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Adm2304 Assignment 2

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Question 1: Available on MyStatLab. Question 2: Two polls (each based on a random sample of 1000) released just before the recent Ontario provincial election seemed to suggest conflicting outcomes. The breakdown in percentages for the two polls were: LiberalPC NDP Green Angus Reid 33% 36% 26% 5% Ekos 37. 7% 31. 5% 23. 3% 5. 9% (a) Test at the . 05 level of significance whether the data constitute evidence of a real overall disagreement between the two polls.

(b) In particular, test whether the data constitute evidence of a real disagreement between the levels of support for the PC party between the two polls.

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Use the . 05 level of significance. AnswerOij| Liberal| PC| NDP| Green| Row Totals| Angus Reid| 330| 360| 260| 50| 1000| Ekos| 377| 315| 233| 59| 984| Column Totals| 707| 675| 493| 109| 1984| a) Test of homogeneity H0: No real overall disagreement between the two polls HA: H0 is not true. There is a real overall disagreement between two polls ? = 0. 05 df = (r-1)*(c-1) = (2-1)(4-1) = 1*3 = 3 For df = 3; chi-square, based on 0. 05 level = 7. 8417 Expected observation = Row Total*Column Total/Grand Total Eij| Liberal| PC| NDP| Green| Row Totals|

Angus Reid| 356. 3508| 340. 2218| 248. 4879| 54. 93952| 1000| Ekos| 350. 6492| 334. 7782| 244. 5121| 54. 06048| 984| Column Totals| 707| 675| 493| 109| 1984| (Oij – Eij)| Liberal| PC| NDP| Green| Row Totals| Angus Reid| -26. 3508| 19. 77823| 11. 5121| -4. 93952| 0| Ekos| 26. 35081| -19. 7782| -11. 5121| 4. 939516| 0| Column Totals| 0| 0| 0| 0| 0| Chi-Square Test: Liberal, PC, NDP, Green Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Liberal PC NDP Green Total 330 360 260 50 1000 356. 35 340. 22 248. 49 54. 94 1. 949 1. 150 0. 533 0. 444 2 377 315 233 59 984 350. 65 334. 78 244. 51 54. 06 1. 980 1. 168 0. 542 0. 451 Total 707 675 493 109 1984 Chi-Sq = 8. 218, DF = 3, P-Value = 0. 042 Chi-square = ? (O-E)2/E = 8. 218 P-value = 0. 0417 Since p-value is less than alpha, we reject the null hypothesis. Conclude that there is a real disagreement between the two polls. b) Two – proportion Z – Test H0: p1 – p2 = p0 HA: p1 – p2 ? p0 = 0. 05 Z = (p1-hat – p2-hat) – p0 / SE (p1-hat – p2-hat) P = (p1*n1 + p2*n2) / (n1 + n2) = [(0. 36*1000) + (0. 315*984)] / (1000 + 984) = 669. 96/1984 == 0. 338 SE = sqrt{p*q*[(1/n1)+(1+n2)]} = sqrt[0. 338*0. 662*(1/1000+1/984)] = sqrt [0. 00045] = 0. 021 Z = (p1-p2) / SE = (0. 36-0. 315) / 0. 021 = 2. 14 p-value = P(Z < -2. 14) + P(Z > 2. 14) = 0. 0224*2 = 0. 0448 Since p-value is less than alpha, we reject the null hypothesis. We conclude that there is a real disagreement between the levels of support for the PC party. Question 3:

A sample of moderately aggressive stocks was followed over a six-month period. The data are displayed below as percentages. Returns: -2 -29 18 -25 -30 -58 -6 -56 (a) Perform a Wilcoxon test manually to see if the median return is indeed negative. Use the . 05 level of significance. (b) Use the large-sample version of the Wilcoxon test to calculate a z-statistic for your result. What is the p-value? (c) Do the test on Minitab. How do the Minitab results support the earlier results in (a) and (b)? Answer: a) H0: median = 0 HA: median < 0 ? = 0. 05

Absolute Values| 2| 29| 18| 25| 30| 58| 60| 56| Ranked Values| 2| 6| 18| 25| 29| 30| 56| 58| Ranks| 1| 2| 3| 4| 5| 6| 7| 8| Signs| -| -| +| -| -| -| -| -| n = 8 T- = 33 T+ = 3 T = T+ = 3 For < alternatives, we expect T+ to be small T – alpha = 6 Since T = 3 < 6, we reject the null hypothesis. Conclude that the median return is indeed negative. b) Z = [T – E(T)] / sq. root (var T) where, E(T) = n*(n+1)/4 8*(8+1)/4 8*9/4 72/4 18 Var(T) = n*(n+1)*(2n+1)/24 = [8*9*17]/4 = 1224/24 = 51 Z = [3-18]/sq. root (51) = -15/7. 14 = -2. 10 P-value = P[Z

Cite this Adm2304 Assignment 2

Adm2304 Assignment 2. (2019, May 02). Retrieved from https://graduateway.com/adm2304-assignment-2-191/

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