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final_2005

# final_2005 - MATH 203 Final Examination December 6 2005...

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MATH 203 Final Examination December 6, 2005 Student Name: Student Number: Faculty of Science FINAL EXAMINATION MATHematics 203 Principles of Statistics I Tuesday, December 6th 9 a.m. - 12 Noon Answer directly on the test (use front and back if necessary). Calculators are allowed. One 8.5” × 11” two-sided sheet of notes is allowed. Language dictionaries are allowed. There are 16 pages to this exam and 3 pages of tables. The total number of marks for the exam is 100. Examiner: Professor Russell Steele Associate Examiner: Professor Masoud Asgharian 1

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MATH 203 Final Examination December 6, 2005 Question 1: (8 points) The amount of lead in a certain type of soil, when released by a standard extraction method, averages 86 parts per million (ppm). A new extraction method is tried on 40 specimens of the soil, yielding a mean of 83 ppm lead and a standard deviation of 10 ppm. Test the hypothesis that the new method yields less lead than the old method (choose any reasonable significance level). (8 points) Answer: This is a test of population mean for one sample. The hypotheses of interest are: H 0 : μ 86 H A : μ < 86 The test statistic is Z = 83 - 86 10 / 40 = - 1 . 90 . If we choose α = 0 . 01, then we would reject for any value of the test statistic less than - z 0 . 05 = - 2 . 33. Therefore we would not reject for this level of α . The p-value of for this particular z-statistic is 0.028, so we would reject for any α bigger than this p-value. 2
MATH 203 Final Examination December 6, 2005 Question 2: (6 points) Assume that Y is normally distributed with mean 0.05 and variance 0.25. Calculate the follow- ing probabilities: (a) Pr ( Y > - 0 . 05) = (2 points) Answer: Pr ( Y > - 0 . 05) = Pr ( Y - 0 . 05 0 . 25 > - 0 . 05 - 0 . 05 0 . 25 ) = Pr ( Z > - 0 . 2) = 1 - Pr ( Z ≤ - 0 . 2) = 0 . 5793 (b) Pr ( - 0 . 40 < Y < 0 . 50) = (2 points) Answer: Pr ( - 0 . 40 < Y < 0 . 50) = Pr ( - 0 . 40 - 0 . 05 0 . 25 < Y - 0 . 05 0 . 25 < 0 . 50 - 0 . 05 0 . 25 = Pr ( - 0 . 9 < Z < 0 . 9) = 0 . 8159 - 0 . 1841 = 0 . 6318 (c) Pr ( Y = 0 . 05) = (2 points) Answer: 0 because Normal random variables are continuous variables. 3

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MATH 203 Final Examination December 6, 2005 Question 3: (16 points) A Master’s student in Environmental Engineering was studying the e ff ect of two types of treat- ment on bacterial growth in drinking water. The two treatments were UV40 (ultraviolet light) and PAA+UV20 (phenylacetic acid and ultravoiolet light). Each day the student sampled water from the same source, poured it into two containers and then randomly assigned a treatment to each container. The log bacterial growth was then measured after three hours of being exposed to a light source for each container. The student did this on 33 separate days. Assume the days were far enough apart so that the samples were independent and di ff erent for di ff erent days. The goal of the analysis is to see if there is a di ff erence between treatments with respect to suppressing bacterial growth. UV40 PAA2+UV20 1.0 0.5 0.0 0.5 1.0 1.5 Log Growth by Treatment 0.5 1.0 1.5 Difference btw Trts by Day UV40 PAA+UV20 Di ff by day (UV40 - PAAUV20) Min -0.7 -0.9 0.10 25%ile 0.8 -0.2 0.8 50%ile 1.1 0.2 1.0 75%ile 1.4 0.3 1.4 Max 1.8 0.9 1.9 Mean 1.06 0.04 1.02 StDev 0.51 0.43 0.46 4
MATH 203 Final Examination December 6, 2005 (a) Test the hypothesis that there is no mean bacterial growth after 3 hours for each treatment individually. (8 pts) Answer:

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