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A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company produces. He knows from numerous previous samples that this service life is normally distributed with a mean of 500 hours and a standard deviation of 20 hours. On three recent production batches, he tested service life on random samples of four headlamps, with these results below.
Sample Service Life hours
1 495 500 505 500
2 525 515 505 515
3 470 480 460 470
If he uses upper and lower control limits of 540 and 460 hours. What is his risk alpha of concluding service life is out of control when it is actually under control Type I error?
a. 0.6826
b. 0.0456
c. 0.0026
d. 0.3174
e. 0.9544

Respuesta :

ogorwyne

Here is the correct question.

If he uses upper and lower control limits of 520 and 480 hours, what is his risk (alpha) of concluding that service life is out of control when it is actually under control (Type I error)?

Answer:

B. 0.045

Explanation:

520 = 500+Z*(20/√4)

= 520 = 500+z(10)

= 20 = 10z

Z = 2

480 = 500+Z*(20/√4)

480 = 500-z(10)

Z = 2

When we go to the standard normal table, a z value of 2 has a value of 0.4772 in the table.

2 x 0.4772 = 0.9544

1-0.9544 = 0.0456

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