098005A - 8005A SEMESTER 2 2009 THE UNIVERSITY OF SYDNEY...

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Unformatted text preview: 8005A SEMESTER 2 2009 THE UNIVERSITY OF SYDNEY FACULTIES OF ARTS, ECONOMICS, EDUCATION, ENGINEERING AND SCIENCE MATH1005 STATISTICS November 2009 LECTURERS: M Stewart E Seneta TIME ALLOWED: One and a half hours T his. examination has three printed components: the Question Paper (this booklet), a Formula Sheet (yellow) and a Multiple Choice Answer Sheet (white). THESE MUST NOT BE REMOVED FROM THE EXAMROOM. The exam has two sections: Multiple Choice and Extended Answer. The Multiple Choice Section is worth 50% of the total examination; there are 25 questions of equal value; attempt all questions. Answers to the Multiple Choice questions must be coded onto the Multiple Choice Answer Sheet. ' The Extended Answer Section is worth 50% of the total examination; there are 3 questions; all questions may be attempted; questions are of equal value; working must be shown. Calculators and statistical tables will be supplied. A formula sheet appears after the last question in this booklet. PAGE 1 OF 26 8005A SEMESTER 2 2009 PAGE 11 OF 26 Extended Answer Sect ion Answer these questions in the spaces provided 1. [15 marks] E. Cam's is a tick—borne disease of dogs which is sometimes contracted by humans. According to Vital Health Statistics(1982) the mean White blood cell count in the general human population is 72.5 (in ’OOO/mm3). It is believed that infection by E. Cam's lowers the average White blood cell count. To investigate this, the White blood cell counts, {36,-}, of a sample of 20 infected humans was obtained: 39, 54, 50, 53, 74, 77, 78, 51, 69, 58, 87, 60, 67, 65, 70, 87, 57, 45, 61, 71. (a) [3 marks] Calculate the sample mean and sample standard deviation for these in— fected humans, given Emir,- = 1273, 2,2193? 2 84349. 1,: (b) [2 marks] Prepare a stem—and—leaf plot (single stem version), and calculate the me— dian. * ' ' 8005A SEMESTER 2 2009 PAGE 12 OF 26 YOU MAY USE THIS PAGE IF YOU NEED MORE ROOM FOR YOUR ANSWERS 8005A SEMESTER 2 2009 PAGE 13 OF 26 (C) [6 marks] Set up appropriate null and alternative hypotheses, and perform a test of significance which assumes normality. 8005A SEMESTER 2 2009 PAGE 14 OF 26 YOU MAY USE THIS PAGE IF YOU NEED MORE ROOM FOR YOUR ANSWERS 8005A SEMESTER 2 2009 PAGE 15 OF 26 (d) [4 marks] A follow—up white blood cell count was obtained for each person in the sample one year later when the infection was gone. The purpose was to test the theory that the white blood cell counts increase on average once the infection has gone. There were 15 instances where the white blood cell count increased, and 5 where it decreased. Given the following R—output: > sum(dbinom(15:20, 20, 0.5)) [1] 0.02069473 set up appropriate null and alternative hypotheses, and perform a test of significance. Explain the meaning of your P—value. 8005A SEMESTER 2 2009 PAGE 16 OF 26 YOU MAY USE THIS PAGE IF YOU NEED MORE ROOM FOR YOUR ANSWERS 8005A SEMESTER 2 2009 PAGE 17 OF 26 2. [15 marks] A newly developed cream for treating a skin disorder has an advertised success rate of “ 4 out of 5” successes. A doctor believes that the cream is not as good as advertised. In the medical centre where he works, he discovers that there are records on 40 unrelated patients who have been treated using the cream. He checks these records looking for evidence against the advertised claim. (a) [8 marks] Writing 1) for the probability of success using the cream, set up the null hypothesis H0 : p = 0.8 and the appropriate (one-sided) alternative hypothesis, and perform the test of significance, if records show that there were 29 successes for the 40 patients. [Hint Use the normal approximation] 8005A SEMESTER 2 2009 PAGE 18 OF 26 YOU MAY USE THIS PAGE IF YOU NEED MORE ROOM FOR YOUR ANSWERS 8005A SEMESTER 2 2009 ' PAGE 19 OF 