3 54.79 28 49.85 3 4 77.19 5 50.02 4 5 50.02 18 52.62 5 6 61.46 22 52.84 6 7 66.64 3 54.79 7 8 64.86 12 56.30 8 9 49.77 27 57.17 9 10 63.86 15 59.37 10 11 66.69 20 59.62 11 12 56.30 24 60.02 12 13 60.99 16 60.31 13 14 66.77 30 60.88 14 15 59.37 13 60.99 15 16 60.31 25 61.02 16 17 70.09 6 61.46 17 18 52.62 21 61.74 18 19 71.07 26 63.00 19 20 59.62 23 63.81 20 21 61.74 10 63.86 21 22 52.84 8 64.86 22 23 63.81 7 66.64 23 24 60.02 11 66.69 24 25 61.02 14 66.77 25 26 63.00 17 70.09 26 27 57.17 19 71.07 27 28 49.85 1 76.05 28 29 49.60 4 77.19 29 30 60.88 2 77.70 30

4 From the sample set, Minitab provides:

Descriptive Statistics: Break Point (kN/m2)

Sum of Variable N N* Mean SE Mean StDev Variance Sum Squares Minimum Break Point (kN/m2) 30 0 61.54 1.44 7.87 61.90 1846.10 115398.06 49.60 Variable Q1 Median Q3 Maximum Range IQR Mode Break Point (kN/m2) 55.92 61.01 66.65 77.70 28.10 10.73 *

Calculations as explanation follow

a) Calculate the measures of central tendency. Mean = (Sum of samples 1 to 30) / 30

= 1846.08/30 = 61.536

Mean as calculated = 61.536 Median = [(30/2)th sample + (30/2 + 1)th sample]/2

= (ordered samples 15 + 16)/ 2 Samples by order, 15 and 16 are: 60.99 & 61.02 Average of 15th and 16th samples = (60.99 + 61.02) / 2 = 61.005

5 Median as calculated = 61.005 ie, 50% of samples are each above and below this value. Mode of the data set is the most commonly recurring number

The sorted Break Point column of samples shows NO recurrences; there is no mode for this sample set.

b) Calculate the measures of variation. Range: the largest value - smallest value Range = 77.70 - 49.60

= 28.10

The Range of the sample is 28.10

Original Data Set Values for standard deviation calculation Sample No.

Break Point (kN/m2) Sample - Mean

Real value of sample-mean

Sample - Mean Squared

1 76.05 -11.94 11.93854419 142.5288375

2 77.70 -11.76 11.76245408 138.3553259

3 54.79 -11.69 11.68921611 136.6377733

4 77.19 -11.52 11.51869435 132.6803196

5 50.02 -8.91 8.913068514 79.44279034

6 61.46 -8.69 8.693604468 75.57875865

7 66.64 -6.75 6.747386485 45.52722438

8 64.86 -5.24 5.236437871 27.42028158

9 49.77 -4.37 4.367602007 19.07594729

10 63.86 -2.16 2.161951203 4.674033003

11 66.69 -1.91 1.912166958 3.656382474

12 56.30 -1.52 1.520442765 2.3117462

13 60.99 -1.23 1.227964103 1.507895839

14 66.77 -0.66 0.6606200302 0.4364188243

15 59.37 -0.55 0.5481399747 0.3004574318

16 60.31 -0.51 0.5118664636 0.2620072766

17 70.09 -0.08 0.07885353642 0.006217880206

18 52.62 0.21 0.205323754 0.04215784395

19 71.07 1.46 1.460194853 2.132169009

20 59.62 2.27 2.271944737 5.161732887

21 61.74 2.32 2.322901929 5.395873373

22 52.84 3.32 3.319314784 11.01785063

23 63.81 5.11 5.108739655 26.09922087

24 60.02 5.15 5.153146044 26.55491415

25 61.02 5.24 5.23663726 27.42236979

26 63.00 8.56 8.556808314 73.21896853

27 57.17 9.53 9.529831903 90.81769609

28 49.85 14.51 14.51217659 210.6032693

29 49.60 15.66 15.6553456 245.0898457

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30 60.88 16.16 16.1597339 261.1369998

Sum 178.9811124 1795.095485 n = 30 Quartiles = 30 / 4

= 7.25 or 8 in whole numbers Q1 = 56.3 Q3 = 66.4

Interquartile Range = 66.4-56.3 = 10.1 S = the square root of (1795.095485 / 30-1)

= 7.867645410816684

The Standard Deviation of the sample is 7.87 Mean Absolute Deviation = (Sum of the real value each sample value less the mean)/number of samples

= 178.9811124/30 = 5.96603708

The Mean Absolute Deviation is 5.97

c) If the shearpins required a tensile strength of 50kN/m2, what percentage of the

sample failed the test (assuming the samples were from a population normally distributed)

Samples 1, 2, 3 from the sorted column are below 50kN/m2.

