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WK 4 - Apply: Signature Assignment: Globalization and Information Research
ProtocolType QueueTime ServiceTime
PE 15 39 ProtocolType: indicates protocol type, either PT or PE
PE 27 38 QueueTime: Time in Queue, in seconds
PT 27 284 ServiceTime: Service Time, in seconds
PE 29 36 Perform a test of hypothesis to determine whether the average TiQ is lower than the industry standard of 2.5 minutes (150 seconds). Use a significance level of α=0.05.  
PE 29 47 Evaluate if the company should allocate more resources to improve its average TiQ. 
PT 29 277 Perform a test of hypothesis to determine whether the average ST with service protocol PE is lower than with the PT protocol. Use a significance level of α=0.05.  
PE 31 206 Assess if the new protocol served its purpose. (Hint: this should be a test of means for 2 independent groups.) 
PE 31 46 Submit your calculations and a 175-word summary of your conclusions. 
PT 31 315
PE 35 188
PE 37 40
PE 37 44
PT 41 274
PT 41 275 POST IN ANNOUNCEMENT SECTION FOR STUDENTS
PT 41 288
PT 41 310
PE 43 976
PE 43 43
PT 43 350
PT 45 369
PE 47 214
PE 47 47
PT 47 295
PT 49 345
PE 51 38
PE 51 44
PT 53 246
PE 55 46
PE 55 47
PT 55 219
PT 55 269
PE 57 35
PE 57 47
PT 57 310
PE 59 41
PT 59 203
PT 59 347
PE 61 36
PE 61 37
PE 63 45
PE 65 387
PE 65 50
PE 67 243
PT 67 271
PT 67 406
PE 69 36
PE 71 49
PE 75 119
PT 77 479
PE 79 42
PE 83 33
PE 83 49
PT 83 362
PT 85 335
PE 87 34
PT 87 206
PT 87 479
PE 89 37
PT 89 249
PT 89 283
PE 91 34
PT 93 221
PE 95 94
PE 97 204
PE 99 41
PE 99 46
PT 99 265
PE 105 92
PE 105 47
PT 105 285
PE 107 44
PT 111 230
PE 119 207
PT 123 269
PE 127 129
PE 131 83
PE 131 110
PT 131 403
PE 133 42
PT 133 352
PE 135 112
PT 135 485
PE 137 34
PE 139 45
PE 143 110
PT 143 234
PT 143 399
PE 145 45
PT 145 239
PE 147 37
PE 151 33
PT 161 246
PE 173 45
PT 175 378
PE 181 34
PT 183 380
PT 189 235
PE 195 47
PT 201 216
PT 209 207
PE 211 40
PT 213 278
PT 217 379
PE 219 239
PE 219 49
PE 225 36
PE 225 42
PT 253 369
PT 267 276
PE 277 140
PT 279 404
PT 281 265
PT 305 369
PT 311 348
PT 331 283
PT 333 335
PT 347 215
PT 375 370
PT 427 208
PT 481 237
PT 481 249
PT 489 273
PT 489 323
PT 511 253
PT 527 222
PT 585 325
PT 715 329
PT 763 311
PT 767 316

Sheet5

ProtocolType QueueTime ServiceTime for PE ProtocolType QueueTime ServiceTime for PT
PE 15 39 PT 27 284
PE 27 38 ServiceTime PT 29 277 ServiceTime
PE 29 36 PT 31 315 Perform a test of hypothesis to determine whether the average ST with service protocol PE is lower than with the PT protocol. Use a significance level of α=0.05.  
PE 29 47 Mean 90.7936507937 PT 41 274 Mean 303.4393939394 Assess if the new protocol served its purpose. (Hint: this should be a test of means for 2 independent groups.) 
PE 31 206 Standard Error 16.7600633308 PT 41 275 Standard Error 8.5243140196
PE 31 46 Median 45 PT 41 288 Median 284.5
PE 35 188 Mode 47 PT 41 310 Mode 369
PE 37 40 Standard Deviation 133.0288785929 PT 43 350 Standard Deviation 69.2518544684 Ho ST for PE ≥ ST for PT Ha ST for PE < ST for PT
PE 37 44 Sample Variance 17696.6825396825 PT 45 369 Sample Variance 4795.8193473193
PE 43 976 Kurtosis 32.2452670166 PT 47 295 Kurtosis 0.0702590009 a = .05
PE 43 43 Skewness 5.1540679489 PT 49 345 Skewness 0.6855788254 Rej ho Accept Ho
PE 47 214 Range 943 PT 53 246 Range 282 t z test because n1 = 63, and n2=66 > 30
PE 47 47 Minimum 33 PT 55 219 Minimum 203
PE 51 38 Maximum 976 PT 55 269 Maximum 485 fdr: Reject Ho if z < -1.645 -1.645
PE 51 44 Sum 5720 PT 57 310 Sum 20027 =NORM.INV(0.05,0,1)
PE 55 46 Count 63 PT 59 203 Count 66 S -212.65 = -11.31 -1.645
PE 55 47 PT 59 347 18.8
PE 57 35 PT 67 271
PE 57 47 PT 67 406
PE 59 41 PT 77 479
PE 61 36 PT 83 362 D Since z = -11.31 < -1.645 , reject Ho
PE 61 37 PT 85 335
PE 63 45 PT 87 206 Conclusion St for PE is < ST for PT
PE 65 387 PT 87 479
PE 65 50 PT 89 249 Recommendation Yes, new protype (PE) serves its purpose. go with the new protype
PE 67 243 PT 89 283
PE 69 36 PT 93 221
PE 71 49 PT 99 265
PE 75 119 PT 105 285
PE 79 42 PT 111 230
PE 83 33 PT 123 269
PE 83 49 PT 131 403
PE 87 34 PT 133 352
PE 89 37 PT 135 485
PE 91 34 PT 143 234
PE 95 94 PT 143 399
PE 97 204 PT 145 239
PE 99 41 PT 161 246
PE 99 46 PT 175 378
PE 105 92 PT 183 380
PE 105 47 PT 189 235
PE 107 44 PT 201 216
PE 119 207 PT 209 207
PE 127 129 PT 213 278
PE 131 83 PT 217 379
PE 131 110 PT 253 369
PE 133 42 PT 267 276
PE 135 112 PT 279 404
PE 137 34 PT 281 265
PE 139 45 PT 305 369
PE 143 110 PT 311 348
PE 145 45 PT 331 283
PE 147 37 PT 333 335
PE 151 33 PT 347 215
PE 173 45 PT 375 370
PE 181 34 PT 427 208
PE 195 47 PT 481 237
PE 211 40 PT 481 249
PE 219 239 PT 489 273
PE 219 49 PT 489 323
PE 225 36 PT 511 253
PE 225 42 PT 527 222
PE 277 140 PT 585 325
PT 715 329
PT 763 311
PT 767 316
Perform a test of hypothesis to determine whether the average ST with service protocol PE is lower than with the PT protocol. Use a significance level of α=0.05.  
Assess if the new protocol served its purpose. (Hint: this should be a test of means for 2 independent groups.) 
Submit your calculations and a 175-word summary of your conclusions. 

