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hw6+fall+2011+solutions

# hw6+fall+2011+solutions - ECI 114 Fall 2011 University of...

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ECI 114 Fall 2011 University of California, Davis Department of Civil and Environmental Engineering Homework Assignment 6 (67 points) Turn in to Wei Zou by 10:45 a.m. Wednesday, Nov. 30, 2011 – NO EXTENSIONS Show all work and explain your answers; no credit will be given for solutions only (especially true for problems in the back of the book and on the accompanying CD-ROM). Write your name on the front of the first page and back of the last page of your assignment; do not fold it. A two-point penalty will be assessed for an inability to follow these instructions. A penalty of up to three points may also be assessed for lack of neatness. Preliminary note: With any hypothesis test asked for by these problems, you need to provide the null and alternative hypotheses, the test statistic, the rejection region, the outcome of the test, and an interpretation of the outcome (i.e. not just “reject” or “FTR”, but what that means in the context of the problem). 1. (3+2+3 = 8 pts) Problem 9-96, p. 332. In addition: (c) Explain what you would do if you were not given the information “with a mean of 1.2”. Solution: (a) Given that E[X] = λ= 1.2 P 0 = Pr (X=0) = = 0.3012 P 1 = Pr (X=1) = = 0.3614 P 2 = Pr (X=2) = = 0.2169 P 3 = Pr (X=3) = = 0.0867 P 4 = P (X 4) = 1- P 0 -P 1 -P 2 = 1 - 0.3012 - 0.3614 - 0.2169 - 0.0867 = 0.0338 E i = nP i where n = 100 Value of X Number of Observations Taking this Value Probability of Taking this Value Expected Frequency of Taking this Value out of a Sample of 100 Observations 1

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0 24 0.3012 30.12 1 30 0.3614 36.14 2 31 0.2169 21.69 3 11 0.0867 8.67 4 4 0.0338 3.38 d.f. = k – p – 1 = 5 – 0 – 1 = 4 1. Question of interest: The question of interest is the form of the distribution of the observations. 2. Null Hypothesis: H 0 -- The observation follows the form of the Poisson distribution with a mean of 1.2. 3. Alternative Hypothesis: H 1 – The observation doesn’t follow the form of the Poisson distribution with a mean of 1.2. 4. Test Statistic: 5. Reject H 0 if the P-value is less than 0.05 6. Computations = 7. Conclusion The critical value , so we FTR H 0 , and conclude that distribution of the data is consistent with that of a Poisson distribution with mean 1.2. (b) Therefore 0.05 < p-value < 0.1 (the exact value, obtained from Excel, is 0.094). (c) If “a mean of 1.2” is not given: 1. We need to estimate the mean by using the sample mean: (24*0 + 30*1 + 31*2 + 11*3 + 4*4) / 100 = 1.41. 2. DOF = k-p-1 = 5-1-1 = 3. 2
3. The last two columns of the table above would have to be recalculated, using the estimate for λ. 2. (3+3+2 = 8 pts) Using the data of 9-108, p. 336: (a)Under the assumption of independence, calculate the estimated expected number of students in each of the cells of the table. (b)Using α = 0.05, test the hypothesis that opinion on the change is independent of class

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hw6+fall+2011+solutions - ECI 114 Fall 2011 University of...

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