As a percentage of a normal distribution where the standard deviation is 7.87, 50 is 61.536-50 = 11.536 kN/m2 below the mean 11.536/7.87 = 1.4658 Standard deviations below the mean Or, to obtain the z score: z = (50 - 61.536)/7.87 = -1.47 Probability = 0.5-0.4292 (taken from table 2) = 0.0708 7.08% of the population’s shear pins failed the test.

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d) The IQ’s of the army recruits in a given year are normally distributed with a µ= 110 and σ = 8. The army wants to give special training to the 10% of those recruits with the highest IQ scores. What is the lowest IQ score acceptable for this special training?

The z score for the top 10% is where P = 0.1 From Table 2, Where area =0.4000, z = 1.285 X = 110 + 8 x 1.285 X = 110 + 10.28 X = 120.28 IQ can be a sensitive subject, the army should authoritatively use whole numbers when providing policy that affects people’s career so let’s round the recommendation down to 120.

Training should be given to recruits with a measured IQ above 120.

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Question 2 A method for determine height of individuals was proposed by measuring various parts of the body. Data of 55 people was collected and is summarised in the Assignment Excel file. Use Minitab ’s Best Subset Regression method in determining which variables you would include in your model where the independent (Y variable) is Height and the dependent X variables are LeftArm, RtArm, LeftFoot, RtFoot, LeftHand, RtHand, HeadCirc, and nose. Insert the relevant Minitab output below. State, with reasons, why you have selected your independent variables in the final model. Also examine the model comparing males to females (that is, would you use a different model for males versus females). Provide explanations for your reasoning. ANSWER

MTB > Regress; SUBC> Response 'Height'; SUBC> Nodefault; SUBC> Continuous 'LeftArm' 'RtArm' 'LeftFoot' 'RtFoot' 'LeftHand' 'RtHand' & CONT> 'HeadCirc' 'nose'; SUBC> Terms LeftArm RtArm LeftFoot RtFoot LeftHand RtHand HeadCirc nose; SUBC> Constant; SUBC> Unstandardized; SUBC> Tmethod; SUBC> Tanova; SUBC> Tsummary; SUBC> Tcoefficients; SUBC> Tequation; SUBC> TDiagnostics 0. Regression Analysis: Height versus LeftArm, RtArm, LeftFoot, RtFoot, LeftHand, RtHand, ...

9 Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Regression 8 827.78 103.472 20.97 0.000 LeftArm 1 27.04 27.040 5.48 0.024 RtArm 1 0.02 0.022 0.00 0.947 LeftFoot 1 8.98 8.984 1.82 0.184 RtFoot 1 5.40 5.395 1.09 0.301 LeftHand 1 0.90 0.900 0.18 0.671 RtHand 1 0.02 0.022 0.00 0.947 HeadCirc 1 1.35 1.348 0.27 0.604 nose 1 0.50 0.502 0.10 0.751 Error 46 226.97 4.934 Total 54 1054.75 Model Summary S R-sq R-sq(adj) R-sq(pred) 2.22129 78.48% 74.74% 67.57% Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 17.95 8.21 2.19 0.034 LeftArm 0.769 0.328 2.34 0.024 5.81 RtArm -0.024 0.359 -0.07 0.947 6.00 LeftFoot 0.563 0.417 1.35 0.184 11.89 RtFoot 0.429 0.410 1.05 0.301 12.68 LeftHand 0.306 0.717 0.43 0.671 8.94 RtHand -0.047 0.711 -0.07 0.947 9.36 HeadCirc 0.081 0.156 0.52 0.604 1.34 nose -0.161 0.505 -0.32 0.751 1.16 Regression Equation Height = 17.95 + 0.769 LeftArm - 0.024 RtArm + 0.563 LeftFoot + 0.429 RtFoot + 0.306 LeftHand

