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2010/04/20

2010/04/20. Pei-Yu Chueh. Data. benthic d18O : Lisiecki, L. E., and M. E. Raymo (2005), A Pliocene-Pleistocene stack of 57 globally distributed benthic d18O records, Paleoceanography,20, PA1003, doi:10.1029/2004PA001071.

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2010/04/20

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  1. 2010/04/20 Pei-Yu Chueh

  2. Data • benthic d18O : Lisiecki, L. E., and M. E. Raymo (2005), A Pliocene-Pleistocene stack of 57 globally distributed benthic d18O records, Paleoceanography,20, PA1003, doi:10.1029/2004PA001071. • Vostok : Petit J.R. et al., Climate and Atmospheric History of the Past 420,000 years from the Vostok Ice Core, Antarctica, Nature, 399, pp.429-436. • Orbital Variations and Insolation Database : Berger A. and Loutre M.F., 1991. Insolation values for the climate of the last 10 million years. Quaternary Sciences Review, Vol. 10 No. 4, pp. 297-317, 1991.

  3. Mean= -4.1871 Std= 0.4411 Mean= -4.1879 Std= 0.4486 Mean= -4.1937 Std= 0.4391

  4. 兩段時間平均數差異統計檢定 • 利用t test 去做檢定,查詢 matlab 的 function • H = TTEST2(X,Y) performs a T-test of the hypothesis that two independent samples, in the vectors X and Y, come from distributions with equal means, and returns the result of the test in H. H=0 indicates that the null hypothesis ("means are equal") cannot be rejected at the 5% significance level. H=1 indicates that the null hypothesis can be rejected at the 5% level. The data are assumed to come from normal distributions with unknown, but equal, variances. X and Y can have different lengths. • 利用我取的兩段時間做出來 H = TTEST2(TS1,TS2)=0 • 所以表示 the null hypothesis cannot be rejected at the 5% level

  5. 兩段時間變異數差異檢定 • 利用F test 去做檢定,查詢 matlab 的 function • H = VARTEST2(X,Y) performs an F test of the hypothesis that two independent samples, in the vectors X and Y, come from normal distributions with the same variance, against the alternative that they come from normal distributions with different variances. The result is H=0 if the null hypothesis ("variances are equal") cannot be rejected at the 5% significance level, or H=1 if the null hypothesis can be rejected at the 5% level. X and Y can have different lengths. • 利用我取的兩段時間做出來 F = VARTEST2(TS1,TS2)=0 • 所以表示 the null hypothesis cannot be rejected at the 5% level

  6. 相關分析

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