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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 4 - Answers Due at noon on Thursday, April 5, 2007 1. A cellular phone company advertises that the average age of its customers is fewer than 30. It is known that the customer ages follow a normal distribution. A random sample of 16 of this company’s customers responded as follows, when asked about their age: 35, 19, 25, 33, 32, 51, 65, 43, 44, 29, 33, 21, 34, 30, 25, 57. (30 points) a. Calculate the sample mean and sample variance. (5 points) 36 16 576 = = = n X X i 91 . 12 67 . 166 1 16 2500 1 ) ( 2 2 = = - = - - = s n X X s i b. Develop an appropriate hypothesis to test the company’s claim at the 5% level of significance. (5 pts) H 0 : μ < 30 H 1 : μ > 30 Reject H 0 if estimated t > 1.753 859 . 1 2275 . 3 6 4 91 . 12 6 16 91 . 12 30 36 = = = - = - = n s X t μ and 1.859 > 1.753 Therefore, we reject 0 H and we conclude that the average age of the company’s clients is more than 30.

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Estimate and interpret the p-value of the calculated test statistic in (b) above. (4 points) P-value = P(t 15 > 1.859) = about 4.3% The p-value here indicates that the significance level at which we could reject the null hypothesis is 4.3%, instead of the 5% used in (b) above. This makes our conclusions somewhat weak. d. If the same sample results had been obtained from a random sample of 25 customers, could the company’s claim be rejected at a lower level of significance than in part (b)? (4 points) H 0 : μ < 30 H 1 : μ > 30 Reject H 0 if estimated t > 1.711 323 . 2 582 . 2 6 5 91 . 12 6 25 91 . 12 30 36 = = = - = - = n s X t μ Yes. If the same result had come from a sample of 25, the estimated t-ratio would be 2.323, which corresponds to a P-value of about 1.9%. This means that we could reject the null hypothesis at the 1.9%, level of significance instead of the 5% used in (b) or the one we found in (c) above. In such a case, the conclusion of the test would be stronger. e. Suppose that the alternative hypothesis has been one-sided and that it was set as H a : μ < 30. Make a graph to visualize the problem and state, without doing the calculations, whether the p-value of the test (the level of significance needed to reject the null hypothesis) would be higher than, lower than, or the same as found in (c). (4 points) H 0 : μ = 30 P-value H 1 : μ < 30 Reject H 0 if estimated t < 1.753 30 36 The alternative hypothesis is on the leftward (negative) side of the t-Distribution (which means that the rejection area is also on that side). On the other hand, our present sample mean is found on the rightward (positive) side of the distribution of sample means. Therefore, the p-value of our present mean must be greater than 50%. 2
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