- 0.047 RtHand + 0.081 HeadCirc - 0.161

10 nose Fits and Diagnostics for Unusual Observations Std Obs Height Fit Resid Resid 2 79.00 72.90 6.10 3.20 R 3 75.00 74.46 0.54 0.36 X 6 79.00 74.66 4.34 2.18 R R Large residual X Unusual X MTB > R-squared is 78.48%, suggesting that at least some of the variables correlate with Height. R-squared adjusted is not far below at 74.74% with 8 variables in the equation. To find out which variables are the most significant contributors to the correlation, A best subset regression follows: Best Subsets Regression: Height versus LeftArm, RtArm, ... Response is Height L L H L e e e e f R f R a f R t t t t d t t F F H H C n A A o o a a i o R-Sq R-Sq Mallows r r o o n n r s Vars R-Sq (adj) (pred) Cp S m m t t d d c e 1 67.1 66.4 64.5 19.4 2.5599 X 2 77.2 76.4 73.6 -0.3 2.1492 X X 3 77.9 76.6 72.9 0.3 2.1382 X X X 4 78.3 76.6 72.7 1.4 2.1395 X X X X 5 78.4 76.2 71.3 3.1 2.1550 X X X X X 6 78.5 75.8 70.4 5.0 2.1747 X X X X X X 7 78.5 75.3 68.4 7.0 2.1976 X X X X X X X 8 78.5 74.7 67.6 9.0 2.2213 X X X X X X X X

Consideration of variables: R-squared The best single variable is Left Foot, which yields an R-squared value of 67.1. The combination of Left Foot and Left Arm in combination yield an R-squared of 77.2, a

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significant increase. Adding Rt Foot increases R-squared by a smaller amount to 77.9. Putting together the top 4 variables by adding LeftHand gives an R-squared of 78.3 but adding any more variables only increases the value of R-squared marginally. As a point of interest, RtArm and RtHand provide the lowest R-Squared contribution and as it is common knowledge that 90% of most populations are right handed it might be a worthwhile suggestion to control for left handedness in future surveys. Consideration of variables: R-squared (adj) LeftFoot alone contributes an R-squared of 66.4. R-squared goes to a near high of 76.4 when including LeftArm and addition of the next two variables brings it to a uniform maximum of 76.6. This further suggests that a combination of these variables will give the best prediction. R-squared (pred) similarly provides the highest values among the four variables. Consideration of variables: Mallows C-p The Mallows C-p statistic suggests that the top scorer for R-squared, LeftFoot at 19.4, may not be among the best variables for a good prediction by itself. The very low values of -0.3 and 0.3 are good indications for including LeftArm and RtFoot. For 8 variables, a value closer to 8 would be a better indication. The best combination provided by this interpretation of Mallows C-p is the combination of all 7 (7.0) or 8 (9.0) variables. All Variables Based on R-squared and Mallows C-p, the models to be tested are: 1 Left Foot and Left Arm 2 Left Foot, Left Arm and Rt Foot 3 Left Foot, Left Arm, Rt Foot and Left Hand Results of the 3 models as multiple regressions are in the table below:

Model Variables included p-value in regression R-squared

1 Left Foot Left Arm

0.000 0.000

77.23%

2 Left Foot Left Arm Rt Foot

0.089 0.000 0.221

77.89%

3 Left Foot Left Arm Rt Foot Left Hand

0.000 0.111 0.206 0.338

78.30%

12 Only model 1 has p values under 0.05 and the R-squared (77.23%) is not very much lower than 2 or 3. This makes it the most suitable model for predicting height. The Minitab output for Model 1 is as follows:

Regression Analysis: Height versus LeftFoot, LeftArm Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Regression 2 814.56 407.281 88.18 0.000 LeftFoot 1 224.35 224.349 48.57 0.000 LeftArm 1 107.14 107.143 23.20 0.000 Error 52 240.18 4.619 Lack-of-Fit 44 175.14 3.980 0.49 0.937 Pure Error 8 65.04 8.130 Total 54 1054.75 Model Summary S R-sq R-sq(adj) R-sq(pred) 2.14916 77.23% 76.35% 73.65% Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 21.86 3.58 6.10 0.000 LeftFoot 1.023 0.147 6.97 0.000 1.57 LeftArm 0.796 0.165 4.82 0.000 1.57 Regression Equation Height = 21.86 + 1.023 LeftFoot + 0.796 LeftArm Fits and Diagnostics for Unusual Observations Std Obs Height Fit Resid Resid 2 79.000 73.804 5.196 2.49 R 6 79.000 74.771 4.229 2.10 R 25 73.000 69.770 3.230 1.70 X

