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# hw2ans - STAT 500 HW 2 Solution All parts of Q1 and Q2 are...

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STAT 500 — HW 2 — Solution All parts of Q1 and Q2 are one point each except for 2b that is 2 points. Q3 is 4 points. Q4 is 1 point. 1. Volatility on stock exchanges (a) No, this study is not a randomized experiment. We are observing the volatility in two major stock exchanges. Treatments (exchange) were not randomly assigned to stocks. (b) Yes, these data can be used to make inferences about the mean volatility of all stocks on the NYSE during the week of 12-16 Feb 2007. Because a random sample was selected from all stocks listed on the exchange that week. By randomly choosing stocks, the sample is representative (on average) of the whole exchange. (c) S p = s ( n 1 - 1) S 2 1 + ( n 2 - 1) S 2 2 n 1 + n 2 - 2 = r (40 - 1) × (2 . 23%) 2 + (40 - 1) × (3 . 37%) 2 40 + 40 - 2 = 2 . 85% df = 40 + 40 - 2 = 78 (d) H 0 : μ 1 = μ 2 H a : μ 1 6 = μ 2 T = x 1 - x 2 r S 2 p 1 n 1 + 1 n 2 = 2 . 91 - 4 . 39 q (2 . 85%) 2 ( 1 40 + 1 40 ) = - 2 . 31 df = 78, so p -value=0.024. Using a table, you should find p < 0.05. There is evidence of difference in mean volatility between the two exchanges. Note: Different tables have different d.f. That shouldn’t change the conclusion. (e) No, I do not agree. Because the consultant’s argument requires a causal conclusion. This study is not a randomized experiment. (f) T = x 1 - x 2 r S 2 p 1 n 1 + 1 n 2 = 2 . 91 - 4 . 39 q (2 . 85%) 2 ( 1 20 + 1 20 ) = - 1 . 64 df = 20 + 20 - 2 = 38, so p -value=0.11. (g) T = x 1 - x 2 r S 2 p 1 n 1 + 1 n 2 = 2 . 91 - 4 . 39 q (2 . 85%) 2 ( 1 100 + 1 100 ) = - 3 . 66 df = 100 + 100 - 2 = 198, so p -value=0.0003. (h) As the sample size increases, p-value becomes smaller and we are more likely to reject the null hypothesis. This is reasonable because a larger sample gives more information about the population which reduces the uncertainty about the mean volatility difference.

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hw2ans - STAT 500 HW 2 Solution All parts of Q1 and Q2 are...

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