HW 6
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Auburn University *
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Course
5720
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
doc
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Uploaded by ChancellorFlagFinch34
4-39 Vendor quality control is an integral part of a total quality system. A soft
drink bottling company requires its vendors to produce bottles with an internal
pressure strength of at least 300 kg/cm2. A vendor claims that its bottles have a
mean strength of 310 kg/cm2 with a standard deviation of 5 kg/cm2. As part of a
vendor surveillance program, the bottling company samples 50 bottles from the
production and finds the average strength to be 308.6 kg/cm2.
(a) What are the chances of getting that sample average that was observed, or
even less, if the assertion by the vendor is correct?
Sample Standard Deviation = 5/√50 = 0.707
Z = (308.6-310)/0.707= -1.98
P=2.39%
(b) If the standard deviation of the strength of the vendor’s bottles is 8 kg/cm2,
with the mean (as claimed) of 310 kg/cm2, what are the chances of seeing what
the bottling company observed (or an even smaller sample average)?
Sample Standard Deviation = 8/√50 = 1.131
Z = (305.6-310)/1.131 = -1.24
P=10.75%
4-44 A company that dumps its industrial waste into a river has to meet certain
restrictions. One particular constraint involves the minimum amount of
dissolved oxygen that is needed to support aquatic life. A random sample of 10
specimens taken from a given location gives the following results of dissolved
oxygen (in parts per million, ppm):
9.0 8.6 9.2 8.4 8.1
9.5 9.3 8.5 9.0 9.4
(a) Find a two-sided 95% confidence interval for the mean dissolved oxygen and
interpret it.
Mean= (9.0+8.6+9.2+8.4+8.1+9.5+9.3+8.5+9.0+9.4)/10=8.9
Standard deviation= 0.4738
+ t (α/2, n-1) S/√(n)=8.9-t (0.025,10-1) *0.4738/√10) =8.5611
X
̅
- t (α/2, n-1) S/√(n)=8.9+t (0.025,10-1) *0.4738/√10) =9.2389
X
̅
95% Confidence interval is (8.5611,9.2389)
There are 95% probability that river have the dissolved oxygen between is
(8.5611,9.2389).
(b) What assumptions do you need to make to answer part (a)?
Among these assumptions, the data must be randomly sampled from the population of
interest and the data variables must follow a normal distribution.
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