13 R Large residual X Unusual X The upper end of height value shows unusual fit observations at Height = 79 and 73. The tallest female is 72 so it might be a relevant caution to examine results for male and female separately. The following Best Subsets Regression are for Male data only then female data only Best Subsets Regression: Height versus LeftArm, RtArm, ... Response is Height L L H L e e e e f R f R a f R t t t t d t t F F H H C n A A o o a a i o R-Sq R-Sq Mallows r r o o n n r s Vars R-Sq (adj) (pred) Cp S m m t t d d c e 1 30.1 27.1 15.4 7.2 2.8533 X 2 52.7 48.4 35.6 0.1 2.4011 X X 3 54.9 48.4 23.9 1.2 2.3999 X X X 4 57.7 49.3 19.7 2.1 2.3801 X X X X 5 59.6 49.0 14.0 3.3 2.3877 X X X X X 6 60.2 47.0 11.4 5.1 2.4333 X X X X X X 7 60.4 44.1 0.0 7.0 2.4985 X X X X X X X 8 60.4 40.6 0.0 9.0 2.5754 X X X X X X X X MTB > BReg 'Height' 'LeftArm' 'RtArm' 'LeftFoot' 'RtFoot' 'LeftHand' & CONT> 'RtHand' 'HeadCirc' 'nose' ; SUBC> NVars 1 8; SUBC> Best 1; SUBC> Constant.

14 Best Subsets Regression: Height versus LeftArm, RtArm, ... Response is Height L L H L e e e e f R f R a f R t t t t d t t F F H H C n A A o o a a i o R-Sq R-Sq Mallows r r o o n n r s Vars R-Sq (adj) (pred) Cp S m m t t d d c e 1 51.5 49.4 43.1 -0.4 1.9381 X 2 61.0 57.5 51.1 -2.4 1.7766 X X 3 61.9 56.5 46.8 -0.8 1.7978 X X X 4 62.2 54.7 33.8 1.1 1.8347 X X X X 5 62.3 52.3 22.5 3.0 1.8812 X X X X X 6 62.3 49.7 19.8 5.0 1.9317 X X X X X X 7 62.3 46.8 14.2 7.0 1.9870 X X X X X X X 8 62.3 43.5 0.0 9.0 2.0479 X X X X X X X X For both tests, Left Arm and Left Foot appear to be among the best predictors but Rt Arm has improved. Left Arm for Males has an unacceptably high Mallows C-p at 7.2, suggesting that the data from Males is what discluded a model using only the left arm as a predictor in the mixed gender analysis. A Fit Regression Model where Sex was a categorical predictor gave the following output from Minitab: Regression Analysis: Height versus LeftFoot, LeftArm, Sex Method Categorical predictor coding (1, 0) Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Regression 3 818.32 272.772 58.84 0.000 LeftFoot 1 100.92 100.917 21.77 0.000 LeftArm 1 94.92 94.924 20.48 0.000 Sex 1 3.75 3.753 0.81 0.372 Error 51 236.43 4.636 Lack-of-Fit 44 173.39 3.941 0.44 0.955 Pure Error 7 63.04 9.006 Total 54 1054.75

15 Model Summary S R-sq R-sq(adj) R-sq(pred) 2.15311 77.58% 76.27% 73.35% Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 25.20 5.16 4.88 0.000 LeftFoot 0.908 0.195 4.67 0.000 2.75 LeftArm 0.765 0.169 4.53 0.000 1.64 Sex Male 0.844 0.938 0.90 0.372 2.59 Regression Equation Sex Female Height = 25.20 + 0.908 LeftFoot + 0.765 LeftArm Male Height = 26.04 + 0.908 LeftFoot + 0.765 LeftArm Fits and Diagnostics for Unusual Observations Std Obs Height Fit Resid Resid 2 79.000 73.796 5.204 2.49 R 6 79.000 74.800 4.200 2.08 R 25 73.000 70.331 2.669 1.49 X R Large residual X Unusual X The P-value for Sex is 0.372 suggesting it is not a good variable for predicting height. The Regression Equations for Male and Female are different only by the constant where Male is 26.04 and Female is 25.20. Overall it appears that the L/R side of the body may contribute to the predictability of different measurements and in future studies it might be interesting to control for left handedness. this could point to the accuracy of different measurements being negatively affected by phenotypical development.

