Practice_Midterm_2010

Practice_Midterm_2010 - AMS572.01 Practice Midterm Exam...

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AMS572.01 Practice Midterm Exam Fall, 2010 ♠♣♥♦ Name: ________________________________ ID: _____________________ Signature: _________________________ Instruction: 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. The exam goes from 12:50-2:10pm. Good luck! (*The real midterm will be given on Monday, 10/25/2010, in class.) 1 (for all students in class). In a study of hypnotic suggestion, 10 male volunteers were randomly allocated to an experimental group and a control group. Each subject 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. For subjects in the experimental group, hypnosis was induced between the first and second phases; thus, the suggestion to imagine muscular work was “hypnotic suggestion” for experimental subjects and “waking suggestion” for control subjects. The accompanying table shows the measurements of total ventilation (liters of air per minute per square meter of body area) for all 10 subjects. Experimental Group Control Group Subject Rest Work Subject Rest Work 1 6 6 6 6 5 2 7 9 7 5 5 3 5 8 8 5 5 4 7 12 9 6 6 5 6 7 10 5 4 (a) Use suitable tests to investigate (Use α =.05 for each test. Please report the p-value for each test and state the assumption(s) of the test.) (i) the response of the experimental group to suggestion; (ii) the response of the control group to suggestion; (iii) the differences between the responses of the experimental and control groups. (b) 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. Answer: (a) Response = Work - Rest (i) Inference on one population mean. Small sample. , 2 . 2 1 = x , 9 . 1 1 = s 5 1 = n 0 : 1 0 = μ H vs 0 : 1 a H 56 . 2 5 / 9 . 1 0 2 . 2 / 0 1 1 1 0 - = - = n s x t Since 132 . 2 56 . 2 05 . 0 , 4 0 = t t we reject H 0 at the significance level 0.05. α = Since 132 . 2 56 . 2 776 . 2 05 . 0 , 4 0 025 . 0 , 4 = = t t t we can infer that 05 . 0 025 . 0 < - < value p . The assumption is that the response from the experimental group is normally distributed. Note: if the normality assumption is not true, we will perform the nonparametric test – either the sign test or the signed-rank test.
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Practice_Midterm_2010 - AMS572.01 Practice Midterm Exam...

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