Tentamentsskrivning: Mathematisk statistik TMS Tentamentsskrivning i Mathematisk statistik TMS061

Transcription

1 Tentamentsskrivning: Mathematisk statistik TMS061 1 Tentamentsskrivning i Mathematisk statistik TMS061 Tid: Tisdagen den 25 maj, 2010 kl Examinator och jour: Serik Sagitov, tel , mob , MV-huset rum H3026. Hjälpmedel: valfri räknare, egen formelsamling (4 sidor på 2 blad A4) samt utdelade tabeller. There are five questions with the total number of marks 30. Attempt as many questions, or parts of the questions, as you can. Preliminary grading system: grade 3 for 12 to 17 marks, grade 4 for 18 to 23 marks, grade 5 for 24 and more marks. 1. (6 marks) Australian males have heigths with mean 178 cm and standard deviation 6 cm. Australians have IQs with mean 100 and standard deviation 15. a. An Australian male chosen at random is 2m tall, and has an IQ of 83. What is more unusual, his height or his IQ? Explain. b. A sample of 100 Australian men is taken. Apply the central limit theorem to find the probability that the sample mean exceeds 180 cm. Justify your calculations by clearly specifying the assumptions you make on the sampling design. a. His height is more unusual. It s 3.67 standard deviations above the mean, while the IQ is only 1.13 standard deviations from the mean. b. Let X 1,...,X 100 be the heights in the sample and X = i=1 X i the sample mean. If the sampling was performed by independently picking an Australian man uniformly at random 100 times, the X i s are independent and we have E X = 178 and Var X = Thus, by the central limit theorem, X is approximately normally distributed with mean 178 cm and standard deviation = 0.6 and we have 6 10 ( ) P( X ) = 1 Φ % (6 marks) In the popular game show Who Wants To Be a Millionaire? contestants are asked trivia questions with four possible answers (labeled A,B,C and D). If the contestant does not know the correct answer, one of the things he or she can do is ask the audience members to vote for what they think the

2 Tentamentsskrivning: Mathematisk statistik TMS061 2 correct answer is. Conventional wisdom holds that the audience is almost always right. Suppose there are 100 people in the audience and 10 of them know the answer (say A). We assume that each one either votes for the correct answer A (if they know it) or chooses randomly one of the four responses with equal probability (if they do not know the answer). a. Denote X A, X B, X C, X D the numbers of votes for four alternatives. Justify the formula ( ) 90 P(X A = 35, X B = 30, X C = 20, X D = 15) = , 25, 20, 15 b. The correlation coefficient between any pair of random variables X A, X B, X C, X D is Why it is not surprising that the correlation is negative? c. Find the expectation and variance of X A. a. The variables X A 10, X B, X C, X D follows a multinomial distribution with parameters n = 90 trials and class probabilities p A = p B = p C = p D = 1/4. b. The more persons voting on say alternative A, the less number of persons are left to vote on the remaining alternatives, hence large values on X A should imply smaller values on the other variables. c. Y := X A 10 is binomially distributed: Y Bin(90, 1/4). Hence, E[X A ] = /4 = 22.5 and Var X A = 90 1/4 3/4 = (6 marks) A trucking firm suspects that the average lifetime of miles claimed for certain tires is too high. To check the claim, the firm puts 40 of these tires on its trucks and gets a mean lifetime of miles and a standard deviation of 1348 miles. a. What can the firm conclude at the 0.01 level of significance, if it tests the null hypotheis µ = against an appropriate alternative? b. What do you think are the major factors that cause variation among the lifetimes of tires? Speculate on a proper sampling procedure in this case. a. We want to test the null hypothesis H 0 : µ = against the alternative H 1 : µ < Since the variance is unknown we perform a t-test (the number of samples is large enough to justify the normal approximation of the sample

3 Tentamentsskrivning: Mathematisk statistik TMS Figure 1: A scatter plot for x = production and y = concentration. mean). The t-statistic is: t = X s/ 40 = /sqrt We should reject H 0 if t < t where t 0.01 comes from a t-distribution with 39 degrees of freedom. Hence we cannot reject H 0 at the 0.01 significance level. b. Production, driver, road, truck, etc. For a proper sampling procedure it s important not to put all wheels on for instance two trucks. They have to be distributed randomly on all drivers and trucks for instance. 4. (6 marks) An article in the Tappi Journal (March 1986) presented data on green liquor Na 2 S concentration (in grams per liter) and paper machine production (in tons per day). The data is shown in the Figure 1 with x = production and y = concentration. Here are some summary statistics: correlation coefficient = , mean for x = 939, standard deviation for x = , mean for y = , standard deviation for y = a. Find the mean green liquor Na 2 S concentration when the production rate is 950 tons per day.

4 Tentamentsskrivning: Mathematisk statistik TMS061 4 b. Estimate the variance of the green liquor Na 2 S concentration when the production rate is 950 tons per day. What assumptions are you making? a. We use least squares to fit the data to the linear model y i = β 0 +β 1 x i +ξ i where ξ 1,..., ξ n are assumed i.i.d. normal with mean 0 and variance σ 2. We have S XY = r S XX S Y Y = Linear regression then gives β 1 = SXY S XX = and β 2 0 = ȳ β 1 x = Thus we estimate the mean of Y when X = 950 by β 0 + β b. The variance σ 2 is assumed to be independent of X and can be estimated by ˆσ 2 = SSE = (1 r2 )SST = (1 r2 )S Y Y = (1 r2 )(n 1)s 2 Y = ( ) (6 marks) Alcohol abuse has been described by college presidents as the number one problem on campus, and it is an important cause of death in young adults. How common is it? A survey of students in U.S. four-year colleges collected information on drinking behavior and alcohol related problems. The researchers defined frequent binge drinking (binge = supfest) as having five or more drinks in a row three or more times in the past two weeks. According to this definition, 3314 students were classified as frequent binge drinkers. a. Let p be the population proportion of frequent binge drinkers. What statistical model for the observed data X = 3314 leads to the likelihood function L(p) = ( ) p 3314 (1 p) 13782? b. The maximum likelihood estimate of p is then the sample proportion. Verify this by finding p which maximizes the log-likelihood function l(p) = lnp ln(1 p). c. Compute a 95% confidence interval for p. d. The interval that you are supposed to compute in c) will either cover the true value of p or not. Why do we call this a 95% confidence interval? a. X Bin(17096, p).

5 Tentamentsskrivning: Mathematisk statistik TMS061 5 b. Extreme points are found when l (p) = p p = 0 p = Since l (p) = 3314 p (1 p 2 ) < 0 it is a global maxima. c. p = ˆp±z ˆp(1 ˆp) n ( ) %± %±0.593% d. Because this random interval (which depends on the random sample) will cover the true value p with probability 95 %. Statistical tables supplied: 1. Normal distribution table 2. t-distribution table Svara gärna på svenska. Lycka till!