Practice Midterm Exam (Statistics)

AMS572. 01 Practice Midterm Exam Fall, 2007 Instructions This is a close book trial run. Anyone who cheats in the exam shall receive a grade of F. satisfy provide complete solutions for full credit. Good luck 1 (for all students in class). In a study of hypnotic suggestion, 5 male volunteers participated in a two-phase experimental session. In the first phase, respiration was measured while the subject was awake and at rest.In the second phase, the subject was told to imagine that he was performing muscular work, and respiration was measured again. Hypnosis was induced amidst the first and second phases thus, the suggestion to imagine muscular work was hypnotic suggestion for these subjects. The accompanying put over shows the measurements of total ventilation (liters of air per minute per squ atomic number 18 meter of body area) for all 5 subjects. observational Group Subject Rest Work 1 6 6 2 7 9 3 8 9 4 7 10 5 6 7 (1) drug abuse suitable test to investigate whether th ere is any difference between the two experimental phases in terms of total ventilation.Please state the assumption(s) of the test and report the p-value. At the significance level of 0. 05, what is your close? (2) Please write up the entire SAS program necessary to answer questions raised in (a). Please include the data step as well as tests for testing for various assumptions. Solution (1) wear upon that the difference picis normal. pic and pic The hypotheses are pic v. s pic. The test statistic is pic Since pic and pic, we can not reject pic at pic. pic (2) The SAS code is as follows data hypnosis input subject rest work iff=work-rest datalines 1 6 6 2 7 9 3 8 9 4 7 10 5 6 7 run proc univariate data=hypnosis normal var diff run 2 (for all students in class). John Pauzke, chairwoman of Cereals Unlimited, Inc. , wants to be very certain that the mean value weight ? of packages satisfies the package label weight of 16 ounces. The packages are filled by a machine that is set to f ill each package to a specify weight. However, the machine has haphazard variability measured by ? 2. John would like to have strong cause that the mean package weight is above 16 ounces.George Williams, quality control manager, advises him to examine a random sample of 25 packages of cereal. From his past experience, George knew that the weight of the cereal packages follows a normal distribution with standard aberrancy 0. 4 ounce. At the significance level ? =. 05, (1) What is the decision rule (rejection region) in terms of the sample mean pic? Please derive the general formulation using the concept of Type I error rate. (2) What is the exponent of the test when ? =16. 2 ounces? Please derive the general formula for power calculation first. 3) What is the sample sizing necessary to ensure a power of 80% when ? =16. 2 ounces? Please derive the general formula for sample size calculation based on the Type I and II error pass judgment first. Solution (1) pic pic. pic. pic. pi c Hence, the rejection region is pic. (2) pic pic (3) pic picpic. pic Hence, about 25 packages of cereal should be sampled to achieve a power of 80% when (=16. 2 ounces. 3a (for all except AMS PhD students). Inference on one cosmos mean when the creation is normal, and the population variance is known.Let pic, be a random sample from the given normal population. Please take the stand that 1) pic. 2) pic. Solution (1) pic Thus, pic (2) pic Thus, pic 3b (for AMS PhD students ONLY). For a random sample from any population for which the mean and variance exist. Please put forward that 1) The sample mean and sample variance are unbiased estimators of the population mean and variance respectively. 2) When the population is normal, we have learned that the sample mean and the sample variance, are indeed, independent.Please prove this for n = 2. That is, for a random sample of size 2 only. Solution (1) pic pic (2) When n=2, pic, pic If we can show that pic and pic are independent, past picand picare independent. This can be done easily using the mgf technique pic 4 (extra credit for all). An expert knowledge in a paternity suit testifies that the length (in days) of pregnancy (that is, the time from impregnation to the delivery of the child) is rough normally distributed with parameter pic and pic.The defendant in the suit is able to prove that he was out of the dry land during a period that began 290 days before the birth of the child and ended 240 days before the birth. If the defendant was, in fact, the father of the child, what is the probability that the mother could have had the very long or very rook pregnancy indicated by the testimony? Solutionlet picpic and picpic pic(the woman had a very long or very short pregnancy) pic pic Happy Halloween