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Concordia University Department of Economics ECON 221/4 – Sections AA, B, E Instructors: M. DaPalma, V. Terekhov, G. Tsoublekas Winter 2007 - ASSIGNMENT 3 - ANSWERS Due at noon on Friday, March 16, 2007 Name: I.D. Section: 1. The search committee of a board is considering six different candidates to complete the positions vacated by three of its oldest members. Excepting for their ages, which run as 41, 39, 35, 35, 34, and 38, the qualifications of the candidates are very similar. Therefore, the three new board members are to be chosen at random. (20 points) [The calculations for (b), (c), and (d) below must be done on Excel and pasted in the spaces provided here.] a. How many samples (possible groups) of three candidates can be made? (3 points) 20 ! 3 ! 3 ! 6 6 3 = = C possible samples (groups) of three candidates b. List all possible samples (three-member groups) with their mean age. (4 points) c. Find the probability function of the sampling distribution of the sample mean and calculate its mean and variance. (5 points)

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From the distribution of the sample means above, we get = = = 37 ) ( * ) ( X P X X E i μ We also get = = - = 1255 . 1 27 . 1 ) ( * ) ( 2 2 X i X X P X σ μ σ d. Calculate the population mean and population standard deviation of the age of all candidates. (4 pts) The mean of the population distribution is 37 6 222 = = = N X i μ And the variance is 5166 . 2 33 . 6 6 38 ) ( 2 2 = = = - = σ μ σ N X i e. Verify the relationship between the standard deviation of the population (found in d) and the standard error of the sampling distribution of the sample mean (found in c). (4 points) The relationship between the variance of the population and the variance of the sample mean (while also considering that the population here is finite and therefore we have to also include in the variance formula the finite population correction factor) is shown by 1 2 2 - - = N n N n X σ σ . 2
Plugging the numbers from (d) above in the formula than links the population standard deviation with the standard error of the sampling distribution we get 1255 . 1 7746 . 0 * 453 . 1 6 . 0 * 732 . 1 5166 . 2 1 6 3 6 3 5166 . 2 1 = = = - - = - - = N n N n X σ σ , which is equal to what we have found in (c) above.

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