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LANCASTER UNIVERSITY
2009 EXAMINATIONS
PART II (Second year)
MATHEMATICS & STATISTICS 2 hours
Math 235: Statistics
You should answer ALL Section A questions and THREE Section B questions. In Section A there are questions worth a total of 50 marks, but the maximum mark that you can gain there is capped at 40. There are statistical tables at the end of this examination paper.
SECTION A
A1. A computer network comprises of m computers. The probability of one of these computers to store illegally downloaded files is 0.3, independent for each computer. In a particular network it is found that exactly one computer contains illegally downloaded files. (a) Write down the likelihood function of m, L(m), and determine the possible values for m. [4] (b) Sketch the likelihood function of m up to 8. [3] (c) Find the maximum likelihood estimator for m. Make sure to justify your answer. [3] (d) Find a 50% relative likelihood interval for m. Make sure to justify your answer. [3] A2. X 1 , X 2 ,... , Xn are independent and identically pareto distributed random variables, each having probability density function
f (x|β) = β^2
β xβ+1^ with^ x >^2 , β >^0. (a) Determine the log-likelihood function, l(β). [3] (b) Find the score function and the maximum likelihood estimator of β. [6] (c) Find the observed information when the observed data are x 1 = 2. 4 , x 2 = 3. 8 , x 3 = 2. 3
. [3] please turn over
SECTION A continued
A3. Observations y 1 , y 2 and y 3 are realised values from the normal linear model Yi = θxi + Z, i = 1, 2 , 3 where x = 1, x 2 = 2, x 3 = 4 and VarZi = σ^2. The least squares estimate of the unknown parameter θ is the realised value of the random variable θˆ = 211 (Y 1 + 2 Y 2 + 4Y 3 ). (a) Give the definition of a response variable and an explanatory variable. [3] (b) Show that E θˆ = θ. [2] (c) Show that Varˆθ = 211 σ^2. [3] Now consider the alternative estimator
θ˜ =^1 7 (Y^1 +^ Y^2 +^ Y^3 ). (d) Show that for this alternative E θ˜ = θ. [2] (e) Show that Var θ˜ is greater that Var θˆ. [3]
please turn over
SECTION B
B1. X 1 , X 2 ,... , Xn constitute a random sample drawn from a population with probability density function f (x|θ) = (^31) θ e−^3 xθ^ with x > 0 , θ > 0. Note that for this density we have E(X) = 3θ and V (X) = 9θ^2.
(a) Write down the likelihood function, L(θ), and the log-likelihood function, l(θ), and show that ∑ni=1 xi is a sufficient statistic for θ, clearly stating your reasons. [4] (b) Find the maximum likelihood estimate for θ and also for φ =
θ. [6] An experiment based on n = 15 observations yields that ∑n i=1 xi^ = 45.9. Use these data to answer the following questions. (c) Find the asymptotic distribution of the maximum likelihood estimator and use it to construct an approximate 95% confidence interval for θ. [6] (d) Give a formula for the deviance, D(θ), for this example and produce a rough sketch of it for θ ∈ (0. 5 , 2 .5). [4]
please turn over
SECTION B continued
B2. It is thought that the probability of a British adult having one university degree is θ 42 , the probability of having more than one degree is θ(1 − θ 2 ) and having no university degree is (1 − θ 2 )^2.
(a) What values of θ are admissible? [2] (b) A survey of 50 British adults yielded the following result:
Number of University degrees 0 1 > 1 Number of British adults 21 23 6 (i) Write down the likelihood, L(θ), and find the maximum likelihood estimate. [5] (ii) Determine the observed information. [3] (c) Suppose now that the survey had simply sought information on whether an adult had a university degree, irrespective of how many. The results are now: University degree No Yes Number of British adults 21 29 (i) Find the maximum likelihood estimate. [5] (ii) Determine the observed information. [3] (d) In (b) and (c) you should have found that the maximum likelihood estimators are markedly different. Which of the estimators is probably closer to the true parameter and why? [2]
please turn over
SECTION B continued
B3. (d) Using the results (you do NOT need to use the data in the table), calculate the regression line estimate of the resistance at temperature 35 degrees, together with its 99% confidence interval. [6] (e) Give the form of an equivalent model in which this prediction would appear as the estimate of a constant term in the regression. Justify your answer. [2] (f) Give the form of an equivalent model in which this prediction would be orthogonal. [2]
please turn over
SECTION B continued
B4. The normal linear model for n observations and p explanatory variables can be expressed in matrix form as follows: y = Xθ + Z. (a) Describe briefly the terms y, X, θ and Z that appear in this model and state the assumptions made regarding the error term. [4] (b) The least squares estimator of θˆ is: θˆ = (X′X)−^1 X′Y. Show that θˆ has expectation θ and variance matrix (X′X)−^1 σ^2 , where σ^2 is the model error variance. [5] (c) Define the sum of squares of error function, S(θ), for this model. [2] (d) In an experiment to determine the acceleration due to gravity, g, the distance that a body had fallen since its release was measured every 0.1 seconds following its release. The results are presented in the table below: Time t since release 0.1 0.2 0.3 0.4 0. (seconds) Distance s fallen 5.1 19.9 44.1 78.8 122. (cm) The normal linear model si =^12 gt^2 i + zi was used to estimate g, resulting in a residual sum of squares S 1 = 0.22785. To double-check the experimental configuration, the experimenter estimated a second model si = c + uti +^12 gt^2 i + zi. The residual sum of squares for this second model was S 2 = 0.11257. (i) Explain why the first model is nested within the second. [2] (ii) Give the values of the estimated residual variances for the two models (assuming the respective models are correct.) [2] (iii) What is the mean reduction in the sum of squares from the first to the second model? [2] (iv) Compare the two models by using the F statistic to test the null hypothesis that the first model is acceptable. Give the value of the test statistic and the result of the test. [3] end of exam
Table 2: The T -distribution
Values of t for which P (| T |> t) = p, where T has a T -distribution with r degrees of freedom.
