C.10 Exercises. Y* =!1 + Yz
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1 C.10 Exercises C.I Suppose Y I, Y,, Y N is a random sample from a population with mean fj. and variance 0'. Rather than using all N observations consider an easy estimator of fj. that uses only the first two observations Y* =!1 + Yz
2 68 C.IO EXERCISES 545 (a) Show that Y* is a linear estimator. (b) Show that Y* is an unbiased estimator. (c) Find the variance of Y*. (d) Explain why the sample mean of all N observations is a better estimator than Y*. C. Suppose that Y 1, Yz, Y is a random sample from a N(I1-, 0- ) population. To estimate 11- consider the weighted estimator Y = - Yt + - Y + - Y 6 (a) Show that Y is a linear estimator. (b) Show that Y is an unbiased estimator. (c) Find the variance of Y and compare it to the variance of the sample mean Y. (d) Is Y as good an estimator as Y? (e) If 0- = 9, calculate the probability that each estimator is within I unit on either side of 11-. C. * The hourly sales of fried chicken at Louisiana Fried Chicken are normally distributed with mean 000 pieces and standard deviation 500 pieces. What is the probability that in a 9-hour day more than 0,000 pieces will be sold? C.4 Starting salaries for Economics majors have a mean of $47,000 and a standard deviation of $8,000. What is the probability that a random sample of 40 Economics majors will have an average salary of more than $50,000? C.5* A store manager designs a new accounting system that will be cost effective if the mean monthly charge account balance is more than $170. A sample of 400 accounts is randomly selected. The sample mean balance is $178 and the sample standard deviation is $65. Can the manager conclude that the new system will be cost effective? (a) Carry out a hypothesis test to answer this question. Use the (t 0.05 level of significance. (b) Compute the p-value of the test. C.6 An econometric professor's rule of thumb is that students should expect to spend hours outside of class on coursework for each hour in class. For a hour per week class, this means that students are expected to do 6 hours of work outside class. The professor randomly selects eight students from a class, and asks how many hours they studied econometrics during the past week. The sample values are 1,, 4,4,6,6, 8, 1. (a) Assuming that the population is normally distributed, can the professor conclude at the 0.05 level of significance that the students are studying on average at least 6 hours per week? (b) Construct a 90% confidence interval for the population mean number of hours studied per week. C.7 Modern labor practices attempt to keep labor costs low by hiring and laying off workers to meet demand. Newly hired workers are not as productive as experienced ones. Assume assembly line workers with experience handle 500 pieces per day. A manager concludes it is cost effective to maintain the current practice if new hires, with a week of training, can process at least 450 pieces per day. A random sample of N 50 trainees is observed. Let Y; denote the number of pieces each handles on a
3 546 REVIEW OF STATISTICAL INFERENCE 69 randomly selected day. The sample mean is y = 460 and the estimated sample standard deviation is (j = 8. (a) Carry out a test of whether or not there is evidence to support the conjecture that current hiring procedures are effective, at the 5% level of significance. Pay careful attention when formulating the null and alternative hypotheses. (b) What exactly would a Type I error be in this example? Would it be a costly one to make? (c) Compute the p-value for this test. C.8* To evaluate alternative retirement benefit packages for its employees, a large corporation must determine the mean age of its workforce. Assume that the age of its employees is normally distributed. Since the corporation has thousands of workers a sample is to be taken. If the standard deviation of ages is known to be (J' 1 years, how large should the sample be to ensure that a 95% interval estimate of mean age is no more than 4 years wide? C.9 Consider the discrete random variable Y that takes the values y = 1,,, and 4 with probabilities 0.1, , and 0.4, respectively. (a) Sketch this pdf (b) Find the expected value of Y (c) Find the variance of Y (d) If we take a random sample of size N = from this distribution, what are the mean and variance of the sample mean, = (Yl + Y + Y)/?
4 APPENDIX C Exercise Solutions 4
5 Appendix C, Exercise Solutions, Principles of Econometrics, e 44 EXERCISE C.1 (a) A linear estimator is one that can be written in the form ay i i where a i is a constant. Rearranging Y* yields, Y + Y Y* = = Y1 + Y = Y 1 i i= Thus, Y* is a linear estimator where a = 1 for i= 1, and a = 0 for i=, 4,, N. i i (b) (c) The expected value of an unbiased estimator is equal to the true population mean. Y1 + Y EY ( *) = E = EY ( 1) + EY ( ) = μ+ μ=μ The variance of Y* is given by Y1 + Y 1 1 var ( Y* ) = var = var Y1 + Y 1 var 1 = + var cov, 1 1 σ 4 4 ( Y ) ( Y ) ( Y Y ) 1 1 = σ + σ =, since ( 1 ) cov Y, Y = 0 (d) The sample mean is a better estimator because it uses more information. The variance of the sample mean is σ N which is smaller than σ when N >, thus making it a better estimator than Y*. In general, increasing sample information reduces sampling variation.
