Question 8.5
Return
to the fatigue crack-initiation test in Exercise 6.7. Fit a Weibull
distribution with a given shape parameter p = 2.
(a)
Compute the ML estimate of q. What is the practical interpretation of this
estimate?
ML
Estimates of Distribution Parameters
|
Distribution
|
Location
|
Shape
|
Scale
|
Threshold
|
|
Weibull
|
|
2.79248
|
64.62794
|
|
The
estimate shows shape parameter of 2.79 approximately 3. It means that the true shape parameter would
be 3.
(b)
Obtain the SE mean =8.12.
(c)
Compute a conservative 95% confidence interval for q.
Descriptive
Statistics
|
N
|
Mean
|
StDev
|
SE Mean
|
95% CI for
μ
|
|
9
|
57.44
|
24.36
|
8.12
|
(38.72, 76.17)
|
μ: mean of Crack Initiation
The
95% confidence is (38.72, 76.17). It
means that we are 95% confidents that the true mean lies between the values
38.72 and 76.17.
(d)
Plot the Weibull estimate of F(t) along with the nonparametric estimate.
Comment on the adequacy of the Weibull distribution to describe the data.
Using
the estimates, weibull distribution is well spread and covers a significant
range of the data.
(e)
What is an estimate of the .I quantile of the time-to-initiation distribution?
PERCENTILE
(C1, 0.1) = 18.
(f)
Compute a conservative 95% confidence interval for the .1 quantile of the
time-to-initiation distribution.
Descriptive
Statistics
|
N
|
Mean
|
StDev
|
SE Mean
|
95% CI for μ
|
|
9
|
57.44
|
24.36
|
8.12
|
(41.53, 73.36)
|
μ: mean of Crack
Initiation
Known standard deviation = 24.36
Known standard deviation = 24.36
Question 8.7
(a) Make
a lognormal probability plot of the failure data.
(b) Compute
ML estimate of the lognormal F(200) and F(1000) and use these to draw a line
representing the lognormal estimate of F(t).
From the graph; F(200)=0.01 and
F(1000) > 0.99% almost 1.
ML Estimates of Distribution
Parameters
|
Distribution
|
Location
|
Shape
|
Scale
|
Threshold
|
|
|
Lognormal*
|
5.87059
|
|
0.24128
|
|
|
* Scale: Adjusted ML estimate
(c) Use
µ and σ to compute an estimate of the mean of the lognormal distribution
(equation in section 4.6). Compare this
with the estimate of the lognormal median.
Comment on the difference.
Descriptive Statistics
|
N
|
N*
|
Mean
|
StDev
|
Median
|
Minimum
|
Maximum
|
Skewness
|
Kurtosis
|
|
5
|
0
|
362.6
|
84.5299
|
369
|
252
|
474
|
-0.0083322
|
-0.375009
|
The true mean=362.6, standard deviation =84.53 and media
=369. Using the estimates, µ
=6.56 and σ =0.534 the media
(d) Compute
t1, the ML estimate of the 0.1 quantile of the life distribution.
PERCENTILE ('Number of Fans', 0.1) =
1
(e) Compute
an approximate 95% confidence interval for t.1. Include this interval in the
plot in part (a). Explain the interpretation of this interval and the
justification for the approximate method that you use.
Descriptive Statistics
|
N
|
Mean
|
StDev
|
SE Mean
|
95% CI
for μ
|
|
5
|
362.6
|
84.5
|
37.8
|
(257.6, 467.6)
|
μ: mean of Intergrated Circuit
The above results show that we
are 95% confident that the interval (257.6, 467.6) captured the true mean
observed failures. It therefore, provides us with a range of considerable
values for the given data. The mean and media fall within the given range.
(f)
Compute an approximate 95% confidence
interval for t Include this interval in the plot in part (a). Explain the
interpretation of this interval and the justification for the approximate
method that you use.