Table 4: The F distribution Values of f for which P (F > f ) = 0.05 (upper values) and P (F > f ) = 0.01 (lower values) where F has an F -distribution with r and s degrees of freedom. r
- 0.20 0.10 0.05 0.01 0. p - 1 3.078 6.314 12.706 63.657 636. - 2 1.886 2.920 4.303 9.925 31. - 3 1.638 2.353 3.182 5.841 12. - 4 1.533 2.132 2.776 4.604 8. - 5 1.476 2.015 2.571 4.032 6. - 6 1.440 1.943 2.447 3.707 5. - 7 1.415 1.895 2.365 3.499 5. - 8 1.397 1.860 2.306 3.355 5. - 9 1.383 1.833 2.262 3.250 4. - 10 1.372 1.812 2.228 3.169 4. - 11 1.363 1.796 2.201 3.106 4. - 12 1.356 1.782 2.179 3.055 4. - 13 1.350 1.771 2.160 3.012 4. - 14 1.345 1.761 2.145 2.977 4. - 15 1.341 1.753 2.131 2.947 4. - 16 1.337 1.746 2.120 2.921 4.
- r 17 1.333 1.740 2.110 2.898 3. - 18 1.330 1.734 2.101 2.878 3. - 19 1.328 1.729 2.093 2.861 3. - 20 1.325 1.725 2.086 2.845 3. - 21 1.323 1.721 2.080 2.831 3. - 22 1.321 1.717 2.074 2.819 3. - 23 1.319 1.714 2.069 2.807 3. - 24 1.318 1.711 2.064 2.797 3. - 25 1.316 1.708 2.060 2.787 3. - 26 1.315 1.706 2.056 2.779 3. - 27 1.314 1.703 2.052 2.771 3. - 28 1.313 1.701 2.048 2.763 3. - 29 1.311 1.699 2.045 2.756 3. - 30 1.310 1.697 2.042 2.750 3. - 40 1.303 1.684 2.021 2.704 3. - 50 1.299 1.676 2.009 2.678 3. - 60 1.296 1.671 2.000 2.660 3. - 70 1.294 1.667 1.994 2.648 3. - 80 1.292 1.664 1.990 2.639 3. - 90 1.291 1.662 1.987 2.632 3.
- 100 1.290 1.66 1.984 2.626 3.
- ∞ 1.282 1.645 1.960 2.576 3.
- 3 10.13 9.55 9.28 9.12 9.01 8.94 8.85 8.79 8. 1 2 3 4 5 6 8 10 ∞
- 34.12 30.82 29.46 28.71 28.24 27.91 27.49 27.23 26.
- 4 7.71 6.94 6.59 6.39 6.26 6.16 6.04 5.96 5.
- 21.20 18.00 16.69 15.98 15.52 15.21 14.80 14.55 13.
- 5 6.61 5.79 5.41 5.19 5.05 4.95 4.82 4.74 4.
- 16.26 13.27 12.06 11.39 10.97 10.67 10.29 10.05 9.
- 6 5.99 5.14 4.76 4.53 4.39 4.28 4.15 4.06 3.
- 13.75 10.92 9.78 9.15 8.75 8.47 8.10 7.87 6.
- 8 5.32 4.46 4.07 3.84 3.69 3.58 3.44 3.35 2.
- 11.26 8.65 7.59 7.01 6.63 6.37 6.03 5.81 4.
- 10 4.96 4.10 3.71 3.48 3.33 3.22 3.07 2.98 2. - 10.04 7.56 6.55 5.99 5.64 5.39 5.06 4.85 3.
- s 15 4.54 3.68 3.29 3.06 2.90 2.79 2.64 2.54 2. - 8.68 6.36 5.42 4.89 4.56 4.32 4.00 3.80 2. - 20 4.35 3.49 3.10 2.87 2.71 2.60 2.45 2.35 1. - 8.10 5.85 4.94 4.43 4.10 3.87 3.56 3.37 2. - 25 4.24 3.39 2.99 2.76 2.60 2.49 2.34 2.24 1. - 7.77 5.57 4.68 4.18 3.85 3.63 3.32 3.13 2. - 30 4.17 3.32 2.92 2.69 2.53 2.42 2.27 2.16 1. - 7.56 5.39 4.51 4.02 3.70 3.47 3.17 2.98 2. - 40 4.08 3.23 2.84 2.61 2.45 2.34 2.18 2.08 1. - 7.31 5.18 4.31 3.83 3.51 3.29 2.99 2.80 1. - 60 4.00 3.15 2.76 2.53 2.37 2.25 2.10 1.99 1. - 7.08 4.98 4.13 3.65 3.34 3.12 2.82 2.63 1.
- 120 3.92 3.07 2.68 2.45 2.29 2.18 2.02 1.91 1. - 6.85 4.79 3.95 3.48 3.17 2.96 2.66 2.47 1.
- ∞ 3.84 3.00 2.60 2.37 2.21 2.10 1.94 1.83 1. - 6.64 4.61 3.78 3.32 3.02 2.80 2.51 2.32 1.
