Asn2Soln - Assignment#2 Solutions Question 2. (12 Marks)...

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1 Assignment#2 Solutions Question 2. (12 Marks) The MiniTab output is given in a separate file but it is only as a double-check! a. (5 Marks) Based on Unequal Variances (1 Mark) S1: H 0 : μ 1 – μ 2 = 0.0 H A : μ 1 – μ 2 ≠ 0.0 (Two-Tailed Test) (1 Mark) Here, 2 2 2 2 1 2 1 2 1 2 34.4893 31.9953 ( ) 14.2919 10 12 s s s X X n n - = + = + = And df = 18 (see the MiniTab output with Unequal Variances.) (1 Mark) S2: t Calc = 1 2 1 2 ( ) 0 (481.2000 466.3333) 0 1.0402 14.2919 ( ) X X s X X - - - - = = - 1.04 With α = 0.05 (1/2 Mark) S3: t Crit = t α/2 (df=18) = t 0.025 (18) = 2.101 N.B. The values of df = 18 and p-Val = 0.312 are taken from the printout and the p-Val will be calculated in part ‘c’. (1 Mark) S4: Since {|t Calc | = 1.0402} < {t Crit = 2.101} b Do not Reject H 0 . (1/2 Mark ) Based on the statistical evidence, we cannot say that the two (population) mean values of the selling prices of properties for the two firms are different. {Or, {p-Val = 0.312} > {α = 0.05} b Do not Reject H 0 .} b. (2 Marks) The (Symmetric) Confidence Interval is given by: * 1 2 1 2 { (18)}{ ( )} X X t s X X - ± - = (481.20-466.33) ± 2.101 x (14.2919) = 14.8667 ± 30.0273
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2 = -15.1606 to 44.8940 (1 1/2 Marks) {The MiniTab printout gives an acceptable CI as: -15.20 to 44.90 } This CI is Consistent with the Conclusion reached in part ‘a’, since it does straddle ‘Zero’. (1/2 Mark) c. (3 Marks) Mann-Whitney Test and CI: Firm1, Firm2 N Median Firm1 10 477.00 Firm2 12 460.00 Point estimate for ETA1-ETA2 is 14.00 95.6 Percent CI for ETA1-ETA2 is (-14.01,47.98) W = 130.0 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.3390 The test is significant at 0.3382 (adjusted for ties) (1/2 Mark for the MiniTab output) -(1/2 Mark) S1: H 0 : Md Pop1 – Md Pop2 = 0 {You could write: 1 2 Md Md - 1 – η 2 = 0 or Eta 1 – Eta 2 = 0 etc} H A : Md Pop1 – Md Pop2 ≠ 0 S2 and S3 will not be calculated. -(1/2 Mark) S4/S5: Since {p-Val = 0.3382} > { α = 0.05} b Do not Reject H 0 . -(1/2 Mark) Based on the statistical evidence, the two population medians are not different. (Total 2 marks for above) (1 Mark) This is a Non-Parametric Test for testing the Difference in Medians of two Populations using two Independent Random Samples. It is known as the Mann-Whitney Test . d. (2 Marks) (1 Mark) The boxplots for the two samples are shown below. The sample for “Firm2” has (suspected) outliers and is therefore not normally distributed. The population from which it is drawn is considered not normally distributed either. Hence non-parametric test for the difference in the population medians is most appropriate and as such Mann- Whitney test done in part ‘c’ is more appropriate.
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This note was uploaded on 03/12/2012 for the course TELFER ADM2304 taught by Professor Phansalker during the Winter '12 term at University of Ottawa.

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Asn2Soln - Assignment#2 Solutions Question 2. (12 Marks)...

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