A sample of 10 joint specimens of a particular type gave a sample mean proportio
ID: 3366830 • Letter: A
Question
A sample of 10 joint specimens of a particular type gave a sample mean proportional limit stress of 8.5 MPa and a sample standard deviation of 0.77 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. 0 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. 0 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. 0 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. What, if any, assumptions did you make about the distribution of proportional limit stress? O We do not need to make any assumptions. O We must assume that the sample observations were taken from a O We must assume that the sample observations were taken from a O We must assume that the sample observations were taken from a (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.) MPa Interpret this bound. O If this bound is calculated for sample after sample, in the long run 95% of these bounds will provide a higher bound for the corresponding future values of the proportional limit stress of a single joint of this type. O If this bound is calculated for sample after sample, in the long run, 95% of these bounds will provide a lower bound for the corresponding future values of the proportional limit stress of a single joint of this type. O If this bound is calculated for sample after sample, in the long run 95% of these bounds will be centered around this value for the corresponding future values of the proportional limit stress of a single joint of this type.Explanation / Answer
Solution:
From the given information,
Let n = 10 represent the size of the sample.
Let x? = 8.5 reprsent the mean of the sample.
Let s = 0.77 represent the standard deviation of the sample.
The significant level ? = 1-0.95 = 0.05
Degrees of freedom for t-distribution:
df = n-1 = 10-1 = 9
Critical value of one sided confidence:
t?,df = t0.05,9
= 1.833
Margin of Error:
E = t?,df * (s/?n)
= 1.833 * (0.77/?10)
= 1.833 * 0.2435
= 0.4463
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a) Hence the 95% lower confidence bound for the true average proportional limit stress of all such joints is,
CI = x? -E
= 8.5 - 0.4463
= 8.0537
With 95% confidence, the value of the true mean proportional limit stree of all such joints lies in the interval (8.054, ?). If this interval is calculated for sample after sample, in the long run 955 of these intervals will include the true mean proportional limit stress of all such joints.
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b) The 95% lower prediction bound for the true average propotional limit stress of single joint of this type is calculated by using the following formula:
PI = x? - t?,df (S ?1+(1/n) )
= 8.5 - 1.833(0.77 ?1+(1/10))
= 8.5 - 1.833(0.8069)
= 8.5 - 1.4790 = 7.0210