Question

The life in hours of a thermocouple used in a furnace is known to be ap-

proximately normally distributed, with standard deviation

σ

= 20 hours. A

random sample of 15 thermocouples resulted in he following data: 553, 552,

567, 579, 550, 541, 537, 553, 552, 546, 538, 553, 581, 539, 529.

Use the 5 steps in the testing procedure to solve following question:

a) Is there evidence to support the claim that mean life exceeds 540 hours?

Use a fixed-level test with

α

= 0

.

05.

b) What is the P-value for this test?

c) Construct a 95% one-sided lower CI on the mean life.

d) Use the CI found above to test the hypothesis.

e) What is the β-value for this test if the true mean life is 560 hours?

Answer 1

a)

= (553 + 552 + 567 + 579 + 550 + 541 + 537 + 553 + 552 + 546 + 538 + 553 + 581 + 539 + 529)/15 = 551.33

The test statistic z = ()/()

                              = (551.33 - 540)/(20/)

                              = 2.19

At alpha = 0.05, the critical value is z_0.95 = 1.645

Since the test statistic value is greater than the critical value(2.19 > 1.645), so we should reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to support the claim that mean life exceeds 540 hours.

b) P-value = P(Z > 2.19)

                  = 1 - P(Z < 2.19)

                  = 1 - 0.9857

                  = 0.0143

c) The lower limit of the 95% one-sided lower confidence interval is

= 551.33 - 8.49

= 542.84

So the 95% one-sided lower confidence interval is (542.84, ).

d) Since the hypothesized value 540 is less than the lower limit of the interval(540 < 542.84), so we should reject the null hypothesis.

e) z_crit = 1.645

or, ( - )/() = 1.645

or, ( - 540)/(20/) = 1.645

or, = 1.645 * (20/) + 540

or, = 548.49

= P( < 548.49)

   = P( - )/() < (548.49 - )/())

   = P(Z < (548.49 - 560)/(20/))

   = P(Z < -2.23)

   = 0.0129