Please give a correct answer with CORRECT NUMBER OF DECIMAL PLACES !!!!! A sampl
ID: 3246532 • Letter: P
Question
Please give a correct answer with CORRECT NUMBER OF DECIMAL PLACES!!!!!
A sample of 20 joint specimens of a particular type gave a sample mean proportional limit stress of 8.43 MPa and a sample standard deviation of 0.74 MPa. (a) Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of all such joints. 8.08 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 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 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 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 observations were taken from a uniformly distributed population. (b) Calculate and interpret a 95% lower prediction 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 for the corresponding future values of the proportional limit stress of a single 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 corresponding future values of the proportional limit stress of a single joint of this type. If this bound is calculated for sample after sample, in the long run 95% of these bonds will provide a higher bound for the corresponding future values of the proportional limit stress of a single joint of this type. You may need to use the appropriate table in the Appendix of Tables to answer this question.Explanation / Answer
sample mean proportional limit stress xbar = 8.43 mpa
sample standard deviation s= 0.74 mpa
sample size = 20
standard error of the sample mean se= s/n = 0.74/20 = 0.1655
Degrees of freedom = 19 and alpha = 0.05
95% lower confidence bound = xbar - t19,0.05 * se
= 8.43 - 1.729 * 0.1655 = 8.144 mpa
(b) Option C is correct.
(c) Option B is correct.
(B) 95% lower prediction bound for proportional limit stress of a single joint
= xbar - t19,0.05 * s * sqrt (1 +1/n)
= 8.43 - 1.729 * 0.74 * sqrt (1 + 1/20)= 7.12 mpa
Option a is corrrect.