A sample of 16 joint specimens of a particular type gave a sample mean proportio

Question

A sample of 16 joint specimens of a particular type gave a sample mean proportional limit stress of 0.44 MPa and a sample standard deviation of C.77 MPa. (a) Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of all such joints. Interpret this bound. With 95% confidence, we can say that the value of the true mean proportional limit stress of al such joints s less than this value. With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints s centered around this value. With 95% confidence., we can say that the value of the true mean proportional limit stress of all such joints s 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 chi-square distributed population. We must assume that the sample observations were taken from a normally distributed population. We do not need to make any assumptions. We must assume that the sample conservators were taken from a uniformly distributed population. (b) Calculate and interpret a 95% lower production bound for proportional limit stress of a single joint of this type. Interpret this bound. If this bound is calculated for sample after sample, in the long run, 95% of these bounds will provide a lower bound or the corresponding future values of the proportional limit stress of a sing e joint of this type. 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 co-responding future values of the proportional limit stress of a single joint of this type. If this bound s 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.

Explanation / Answer

a) std error of mean=std deviation/(n)1/2 =0.1925

for 15 degree of freedom and 95% lower bound; t=1.753

hence lower bound =sample mean -t*std error =0.10

b) for prediction bound

std error of mean =std deviation(1+1/n)1/2 =0.7937

for 15 degree of freedom and 95% lower bound; t=1.753

hence lower bound =sample mean -t*std error =-0.95

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