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Question 3

You wish to predict computer sales for the next month using data collected over the past 2 years (24 months). You also collected software sales data over the same time period.

In the Excel file use the data with column headings ‘Computer Sales’ and ‘Software Sales’

Since there is not a clear trend or seasonal pattern in the data, use double exponential smoothing to predict sales for the 25th month using for both Computer Sales and Software Sales:

a) Minitab (copy & past output below)

Double Exponential Smoothing for Computer Sales

Data Computer Sales

Length 24

Smoothing Constants

α (level) 0.599409

γ (trend) 0.130921

Accuracy Measures

MAPE 9

MAD 25814

MSD 1027385418

Computer

Time Sales Smooth Predict Error

17 Feb 195000 190483 183724 11276.3

Mar 213330 203511 188818 24512.3

Apr 208005 210528 214304 -6298.7

May 249000 235520 215349 33650.6

Jun 237040 239420 242982 -5941.8

Jul 270412 260799 246416 23996.1

Aug 306200 291570 269678 36521.8

Sep 264500 280049 303315 -38814.5

Oct 247800 264203 288748 -40947.6

Nov 260500 264181 269689 -9188.7

Dec 259000 262984 268945 -9945.3

Jan 290870 281295 266968 23902.1

Feb 294453 291529 287155 7298.3

Mar 271400 282040 297962 -26561.8

Apr 303050 296375 286388 16661.6

May 310200 306928 302031 8169.0

Jun 354210 337791 313224 40985.9

Jul 389700 372717 347304 42395.7

Aug 361400 371077 385557 -24156.6

Sep 388700 386025 382021 6678.8

Oct 441050 423601 397493 43557.1

Nov 346750 383499 438488 -91737.9

Dec 356200 370215 391187 -34986.8

Jan 386500 381956 375157 11342.8

Forecasts

Period Forecast Lower Upper

Feb 387788 324546 451030

Double Exponential Smoothing Plot for Computer Sales

MTB >

Saving file as: ‘C:\Users\Stewart\Google Drive\MGSM 960

18 Information and

Decision Analysis\Assignment\Double Exponential Smoothing Plot for Computer

Sales.mgf’

MTB > Name c10 "SMOO4" c11 "FITS4"

MTB > DES 'Software Sales';

SUBC> Smoothed 'SMOO4';

SUBC> Fits 'FITS4';

SUBC> Forecasts 1;

SUBC> Origin 24;

SUBC> Brief 2;

SUBC> Smplot;

SUBC> Month.

Double Exponential Smoothing for Software Sales

Data Software Sales

19 Length 24

Smoothing Constants

α (level) 0.629875

γ (trend) 0.126681

Accuracy Measures

MAPE 10

MAD 15920

MSD 458620068

Software

Time Sales Smooth Predict Error

Feb 80500 77186 71547 8952.9

Mar 88000 83827 76725 11275.1

Apr 89000 89904 91443 -2442.9

May 110520 104427 94059 16461.1

Jun 120050 116292 109895 10154.5

Jul 134000 129769 122570 11430.0

Aug 140500 139190 136960 3540.1

Sep 135600 139695 146663 -11062.7

Oct 140506 142645 146285 -5778.8

Nov 156800 153829 148774 8026.0

Dec 160000 160222 160599 -598.9

Jan 168900 168176 166943 1956.6

Feb 189000 183838 175054 13946.3

Mar 175600 181607 191829 -16228.8

Apr 150000 164177 188302 -38302.4

May 144500 153130 167816 -23316.1

Jun 189000 176382 154909 34091.2

20 Jul 200050 192955 180881 19168.7

Aug 200658 200038 198984 1674.1

Sep 213000 210483 206201 6799.3

Oct 259000 243524 217188 41811.7

Nov 198768 219050 253566 -54797.7

Dec 189700 202661 224719 -35018.6

Jan 200300 202238 205536 -5235.7

Forecasts

Period Forecast Lower Upper

Feb 204695 165692 243697

Double Exponential Smoothing Plot for Software Sales

MTB >

Saving file as: ‘C:\Users\Stewart\Google Drive\MGSM 960 Information and

Decision Analysis\Assignment\Double Exponential Smoothing Plot for Software

Sales.mgf’

21

b) Single Exponential Smoothing using Excel (Damping factor = 0.1) Copy and paste output below. Include a graph of the output.

