Question
In: Statistics and Probability
A sample of 16 joint specimens of a particular type gave a sample mean proportional limit...
A sample of 16 joint specimens of a particular type gave a sample mean proportional limit stress of 8.44 MPa and a sample standard deviation of 0.73 MPa.
(a) Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of all such joints. (Round your answer to two decimal places.)
(b) Calculate and interpret a 95% lower prediction bound for proportional limit stress of a single joint of this type. (Round your answer to two decimal places.)
Solutions
orchestra answered 1 year agoRelated Solutions
A sample of 16 joint specimens of a particular type gave a sample mean proportional limit...
A sample of 16 joint specimens of a particular type gave a
sample mean proportional limit stress of 8.55 MPa and a sample
standard deviation of 0.79 MPa.
a)Calculate and interpret a 95% lower confidence bound for the
true average proportional limit stress of all such joints. (Round
your answer to two decimal places.)
b)Calculate and interpret a 95% lower prediction bound for
proportional limit stress of a single joint of this type. (Round
your answer to two decimal places.)
A sample of 13 joint specimens of a particular type gave a sample mean proportional limit...
A sample of 13 joint specimens of a particular type gave a
sample mean proportional limit stress of 8.56 MPa and a sample
standard deviation of 0.75 MPa.
(a) Calculate and interpret a 95% lower confidence bound for the
true average proportional limit stress of all such joints. (Round
your answer to two decimal places.)
(b) Calculate and interpret a 95% lower prediction bound for
proportional limit stress of a single joint of this type. (Round
your answer to two...
A sample of 18 joint specimens of a particular type gave a sample mean proportional limit...
A sample of 18 joint specimens of a particular type gave a
sample mean proportional limit stress of 8.54 MPa and a sample
standard deviation of 0.71 MPa.
(a) Calculate and interpret a 95% lower confidence bound for the
true average proportional limit stress of all such joints. (Round
your answer to two decimal places.)
Interpret this bound.
With 95% confidence, we can say that the value of the true mean
proportional limit stress of all such joints is less...
#1) A sample of 12 joint specimens of a particular type gave a sample mean proportional...
#1)
A sample of 12 joint specimens of a particular type gave a
sample mean proportional limit stress of 8.57 MPa and a sample
standard deviation of 0.78 MPa.
(a) Calculate and interpret a 95% lower confidence bound for the
true average proportional limit stress of all such joints. (Round
your answer to two decimal places.)
_______ MPa
Interpret this bound. (pick one)
o With 95% confidence, we can say that the value of the true
mean proportional limit stress...
An Izod impact test was performed on 20 specimens of PVC pipe. The sample mean is...
An Izod impact test was performed on 20 specimens of PVC pipe.
The sample mean is x Overscript bar EndScripts equals 1.44 and the
sample standard deviation is s = 0.27. Find a 99% lower confidence
bound on the true Izod impact strength. Assume the data are
normally distributed. Round your answer to 3 decimal places.
less-than-or-equal-to
An Izod Impact Test was performed on 20 specimens of PVC pipe. The sample mean is...
An Izod Impact Test was performed on 20 specimens of PVC pipe.
The sample mean is ? = 1.25, and the sample ??? ??????? ? standard
deviation is ? = 0.25. You need to test if the true mean Izod
Impact Strength is less than 1.5.
a.Write down the Null and Alternative Hypotheses to be
tested
b.What is the appropriate type of statistical test? Explain.
c.Construct a 95% ?????????? ???????? and test the Hypotheses.
Clearly interpret the test result.
d.Test...
A random sample of n1 = 16 communities in western Kansas gave the following information for...
A random sample of n1 = 16 communities in western Kansas gave
the following information for people under 25 years of age.
x1: Rate of hay fever per 1000 population for people under 25 96
92 120 130 91 123 112 93 125 95 125 117 97 122 127 88 A random
sample of n2 = 14 regions in western Kansas gave the following
information for people over 50 years old.
x2: Rate of hay fever per 1000 population for...
A random sample of n1 = 16 communities in western Kansas gave the following information for...
A random sample of n1 = 16 communities in
western Kansas gave the following information for people under 25
years of age.
x1: Rate of hay fever per 1000
population for people under 25
96
88
122
130
90
123
112
93
125
95
125
117
97
122
127
88
A random sample of n2 = 14 regions in
western Kansas gave the following information for people over 50
years old.
x2: Rate of hay fever per 1000
population for...
A random sample of n1 = 16 communities in western Kansas gave the following information for...
A random sample of n1 = 16 communities in
western Kansas gave the following information for people under 25
years of age.
x1: Rate of hay fever per 1000
population for people under 25
100
92
122
129
94
123
112
93
125
95
125
117
97
122
127
88
A random sample of n2 = 14 regions in
western Kansas gave the following information for people over 50
years old.
x2: Rate of hay fever per 1000
population for...
A random sample of n1 = 16 communities in western Kansas gave the following information for...
A random sample of n1 = 16 communities in
western Kansas gave the following information for people under 25
years of age.
x1: Rate of hay fever per 1000
population for people under 25
98
90
120
128
92
123
112
93
125
95
125
117
97
122
127
88
A random sample of n2 = 14 regions in
western Kansas gave the following information for people over 50
years old.
x2: Rate of hay fever per 1000
population for...
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