26 (c) [5 marks] Another doctor has prescribed the cream for 10 of his patients. Unknown to him, the treatment was successful in exactly 8 of the 10 cases. He asks his secretary to contact any 5 of the 10 patients. Assuming the selection of the 5 patients is made randomly, what is the probability that at least 4 of those contacted had a successful result with the cream? [Hintz No patient is contacted more than once] 8005A SEMESTER 2 2009 PAGE 20 OF 26 YOU MAY USE THIS PAGE IF YOU NEED MORE ROOM FOR YOUR ANSWERS 8005A SEMESTER 2 2009 PAGE 21 OF 26 3. [15 marks] The lengths of 1572 cuckoo eggs were measured and rounded to the nearest 0.5mm. A frequency table of the 1572 rounded measurements appears below: > table (cuckoo) cuckoo 19 19.5 20 20.5 21 21.5 22 22.5 23 23.5 24 24.5 25 25.5 26.5 1 3 33 39 156 152 392 288 286 100 86 21 12 2 1 > quantile(cuckoo,type=2) 0% 25% 50% 75% 100% 19.0 22.0 22.5 23.0 26.5 A normal distribution N (11, 02) is fitted to the data, with ,u. and 0‘2 estimated using the mean and stande deviation of the rounded data. After grouping the two lowest values together and the two largest values together into single classes, the resultant 13 classes have respective expected frequencies (using interval probabilities from the fitted normal distribution) of 7.0, 20.8, 59.5, 132.1, 227.0, 302.1, 311.3, 243.3, 153.4, 73.4, 27.2, 7.3, 2.1. (a) [5 marks] [1 mark] Which of the new classes consist of outliers, if we were to draw a boxplot? (ii) [1 mark] What is the expected total number of observations in these (outlier) classes? 8005A SEMESTER 2 2009 PAGE 22 OF 26 YOU MAY USE THIS PAGE IF YOU NEED MORE ROOM FOR YOUR ANSWERS 8005A SEMESTER 2 2009 PAGE 23 OF 26 [1 mark] What is the observed total number of observations in these (outlier) classes? (iv) [2 marks] Explain how your responses here lend support to or provide evidence against the fitted normal distribution. (b) [2 marks] Ignoring the outliers, would a boxplot of this rounded data lend support to or provide evidence against the fitted normal distribution? Explain. 8005A SEMESTER 2 2009 PAGE 24 OF 26 YOU MAY USE THIS PAGE IF YOU NEED MORE ROOM FOR YOUR ANSWERS 8005A SEMESTER 2 2009 PAGE 25 OF 26 (c) [3 marks] Find the value of a: such that P(X3 > as) = 0.01 Where 1/ is the degrees of freedom appropriate for a goodness—of—fit test of the fitted normal distribution to the (rounded) data (including the outliers). (d) [2 marks] The calculated value of the goodness—of—fit statistic is 103.7. What is your conclusion? (e) [1 mark] Calculate the contribution to the goodness—of—fit statistic from the class ' Where the expected value is 227.00. (f) [2 marks] Explain how (6) might be used to help reconcile your conclusion in (d) with your responses in parts (a) and End of Extended Answer Section 80050 SEMESTER 2 2009 NOTES FOR USE IN THE STATISTICS EXAMINATION o Calculation formulae: — For a sample £131,122,. . .,:cn Sample mean f Sample variance 52 — For paired observations ($1,y1),(z:2,y2),-- ,(mn,yn) TL 1 TL 71- Smy Z iii/i — '7; Z 171‘ Z y'i. For the regression line i=1 i=1 i=1 y = a + bx: n 1 ‘2 2 SW Z 3112 _ 7—1 Eli) i=1 i=1 SM ‘ 5(2“) 53y ,- _____._ ‘ /S'mSyy - Some probability results: For any two events A and B P(A U B) = P(A) + P(B) — P(A n B) If A and B are mutually exclusive (me) P(A n B) = 0 and P(A U B) = P(A) + P(B) If A and B are independent P (A n B) = P(A)P(B) and P(A U B) = P(A) + P(B) — P(A)P(B) - If X ~ BM), then M = r) = (2)2711 —p)“—I,r = o, - -- ,n. - Some test statistics and their sampling distributions (under appropriate assumptions and hypotheses) : 17 ( ) N tn$+ny_2, Where + /1 A S? 'n.I 17.1, S; 2 [(III — 1)S§+ (my — + my — 2) ’L Oi - E 2 E: N X122, for appropriate 1/ 1' PAGE 1 OF 0 ...
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098005A - 8005A SEMESTER 2 2009 THE UNIVERSITY OF SYDNEY...

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