Sheet4

QueueTime Perform a test of hypothesis to determine whether the average TiQ is lower than the industry standard of 2.5 minutes (150 seconds). Use a significance level of α=0.05.  
15
27
27 QueueTime
29 Ho u ≥ 150 Ha u < 150
29 Mean 154.6124031008 0.01
29 Standard Error 13.4189188282 a = 0.05 0.05 1.645
31 Median 99 0.1
31 Mode 41 t z test because n = 129 > 30
31 Standard Deviation 152.4096202507 Reject Ho Accept Ho
35 Sample Variance 23228.6923449612 fdr Reject Ho if z < -1.645
37 Kurtosis 4.7300425271
37 Skewness 2.1348081912 S =154.61-150 4.61 -1.645
41 Range 752 =152.41/Ö129 13.42
41 Minimum 15
41 Maximum 767 z = .344
41 Sum 19945 z > 30 norm.inv ()
43 Count 129 D Since Z = .344 > -1.645, Accept Ho
43 t < 30 t.inv()
43 C TIQ is not lower than the industry standard of 150 sec
45
47 R Nothing to report. All the analysis wanted was to determine if
47 TIQ was lower or not.
47
49
51
51
53
55
55
55
55
57
57
57
59
59
59
61
61
63
65
65
67
67
67
69
71
75
77
79
83
83
83
85
87
87
87
89
89
89
91
93
95
97
99
99
99
105
105
105
107
111
119
123
127
131
131
131
133
133
135
135
137
139
143
143
143
145
145
147
151
161
173
175
181
183
189
195
201
209
211
213
217
219
219
225
225
253
267
277
279
281
305
311
331
333
347
375
427
481
481
489
489
511
527
585
715
763
767

.05

Sheet1

Team A Team B
Eric Maricelli Claudia Contreras
Korey Jones Marcus Oehmke
Christian Alvarez Tony Wilson
Increase Profits Decrease Absentism
Ho u < or equal 100 Ha u > 100
n= 36
a 0.1 Accept Reject
0.9
t 0.1 z test because n = 36 > 30
=NORM.INV(0.9,0,1)
1.282
FDR Reject Ho if z > 1.282
5 = 2
S 2.5
D Since z = 2 > 1.282 , reject Ho
C Lowering premiums helped increase profits
R Continue with new premiums

Sheet2

Team = Group II Team = Group 1
Eric Maricelli Claudia Contreras
Korey Jones Marcus Oehmke
Christian Alvarez Tony Wilson
Selling Insurance lower premiums rate ;;; advertising Decrease Absentism established a reward incentive bonus for none absent employees
Increase Profits Current Status
Current status m = 12
m = 100 s = 2
s = 15 Ho: m ≤ 100 Ha: m > 100 Accept Region n = 16 Ho: m ≥ 12 Ha: m < 12
n = 36 a= 0.1 Reject Region t test a= 0.01 Accept Region
z test a=.01 Reject Region
a = .10 test statistics: z test because n = 36> 30 x bar=10 test statistics: t test because n = 16 < 30
x bar=105 =NORM.INV(0.9,0,1)
FDR: Reject Ho if z > crv = 1.282 =NORM.INV(0.9,0,1) FDR: Reject Ho if t < crv = -2.602 =T.INV(0.01,15)
1.282 -2.602
= (105-100) = 5 = 5 = 2.00
Statistics: 15/Ö36 15/6 2.5 Statistics: = (10-12) = -2 = -2 = -4
2/Ö16 2/4 0.5
Decision Since z = 2 > CRV = 1.282, Reject Ho Decision Since t = -4 < CRV= -2.602, Reject Ho
Conclusion Lower premium rates = advertise worked Conclusion Reward incentive bonus works
Recommendation Continue with advertising with lower premium rates Recommendation Continue with reward incentive idea

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