Month Computer Sales Software Sales Computer Sales Smoothed

Software Sales Smoothed

1 195000 80500 * *

2 213330 88000 213330 88000

3 208005 89000 208537.5 88900

4 249000 110520 244953.75 108358

5 237040 120050 237831.375 118880.8

6 270412 134000 267153.9375 132488.08

7 306200 140500 302295.3938 139698.808

8 264500 135600 268279.5394 136009.8808

22

9 247800 140506 249847.9539 140056.3881

10 260500 156800 259434.7954 155125.6388

11 259000 160000 259043.4795 159512.5639

12 290870 168900 287687.348 167961.2564

13 294453 189000 293776.4348 186896.1256

14 271400 175600 273637.6435 176729.6126

15 303050 150000 300108.7643 152672.9613

16 310200 144500 309190.8764 145317.2961

17 354210 189000 349708.0876 184631.7296

18 389700 200050 385700.8088 198508.173

19 361400 200658 363830.0809 200443.0173

20 388700 213000 386213.0081 211744.3017

21 441050 259000 435566.3008 254274.4302

22 346750 198768 355631.6301 204318.643

23 356200 189700 356143.163 191161.8643

24 386500 200300 * *

23

Question 4

The board of regents at a university would like to determine if two variables are independent: employee classification (staff, faculty, and administrator) and whether a union should be the sole collective bargaining agent for employee benefits.

A random sample of 200 employees is taken from employee records and each employee is classified according to both variables and is summarised below.

Opinion Employee Classification Favour

Do Not Favour Undecided Total

Staff 30 15 15 60

Faculty 40 50 10 100

Administration 10 25 5 40

Total 80 90 30 200

a) State the Null and Alternate Hypothesis

HO: There is no significant difference in opinion between Staff, Faculty and Administration on whether a union should be the sole collective bargaining agent for employee benefits.

v HA: There is a significant difference in opinion between Staff, Faculty and Administration on whether a union should be the sole collective bargaining agent for employee benefits.

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b) Use Minitab to determine if there are any relationships between a person’s opinion about collective bargaining and employment status atα = 0.05. Copy and paste the Minitab output and write your conclusions below.

Chi-Square Test for Association: Favour by Employee - Summary Report MTB > XAChiSquare 30 40 10 15 50 25 15 10 5; SUBC> XName "Employee"; SUBC> YName "Favour"; SUBC> NXValues 3; SUBC> NYValues 3; SUBC> XLabels "Staff" "Faculty" "Admin"; SUBC> YLabels "Favour" "Don't Favour" "Undecided"; SUBC> Alpha 0.05. Chi-Square Test for Association: Favour by Employee Chi-Square Test for Association: Favour by Employee - Diagnostic Report

25

A significant number of variations in opinion had a measured p-value of under 0.05, therefore the Null Hypothesis is rejected. Minitab output below: Chi-Square Test for Association: Worksheet rows, Employee Classification Rows: Worksheet rows Columns: Employee Classification Do Not Favour Favour Undecided All 1 30 15 15 60 24 27 9 1.500 5.333 4.000 2 40 50 10 100 40 45 15 0.000 0.556 1.667 3 10 25 5 40 16 18 6 2.250 2.722 0.167

26 All 80 90 30 200 Cell Contents: Count Expected count Contribution to Chi-square Pearson Chi-Square = 18.194, DF = 4, P-Value = 0.001 Likelihood Ratio Chi-Square = 18.708, DF = 4, P-Value = 0.001 Based on the P-Value of 0.001 we can reject the Null Hypothesis that Staff, Faculty and Administration has no predictive indication of opinion on collective bargaining. The board of regents will be informed that there is a significant difference in opinion between Staff, Faculty and Administration on whether a union should be the sole collective bargaining agent for employee benefits