Descriptive
Statistics
|
N
|
Mean
|
StDev
|
SE Mean
|
95% CI
for μ
|
|
5
|
362.6
|
84.5
|
37.8
|
(288.5, 436.7)
|
μ: mean of intergrated Circuit
Known standard deviation = 84.5
Known standard deviation = 84.5
The
finding here shows that we are 95% confident that the true mean falls in the
range (288.5, 436.7). This is more
appropriate since we obtain values closer the the mean than in the part (e)
above.
Question 8.8
(a) Make
a Weibull probability plot of these data. Determine if the data are consistent
with the specified value of p.
The
results below indicate the data are consistent for P>0.250.
Goodness
of Fit Test
|
Distribution
|
AD
|
P
|
|
Weibull
|
0.322
|
>0.250
|
(b)
Compute the ML estimate of the Weibull scale parameter q. Compute and graph the
ML estimate of F(t) on the probability plot constructed in part (a).
ML
Estimates of Distribution Parameters
|
Distribution
|
Location
|
Shape
|
Scale
|
Threshold
|
|
Weibull
|
|
6.63308
|
1223.69912
|
|
(c)
Compute a 95% confidence interval for 7.What is the practical interpretation of
this interval?
Descriptive
Statistics
|
N
|
Mean
|
StDev
|
SE Mean
|
95% CI
for μ
|
|
7
|
1140.0
|
214.1
|
80.9
|
(942.0, 1338.0)
|
μ: mean of prototype high power
This means that we are 95%
confident that the interval (942.0, 1338.0) captured the true mean observed
failures. It therefore, provides us with a range of considerable values for the
given data.
d)
Compute the ML estimate and a 95% confidence interval for F(r) at 1000 hours.
(e) Explain the interpretation of the confidence interval from part (d).
Descriptive
Statistics
|
N
|
Mean
|
StDev
|
SE Mean
|
95% CI
for μ
|
|
7
|
1140.0
|
214.1
|
80.9
|
(981.4, 1298.6)
|
μ: mean of prototype high power
Known standard deviation = 214.087
Known standard deviation = 214.087
The 95 % interval is (981.4,1298.6). It implies that we are 95% confident that the
true mean ranges between 981.4 and 1298.6.
Question 8.9
Redo
Exercise 8.8, but use p = 2. Comment on the differences in the results and the
potential effect of using an incorrect value of p = 3, if the actual Weibull
shape parameter is p = 2.
The
results below indicate the data are consistent for P>0.250which is given as
0.641.
Goodness of Fit Test
|
Distribution
|
AD
|
P
|
|
Weibull
|
0.322
|
>0.250
|
The distribution shows for shape
parameters=2 indicates a more plausible data spread as compared for when we use
shape parameter = 3.
Question
8.11
Use
the diesel locomotive fan failure data in Appendix Table C.1 to do the
following:
(a)
Fit an exponential distribution to the fan data.
(b)
Fit a Weibull distribution to the fan data.
We fit the distribution using two
significant parameters that is; shape and the scale. Shape helps demonstrate how the data is
distributed. In this case we will use shape =3 and a scale of 1. The scale also describes how the data is
spread out. We will use a slightly
larger scale to obtain a more spread out results. The overview of our distribution is as
follows;
The
weibull distribution
(c)
Do a likelihood-ratio test that the Weibull shape parameter is equal to 1. How
would the conclusion affect fan replacement policy?
For shape parameters =1, we obtain a
continuous distributions more like exponential. However the likely hood of
approximately 0.05.
(d)
Compute an approximate 95% confidence interval for rl the time at which 10% of
the fan population will fail, based on Zt,1 ~ NOR(0, I ).
Descriptive Statistics
|
N
|
Mean
|
StDev
|
SE Mean
|
95% CI
for μ
|
|
37
|
1.892
|
1.173
|
0.193
|
(1.514, 2.270)
|
μ: mean of Number of Fans
Known standard deviation = 1.173
Known standard deviation = 1.173
(e)
Compute an approximate 95% ~ confidence interval for t I, based on Zt.1~ NOR (0,
1).
Descriptive Statistics
|
N
|
Mean
|
StDev
|
SE Mean
|
95% CI
for μ
|
|
37
|
1.892
|
1.173
|
0.193
|
(1.501, 2.283)
|
μ: mean of Number of Fans
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