problem 9.5 solution which refers to problem 8.11

160 Expect the Unexpected: A First Course in Biostatistics Problem 9.5. Refer to Problem 8.11. Using these data, is there enough evidence to conclude that the mean yield has increased, compared to the average yield of 2.7 kg per year? Problem 9.6. Percutaneous nephrolithotomy (PN) is a surgical procedure used for removing kidney stones by a small puncture incision through the skin. The authors of [13] studied the efficiency of PN in removing kidney stones. The treatment was defined as successful if stones were eliminated or reduced to less than 2 mm after three months. PN was successful for 289 out of 350 patients that underwent the treatment. The traditional open surgery has a success rate of 78%. (a) Give a point estimate for the success rate of PN in treating kidney stones and give the corresponding estimated standard error. (b) Is there significant evidence that the success rate of PN in treating kidney stones is different than the success rate of open surgery? Formulate a null hypothesis and an alternative hypothesis and find the corresponding p-value. Use a = 0.01. (c) Use a 95% confidence interval for comparing the success rate of PN in treating kidney stones with the success rate of open surgery. Problem 9.7. Refer to Problem 4.7. This medication was given to 2 patients and 17 reported a significant reduction in their pain. Using these data, is there enough evidentconfidence interval for the mean water (e) According to Health Canada (source: www.he-sc.gc.ca), water with a hardness smaller than 100 mg/1 is classified as soft and water with a hard. ness larger than 180 mg/I is classified as hard. If the hardness is between soft and hard, we say that the water is medium hard. With a level of confidence of 95%, can we classify the mean hardness of the water in this district as medium hard? Problem 8.10. This problem refers to the R commands pt and qt. We use the notation X ~ T(n) if X has a T(n)-distribution. Find the following values using R: (a) P(X 2.5) where X ~ T(5); (d) the value { such that P(X () = 0.025 where X ~ T(16); (f) the value / such that P(-t