6 Appendix C, Exercise Solutions, Principles of Econometrics, e 45 EXERCISE C. (a) Y = Y1 + Y + Y = ay i i, where aiare constants for i= 1, and. 6 i= = = = μ+ μ+ μ=μ (b) EY ( ) E Y Y Y EY ( ) EY ( ) EY ( ) (c) var ( Y) = var Y1 + Y + Y = 1 var ( Y1) + 1 var ( Y) + 1 var ( Y), since cov ( Y1, Y) = cov ( Y, Y) = σ = σ + σ + σ = The variance of the sample mean is var ( Y ) σ σ 6σ = = = N 18 which is smaller than the variance of Y. (d) Since var ( Y) > var ( Y), Y is not as good an estimator asy. (e) The probability that the estimatory is within one unit on either side of μ is: 1 Y μ 1 P 1 Y 1 P μ μ+ = var ( Y) var ( Y) var ( Y) 1 1 = P Z [ Z ] = P = 0.46 The probability that the estimatory is within one unit on either side of μ is: 1 Y μ 1 P μ 1 Y 1 P μ+ = var ( Y ) var ( Y ) var ( Y ) 1 1 = P Z [ Z ] = P = 0.407
7 Appendix C, Exercise Solutions, Principles of Econometrics, e 46 EXERCISE C. Let X be the random variable denoting the hourly sales of fried chicken which is normally distributed; X N(000,500 ). The probability that in a 9 hour day, more than 0,000 pieces will be sold is the same as the probability that average hourly sales of fried chicken is greater than 0,000/9, pieces. P X X μ μ > = P > σ N σ N 000 = P Z > = P Z > 500 = PZ [ > 1.] = 0.091
8 Appendix C, Exercise Solutions, Principles of Econometrics, e 47 EXERCISE C.4 Let the random variable X denote the starting salary for Economics majors. Assume it is normally distributed; X N(47000,8000 ). P X X μ μ > = P > σ N σ N = P Z > = PZ [ >.7] = =
9 Appendix C, Exercise Solutions, Principles of Econometrics, e 48 EXERCISE C.5 (a) We set up the hypotheses H : μ 170 versus H : μ > 170. The alternative is H : μ> because we want to establish whether the mean monthly account balance is more than 170. The test statistic, given H 0 is true, is: X 170 t = t σ ˆ N ( 99) The rejection region is t The value of the test statistic is t = = Since t =.46 > 1.649, we reject H 0 and conclude that the new accounting system is cost effective. =.46 = 1 <.46 = (b) p P t(99) P t(99)
10 Appendix C, Exercise Solutions, Principles of Econometrics, e 49 EXERCISE C.6 (a) To decide whether the students are studying on average at least 6 hours per week, we set up the hypotheses H : μ= 6 versus H : μ > The test statistic, given H 0 is true, is X 6 t = t σ ˆ N ( 7) X xi ( ) i= 1 8 = = = 8 1 σ ˆ = var ( X) = ( xi X) = i= t = = At the 0.05 level of significance, the rejection region is t > Since t = < 1.895, we do not reject H 0 and therefore cannot conclude that, at the 0.05 level of significance, the students are studying more than 6 hours per week (b) A 95% confidence interval for the population mean number of hours studied per week is: σˆ X ± t c = N ± = [.5, ]
11 Appendix C, Exercise Solutions, Principles of Econometrics, e 440 EXERCISE C.7 (a) To test whether current hiring procedures are effective, we test the hypothesis that H : μ 450 against 0 H : μ> The manager is interested in workers who can process at least 450 pieces per day. The test statistic, when H 0 is true, is X 450 t = t σ ˆ N ( 49) The value of the test statistic is t = = Using a 5% significance level at 49 degrees of freedom, the rejection region is t > Since > 1.677, we reject H 0 and conclude that the current hiring procedures are effective. (b) A type I error occurs when we reject the null hypothesis but it is actually true. In this example, a type I error occurs when we wrongly reject the hypothesis that the hiring procedures are effective. This would be a costly error to make because we would be dismissing a cost effective practice. (c) p-value = ( 1 Pt ( (49) 1.861) ) < = ( ) = 0.044
12 Appendix C, Exercise Solutions, Principles of Econometrics, e 441 EXERCISE C.8 The interval estimate of a normally distributed random variable is given by Y ± zc σ N, where zc is the corresponding critical value at a 95% level of confidence. z σ N. The length of the interval is therefore ( c ) To ensure that the length of the interval is less than 4, derive N as follows: σ zc < 4 N ( z ) c σ < N ( z ) c σ < 4N ( ) < 4N 4.55 < N A sample size of 44 employees is needed.
13 Appendix C, Exercise Solutions, Principles of Econometrics, e 44 EXERCISE C.9 (a) A sketch of the pdf is shown below pdf y (b) EY ( ) ypy i ( yi) 4 = = = = i= 1 4 (c) var ( Y) = ( yi E( Y) ) P( Y = yi) i= 1 ( ) ( ) ( ) ( ) = = 1 Y + Y + Y Y Y Y EY = E E E E = (d) ( ) var = 1 E Y + 1 E Y + 1 E Y ( ) ( ) ( ) = + + = Y + Y + Y = 1 ( Y ) var = 1 var + 1 var + 1 var ( Y ) ( Y ) ( Y ) = = 9 9 9
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