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ec41f10_hw4_sol

ec41f10_hw4_sol - Kata Bognar [email protected] Economics 41...

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Unformatted text preview: Kata Bognar [email protected] Economics 41 Statistics for Economists UCLA Fall 2010 Homework Assignment 4.- suggested solutions by Kyle Woodward - NOTE: Please show your calculations for Questions 3,5,8-9,12-16. The exercises are from the textbook (Tanis and Hogg: A Brief Course in Mathematical Statistics). Sample mean. 1. The expected value of the sample mean is: (a) always equal to the sample mean (b) sometimes equal to the population mean (c) always equal to the population mean (d) always equal to the sampling procedure Solution: (c), always equal to the population mean Using properties of expectation, we can derive this as in lecture. Suppose that we have n observations { X i } n i =1 of some random variable X . The sample mean of these observations is ¯ X = 1 n ∑ n i =1 X i . We can calculate the sample mean as E ¯ X = E " 1 n n X i =1 X i # = 1 n E " n X i =1 X i # = 1 n n X i =1 E [ X i ] = 1 n n X i =1 E [ X ] = 1 n n X i =1 μ X = 1 n nμ X = μ X Then the expected value of the sample mean is equal to the population mean. This implies that the sample mean is an unbiased estimator for the population mean. 2. As the sample size increases, the standard deviation of the sample mean: (a) increases (b) decreases (c) remains the same (d) none of these Solution: (b), decreases 1 Seeing this will be similar to the algebra used above for expectation; we will use variance since it is easier to manipulate, and since standard deviation is the square root of variance anything we find in one will apply just as well to the other. We generally assume that observations of a random variable are independent (that is, observing X 1 the first time does not affect the probability of observing X 2 the second time); this is a fact we will need. We will rely on properties of variance to prove the answer to this question. Var[ ¯ X ] = Var " 1 n n X i =1 X i # = 1 n 2 Var " n X i =1 X i # = 1 n 2 n X i =1 Var[ X i ] = 1 n 2 n X i =1 Var[ X ] = 1 n 2 n Var[ X ] = Var[ X ] n Since Var[ X ] is constant, we can see that Var[ ¯ X ] is decreasing in N , the number of observations. This is somewhat intuitive: as the number of observations increases, we gain more knowledge as to where the population mean might be, so the standard deviation should be shrinking. 3. Let ¯ X be the mean of a random sample of size 16 from a distribution with p.m.f. f ( x ) = x +2 c for x =- 1 , , 1 , 2 . (a) Find the value of c. Solution: the easiest fact to start with here is that ∑ N i =1 f ( x i ) = 1, or that the probability of X lying within its support is 1. In this case, we have 4 X i =1 f ( x i ) = 4 X i =1 x i + 2 c = 1 c 4 X i =1 x i + 1 c 4 X i =1 2 = 1 c (- 1 + 0 + 1 + 2) + 1 c 8 = 2 c + 8 c = 10 c Then since 1 = ∑ 4 i =1 f ( x i ) = 10 c , we know c = 10....
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ec41f10_hw4_sol - Kata Bognar [email protected] Economics 41...

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