A sample of 20 joint specimens of a particular type gave a sample mean proportio
ID: 3151242 • Letter: A
Question
A sample of 20 joint specimens of a particular type gave a sample mean proportional limit stress of 8.51 MPa and a sample standard deviation of 0.76 MPa.
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.
With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is centered around this value.
With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is less than this value.
With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is greater than this value.
What, if any, assumptions did you make about the distribution of proportional limit stress?
We must assume that the sample observations were taken from a uniformly distributed population.
We do not need to make any assumptions.
We must assume that the sample observations were taken from a chi-square distributed population.
We must assume that the sample observations were taken from a normally distributed population.
Explanation / Answer
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
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
where
alpha = (1 - confidence level) = 0.05
X = sample mean = 8.51
t(alpha) = critical t for the confidence interval = 1.729132812
s = sample standard deviation = 0.76
n = sample size = 20
df = n - 1 = 19
Thus,
Lower bound = 8.216149153 [ANSWER]
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b)
OPTION C: With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is greater than this value. [ANSWER]
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c)
OPTION D: We must assume that the sample observations were taken from a normally distributed population. [ANSWER]