# Practice Midterm Exam (Statistics)

AMS572. 01 Practice Midterm Exam Fall, 2007 Instructions: This is a close book exam. Anyone who cheats in the exam shall receive a grade of F. Please 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 between the first and second phases; thus, the suggestion to imagine muscular work was “hypnotic suggestion” for these subjects. The accompanying table shows the measurements of total ventilation (liters of air per minute per square meter of body area) for all 5 subjects. Experimental Group | |Subject |Rest |Work | |1 |6 |6 | |2 |7 |9 | |3 |8 |9 | |4 |7 |10 | |5 |6 |7 | (1) Use suitable test to investigate whether there is any difference between the two experimental phases in terms of total ventilation.

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Please state the assumption(s) of the test and report the p-value. At the significance level of 0. 05, what is your conclusion? (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) Assume that the difference [pic]is 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, president of Cereals Unlimited, Inc. , wants to be very certain that the mean weight ? of packages satisfies the package label weight of 16 ounces. The packages are filled by a machine that is set to fill each package to a specified weight. However, the machine has random variability measured by ? 2. John would like to have strong evidence 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 deviation 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 formula using the concept of Type I error rate. (2) What is the power of the test when ? =16. 2 ounces? Please derive the general formula for power calculation first. 3) What is the sample size 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 rates first. Solution: (1) [pic] [pic]. [pic]. [pic]. [pic] Hence, the rejection region is [pic]. (2) [pic] [pic] (3) [pic] [pic][pic]. [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 population mean when the population is normal, and the population variance is known.

Let [pic], be a random sample from the given normal population. Please prove 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 prove 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, then [pic]and [pic]are independent. This can be done easily using the mgf technique: [pic] 4 (extra credit for all). An expert witness 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 approximately normally distributed with parameter [pic] and [pic].

The defendant in the suit is able to prove that he was out of the country 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 short pregnancy indicated by the testimony? Solution:let [pic]~[pic] and [pic]~[pic] [pic](the woman had a very long or very short pregnancy) [pic] [pic] Happy Halloween!