Question

answer question 1

ars tat9 7.E 034 A sample of 15 joint specimens of a particular type gave a sample mean proportional limit stress of 8.58 MPa and a sample standard deviation of 0.72 MP. (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. O with 9S% 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. We must assume that the sample observations were taken from a normally distributed population O We must assume that the sample observations were taken from a chi-square distributed population We do not need to make any assumptions. (b) C culate and interpret a 95% lower prediction bound for proportional nmt stress of a single 8.18 of this type. (Round your answer to two decimal places.) X MPa of a single joint of this type. if this bound is caiculated for sample after sample, in the long run 95% of these bounds will be centered around this value for the co esponding future vales of the proportional limit stress of a single joint of this type. i, this bound is calculated for sample after semple, in the long run, 95% of these bounds wd prvide a lo er 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, Need Help?

Answer 1

a) Given $\bar x$ = 8.58, s = 0.72 and n = 15

95% confidence interval = ( $\bar x$ $\pm$ Z * $s/\sqrt{n}$ )

= (8.58-(1.96*0.72/sqrt(15)), 8.58+(1.96*0.72/sqrt(15)))

=(8.58-0.3643, 8.58+0.3643)

=(8.2156, 8.9443)

Therefore, the 95% confidence interval for the true average proportional limit stress of all joints is (8.2156, 8.9443)

To construct the required confidence interval we assume that the population distribution follows normal distribution.