# Statistics

A) What is the response variable and what is the factor?

b) Ho many levels of the factor are being studied?

c) Is there any difference in the average time to failure of the disks from the 3 different suppliers? If so, which ones are different?

d) What is your recommendation to the company and why?

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#### Solution Preview

see the attachment

14.2

A company had received complaints that the disks provided by the company failed to work properly after extended use. The company has decided to investigate the complaint. Data are collected on the time to failure for disks from their current supplier and two alternate suppliers. The data are shown below:

Current supplier Alternative 1 supplier Alternative 2 supplier

486 489 508

490 489 510

491 491 517

491 492 506

494 492 515

494 492 520

496 492 503

498 493 524

498 493 515

502 494 503

504 495 507

505 496 515

506 497 509

507 497 518

508 497 495

510 498 499

514 499 510

515 502 503

527 503 533

498 505 517

a) What is the response variable and what is the factor?

The response variable is the time to failure for disks, and the there is only one factor that is the supplier.

b) Ho many levels of the factor are being studied?

There are 3 levels of the factor are being studied. They are Current supplier; Alternative 1 supplier and Alternative 2 supplier. For each level we have 20 sample units.

c) Is there any difference in the average time to failure of the disks from the 3 different suppliers? If so, which ones are different?

We can get the ANVOA table by Excel as follows:

Anova: Single Factor

SUMMARY

Groups Count Sum Average Variance

Column 1 20 10034 501.7 102.3263

Column 2 20 9906 495.3 20.11579

Column 3 20 10227 511.35 81.50263

ANOVA

Source of Variation SS df MS F P-value F crit

Between Groups 2611.233 2 1305.617 19.20545 4.21E-07 3.158846

Within Groups 3874.95 57 67.98158

Total 6486.183 59

Since the p-value is so small that we can reject the hypothesis that there are no differences between the three levels. That mean there must be some differences between three levels.

Now we need to computer the difference between levels:

Let denote the level mean of Current supplier; denote the level mean of Alternative 1 supplier and denote the level mean of Alternative 2 supplier. So from the ANOVA table we know that =501.7, =495.3 and =511.35.

Let - , - , - . So = - =6.4,

= - =-9.65, = - =-16.05. And =sqrt(67.98158*0.1)=2.61

So we construct the null and alternative hypothesis as follows:

H0: =

Ha: (i j)

As we know ...

#### Solution Summary

The response variable and what is the factor are determined. The levels of the factor are being studied are given.