explain in detail
The amount of lateral expansion (mils) was determined for a sample of n : 7 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s : 2.89 mils. Assuming normality, derive a 95% CI far 02 and far (7. (Round yeur answers to two decimal places.) (:,:)mi. A sample of 11 joint specimens of a particular type gave a sample mean proportional limit stress of 8.54 MPa and a sample standard deviation of [1.76 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.) Interpret this bound. O 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 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 less than this value. What, if any, assumptions did you make about the distribution of proportional limit stress? 0 We must assume that the sample observations were taken from a normally distributed population. 0 We do not need to make any assumptions. 0 We must assume that the sample observations were taken from a uniformly distributed population. 0 We must assume that the sample observations were taken from a chi-square distributed population. (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.) Interpret this bound. Q 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. 0 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. 0 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 singlejoint of this type. Silicone implant augmentation rhinoplasty is used to correct congenital nose deformities. The success of the procedure depends on various biomechanical properties of the human nasal periosteum and fascia. An article reported that for a sample of 20 (newly deceased) adults, the mean failure strain (%) was 26.0, and the standard deviation was 3.4. (a) Assuming a normal distribution for failure strain, estimate true average strain in a way that conveys information about precision and reliability. (Use a 95% confidence interval. Round your answers to two decimal places.) (:lw:l%) (b) Predict the strain for a single adult in a way that conveys information about precision and reliability. (Use a 95% prediction interval. Round your answers to two decimal places.) (:IWEWD) How does the prediction compare to the estimate calculated in part (a)? O The prediction interval is much wider than the confidence interval in part (a). O The prediction interval is much narrower than the confidence interval in pait (a). O The prediction interval is the same as the confidence interval in part (a)
