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Unformatted text preview: N = 175 q =males = .45 males = 93 1tailed test = 0.05 CV = 1.65 ztransformation of binomial because variance Npq = 43.31 (which is greater than 9) μ = NP = 96.25 σ = Npq = 6.58 09 . 2 58 . 6 25 . 96 5 . 82 ) 5 . (==± = Npq Np x z 2.09 > 1.65 ∴ H O rejected, p < 0.05 9.5 9.5 13 6 4.) H O : Sample site distribution = Previous site distribution H 1 : Sample site distribution ≠ Previous site distribution 2tailed α = 0.05 df = K – 1 = 2 CVχ 2 = 5.99 2 χ = ∑ = k i e o e f f f 1 2 ) ( = 5 . 10 ) 8 5 . 10 ( 2+ 44 . 13 ) 21 44 . 13 ( 2+ 06 . 18 ) 13 06 . 18 ( 2= 6.27 6.27 > 5.99 ∴ H O rejected, p < 0.05 10.5 13.44 8 21 13 18.06 Upland bluff Holocene terrace Alluvial fan...
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 Spring '08
 VANGERVEN
 Normal Distribution, Probability theory, alluvial fan, site distribution, Sample site distribution

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