Question

Applied Modeling-.. SQL-Exercise Major Project AWS-Exam Inter Data Mining 16. Sixty-five male students in an introductory psychology course were randomly assigned to one of 5 "solution strategy" groups (with n=13 males per group). The solution strategies, labelled below as A, B, C, D and E, involved the extent to which the individual is taught to be conservative in determining the characteristics involved in the problem, where A group is taught to be the least conservative and E group is taught to be the most conservative. The time in minutes required to correctly complete the problem is recorded for each participant. The means for each strategy group and a partially completed summary ANOVA table are below. Is the effect of group on time to complete the task significant? 110 Solution strategy groupITATUTAB ULCINADIIVATE Mean time to complete task 2.40 2.23 113.79 M15.91 V 4.61 W Summary ANOVA dfLOMSTIF Source Between Error Total SS 105.41 347.45 17. Given the following distribution of scores for Forced Expiratory Volume (a measure of lung function) in a group of asthmatic teens: Subject FEV1 Squared 5.3 4.6 12.3 0.8 8.0 Savau AW FEV1 2.30 2.15 3.50 2.60 2.75 2.82 4.05 2.25 2.68 3.00 4.02 2.85 3.38 38.35 16.4 16.2 13 Sum 11.4 117.79 (a) Calculate the mean. (b) Calculate the standard deviation. (C) Find the interquartile range.

Answer 1

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Soln 16. Given: Solution Strategy A I B Mean time to complete task | 2.4 | 2.23 C D G 13.79 15.91 | 4.61 Summary ANOVA SS d. f I 105.41 T I o Source Between Error Total F M.S T 1 341.45 n = 13 males per solution strategy group Test hypothesis Ho: There is no significant difference between groups in to complete the task In other words, the effect of groups on time to - Complete the task is not significant against Hi The effect of groups on time to - task is significant. complete the Consider the model. yij = ut li + Eij i=1,2,..,5 - j = 1,2, ...,131 yij is the yield or response from the ith unit receiving the ith solution strategy. A is the general mean effect Ti is the effect due to the ith solution strategy - group. Er is the error effect due to chance

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- Eij NN00,oo ). Test of hypothesis I Im notations) HO: Y = P2 = 13 = 44 = 15 against Hi all I's are not equal Sum of square is 513 - Il v=1 Treba se 3( Tio yo) I Total 5.5 (9j - goo) Treatment 55 2/ Between $$ = Error SS = 3 2 lys; - io)? 5X13 is yij yo a lung - III Total SS = Between ss t Error ss. 7 Errorss = = = Total s - Between ss 347.45 - 105.41 242.04 Source Between Error Total Ss 1 d. f 205.415 -1= 4 242.04 65-5=60 347.45 5813-1 =65-1=64 MS 2 6.3525 4.034 F F- MSBetween=6.51 MSE (Mean Sum of square) MS = SS i dif Test statistic = 6.53 25 N F ( 460) F = MS (Between) MS(Error) - 26.3525 4.034

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If Ecateulated > F (4,60), tabulated We reject Ho At d = 50% level of significance, critical value of F1460) is 2.53 Conclusion: Since F = 6.5325 7 2.53. Hence we reject Ho, at sol level of significance. Hence the effect of group on time to complete the task is significant. - 11. Subject | FEVI=Y FEVIZ y2 2.3 5.3 2.15 4.6 3.5 12.3 2.6 I 6.8 2.75 7.6 2.82 4.05 I 16.4 2.25 5.1 I 2.68 7.2 13.00 I I 4.02 I 16.2 2.85 | 8.1 13 3.38 11.4 Sum 38.35 I 117.19. Vi = 38.35 y ² = 117.79 il 1 (9) Mean (FEV1) - Sum (FEVO n=13 . n = 38.35 13 = 2.95

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Standard deviation Variance (FEV1) (18-912 ni 13 IF = 1 y ² - y² where 7 = Mean (FEV1) - 117.79 - 2.95) - - 13 9.0601 - 8.7025 0.3582 Standard deviation (FEVI) = V0.3582 =0.5985 co Inter quartile range, IQR = Q3 -Q1 where Q3: thind quartile of the distribution I QI : First quartile of the distribution Arrange FEVL in ascending order 2:15 2.25 4.02 4.05 2.3 2.6 2.68, 2.76, 2.83, 2.85, 3, 3.38, 3.5 Median is the middle value which is 2.83 Q1 = Median of upper half from row Q3 = Median of lower half, from row I to 6 8 to 13

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Subject 2 FEVL 2.15 2.25 2. 37 Q1 = 2.3+2.6 3 L 2-3 2.62 ] 2 = 2.45 Median of Upper half 6 2.68. 2.16 2.83 2.85. D2 = Median = 83 - 2 7 Median of 1 3 3.387 .5 - 4.02 . 4.05 J Q3 8 3.38 +3.5 2 = 3.44 usti lower 12 13 half Note: - In case of even number of observations, the median is arthematic mean of the middle terms. Interquartile range is geven by IQ R = Q3 - QL = 3.44 - 2.45 = 0.99