Hw10sol - Stat 511 Homework 10 solutions Spring 2011 Due 5pm Tuesday Apr 5 The first two problems present you with a description of a study Each

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Unformatted text preview: Stat 511 Homework 10 solutions Spring 2011 Due: 5pm, Tuesday Apr 5 The first two problems present you with a description of a study. Each has two (or more) sizes of experimental unit. Some of the “experimental” units are not randomly assigned to treatments. Please ignore that and treat them as if they were randomly assigned. Factors of interest are indicated in CAPITOL letters. If possible, your model should include interactions among treatment factors. For each study, write out the skeleton ANOVA table, i.e., write out the sources of variation and their d.f. Also, indicate the appropriate error term (i.e. denominator MS) for the F test of each treatment effect. 1. The ability of an elderly individual to successfully move from sitting down to standing up is a key part of maintaining independence. Folks in the Kinesiology Department at ISU are studying ways to make it easier for subjects to stand. Easy can be quantified by the time required to stand up. One of those this can be made easier is to change the height of the chair. This study compares three GROUPs of subjects: normal elderly, osteoarthritic elderly, and el- derly with successful total knee arthroplasty (a surgical operation). Patients are also classified by SEX. There are 10 men and 10 women in each group. To measure the data, each patient sits in a chair and is asked to stand up. The response is duration, the length of time required to stand. The investigators believe that duration varies with chair HEIGHT. To evaluate this, each patient was measured at four chair HEIGHTs. The order of chair heights was randomly assigned separately each patient (Don’t worry about the order of testing here). The chair is set to the appropriate height then each patient was asked to sit and stand 3 times. These 3 measurements are considered to be subsamples. To recap: there are three treatment factors: GROUP (3 levels), SEX (2 levels), and HEIGHT (4 levels) in a complete factorial structure. Each of the 60 people is measured 3 times at each chair height so there are 720 observations of duration. Answer: Source d.f. Error term for F test Comments Group 2 Person(G*S) Sex 1 Person(G*S) Group*Sex 2 Person(G*S) Person(G*S) 54 Total of 59 df = 60 main plots - 1 Height 3 Split plot error Group*Height 6 Split plot error Sex*Height 3 Split plot error G*S*Height 6 Split plot error Split plot error 162 Total of 239 df = 240 split plots - 1 Subsampling 480 2 df from each of 240 occasions c.total 719 1 Stat 511 Homework 10 solutions Spring 2011 Explanation: (not required) The treatment structure is a 3 way factorial....
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This note was uploaded on 02/11/2012 for the course STAT 511 taught by Professor Staff during the Spring '08 term at Iowa State.

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Hw10sol - Stat 511 Homework 10 solutions Spring 2011 Due 5pm Tuesday Apr 5 The first two problems present you with a description of a study Each

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