solpracfinalS09

# solpracfinalS09 - PSTAT 120C: Solutions to Practice Final...

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PSTAT 120C: Solutions to Practice Final Questions June 5, 2009 1. The outcomes were Lost Won Lost Won Lost Lost Won Lost Won Lost Lost Lost Won Lost Won Lost (a) Calculate a P value for testing the null hypothesis that the outcome of each successive game is independent of the others. The number of Runs is R = 13 out of a sample of 6 Wins and 10 Losses. From Table 871, the P -value is P { R 13 } = 1 - P { R 12 } = 1 - . 99 = . 01. Therefore, P = 0 . 02. (b) We reject the null hypothesis that these outcomes are independent. There is a signiﬁcant negative association between successive outcomes. A win increases the probability that the next outcome will be a loss. 2. (a) The expected value for each cell in this table comes out to be A B C F Total Juniors 14.1 33.0 23.0 2.9 73 Seniors 9.9 23.0 16.0 2.1 51 Total 24 56 39 5 124 The expected number of F’s for both Juniors and Seniors is less than 4 so the normal approximation is not appropriate. The best solution for this table is to aggregate the number of C’s and F’s into one column. The new table is A B C or F Juniors 16 32 25 Seniors 8 24 19 (b) The expected values for this table are A B C or F Juniors 14.1 33.0 25.9 73 Seniors 9.9 23.0 18.1 52 Total 24 56 44 124 Thus the test statistic is X 2 = (16 - 14 . 1) 2 14 . 1 + (32 - 33) 2 33 + (25 - 25 . 9) 2 25 . 9 + + (8 - 9 . 9) 2 9 . 9 + (24 - 23) 2 23 + (19 - 18 . 1) 2 18 . 1 = 0 . 748 (c) There are 2 rows and 3 columns which leads to df = (2 - 1)(3 - 1) = 2. Thus, the critical value for a χ 2 distribution for α = 0 . 01 is 9.21 (from Table 6). 1

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(d) We conclude that there is not signiﬁcant diﬀerence between the grades that Juniors are getting and the grades that the Seniors are getting. 3. The mean of the data is ¯
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## This note was uploaded on 10/02/2009 for the course PSTAT 5E taught by Professor Eduardomontoya during the Spring '08 term at UCSB.

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solpracfinalS09 - PSTAT 120C: Solutions to Practice Final...

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