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Unformatted text preview: ENME392 Spring 2012 Homework 8 Total number of points: 100 The question is a followup to lastweek’s assignment: a) What is the 95% prediction interval for a future number of heads? (8pts) Textbook problems: 1. Chapter 9 : 9.18 (Edition 8) / 9.14 (Edition 9) (6pts) n = 15, ¯ x = 3.7867, s = 0.9709, γ = 1%, and 1  α = 95%, with k = 3.507. So, by calculating 3.7867 ± (3.507)(0.9709) we obtain (0.382, 7.192) which is a 99% tolerance interval that will contain 95% of the drying times. 2. Chapter 9 : 9.25 (Edition 8) / 9.17 (Edition 9) (6pts) n = 20, ¯ x = 11.3, s = 2.45, and t0.025 = 2.093 with 19 degrees of freedom. A 95% prediction interval for a future observation is 11.3 ± (2.093)(2.45) sqrt(1 + 1/20) = 11.3 ± 5.25, which yields (6.05, 16.55). 3. Chapter 9 : 9.30 (Edition 8) / 9.22 (Edition 9) (6pts) (use average of 75.3 kilograms and standard deviation of 5.2 kilograms) d.o.f= 501=49 t 0.05 =1.6775 mean=75.3 kg s.d.=5.2 kg prediction lower bound: (75.3)(1.6775)(5.2)*(1+1/50)(75....
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This note was uploaded on 04/05/2012 for the course ENME enme392 taught by Professor Cukier during the Spring '10 term at Maryland.
 Spring '10
 Cukier

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