Question

The article "Uncertainty Estimation in Railway Track Life-Cycle Cost"† presented the following data on time to repair (min) a rail break in the high rail on a curved track of a certain railway line.

159 120 480 149 270 547 340 43 228 202 240 218

A normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the population distribution of repair time is at least approximately normal. The sample mean and standard deviation are 249.7 and 145.1, respectively.

(a) Is there compelling evidence for concluding that true
average repair time exceeds 200 min? Carry out a test of hypotheses
using a significance level of 0.05.

State the appropriate hypotheses.

*H*_{0}: μ > 200

*H*_{a}: μ = 200*H*_{0}: μ =
200

*H*_{a}: μ <
200 *H*_{0}: μ < 200

*H*_{a}: μ = 200*H*_{0}: μ =
200

*H*_{a}: μ ≠ 200*H*_{0}: μ =
200

*H*_{a}: μ > 200

Calculate the test statistic and determine the *P*-value.
(Round your test statistic to two decimal places and your
*P*-value to four decimal places.)

t | = | |

P-value | = |

What can you conclude?

There is compelling evidence that the true average repair time exceeds 200 min.

There is not compelling evidence that the true average repair time exceeds 200 min.

(b) Using

σ = 150,

what is the type II error probability of the test used in (a)
when true average repair time is actually 300 min? That is, what is
β(300)? (Round your answer to two decimal places.)

β(300) =

Answer #1

Conclusion:

There is not compelling evidence that the true average repair time exceeds 200 min.

b) Here null hypothesis H0 is true. So, there will be no type II error

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specimens with 2% fiber content, the sample mean tensile strength
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x
100
125
125
150
150
200
200
250
250
300
300
350
400
400
y
160
140
190
210
190
320
280
390
420
440
400
590
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670
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2772
2890
3003
2816
2882
Suppose that for a particular application, it is required that
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have been satisfied? State the appropriate hypotheses. (Use
α = 0.05.)
a)H0: μ < 3000
Ha: μ = 3000
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Ha: μ = 3000
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97.0
92.6
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103.5
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86.6
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The normal probability...

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34.2
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