One of major computer maker wishes to see if the average length of switching lap
ID: 3252235 • Letter: O
Question
One of major computer maker wishes to see if the average length of switching laptop in the United States is the same as the average length of switching laptop in Europe. The data (days) of a sample of product life cycle in the excel file. At a=1% level, is there enough evidence to reject the claim?
United States
Europe
729
481
560
724
434
820
329
532
332
357
360
505
450
1776
2315
1122
865
496
330
1224
410
634
1036
230
329
1420
800
326
447
626
600
877
1310
580
652
210
1243
447
605
567
360
252
525
824
926
932
722
600
850
634
310
1124
430
1575
532
565
375
405
1979
2290
710
675
545
454
259
300
470
425
United States
Europe
729
481
560
724
434
820
329
532
332
357
360
505
450
1776
2315
1122
865
496
330
1224
410
634
1036
230
329
1420
800
326
447
626
600
877
1310
580
652
210
1243
447
605
567
360
252
525
824
926
932
722
600
850
634
310
1124
430
1575
532
565
375
405
1979
2290
710
675
545
454
259
300
470
425
Explanation / Answer
n1= 36 , n2 = 32
X1bar = 662.6111 X2bar = 758.875
sigma1^2 = 202383.3 , sigma2^2 = 224795.3
To test hypothesis,
H0 : mu1=mu2 vs H1: mu1 is not equal to mu2
Z calc = (X1bar - X2 bar) / SQRT((sigma1^2/n1)+(sigma2^2/n2))
Z calc = (662.6111-758.875)/ SQRT((202383.3/36)+(224795.3))
Z calc = -96.2639 /SQRT(12646.6115)
Z calc = -0.8560
Z table =Z alpha/2 = Z 0.005 = -2.58
Z calc > Z table
We accept H0.
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