09基礎分析講習会

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1 データ解析の意味を理解しないでパソコンで計算して 序論 誤差解析 何のために も意味がない 以下の本でちゃんと勉強しよう R. A. Millikan ミリカン 水滴の蒸発 大学院生H. Fletcher 水滴を油滴に 博士論文単名 140の観測のうち49個除外 データ削除 実験データを正しく扱うために 化学同人編集部編 油滴実験 Regener がもともとThompsonの実験室(Cambridge Univ.)でお こなっていた F. Ehrenhaft 副電荷 との 論争 勝利 1923年ノーベル物理学賞メンデル できすぎていた実 験 助手の庭師がメンデルの理論に合わせるようにカウ ントしたのかもしれない 科学の罠 過失と不正の科学史 アレクサンダー コーン 著 酒井シヅ 三浦雅弘訳 工作舎 /20 第1版第1刷発行: 2010/3/1 第4刷 加筆改訂を含む 前田 山本 加納著 1 身近なことかも 2 有効数字 こんな時君ならどうする 明日の朝までにデータを出すように言われた 物理的に不可能な状況 Positiveなデータが出ない場合 今後の に多大な 影響がある 忙しい指導教員は 途中の過程をほとんど見ず 結果 だけを重要視する 実験については 自分以外に詳しいものがいない データでおかしな点がある 捨てるべきか否か 再現性がない Best dataのみを採用すべきか否か 統計的な処理をすべきではないのか 3 4

2

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4 Q test Q test

5 t t t-1(z%) t-1(z%) t-1(z%), z%: % 90% 95% 99%

6 x (x 1 x) 2 (x 2 x) 2 (x 5 x) 2 5 (x i x) 2 i=1 u t 1 (90%) u u 2 = i=1 (x i x) 2 1 ± t 1(90%)u u ± σ 23 24

7 t

8 29 30 y x 31 32

9 A (=x) B (=y) = SLOPE(B1:B5,A1:A5) = ITERCEPT(B1:B5,A1:A5)

10 y x

11 41 length / m time / s length / m time / s 43 44

12 45 length / m time / s 46 fxabx fxa bx

13 fx x

14 x ± δx, y ± δy q = q(x, y) =x + y x = 10 ± 2g y = 20 ± 3g q = x + y = 30 g 2 q δq = (δx) x 2 + = = g 2 q (δy) y cm155cm cm 55 56

15 0.8 分布 身長 和の分布 トーテムポール 0.8 低 高 分布 0.6 低 0.4 高 0.2 小針アキ宏 確率 統計入門 岩波書店 身長 59 60

16 q = x y 2 q δq = (δx) x q (δy) y 2 = (δx) 2 +(δy) x ± δx, y ± δy,... q = q(x, y,...) 63 64

17 65 Fig log10 Fig. Harris, Quantitative Chemical Analysis c,! log10 I I0 Transmission: T 透過度 Absorbance: A 吸光度 I I0 = T A 実測しているのは T であり c,! Transmission: Absorbance: Aはその桁であると考えてよい これでいいのか = cl I I0 log10 T A T T*100 / % cl I I0 log10 T 測定誤差T ± "T が Aやcの見積もりに どのように きいてくるのか アレニウスプロット ph など多くの例がある 67 68

18 δc δt F = dc/c dt dc = da c A = dt T ln T log 10 (! " / " ) A!" A A!A T!T 69 70! l Ac 71 72

19 73 74 T ± δt A ± δa log 10 T A log 10 (T ± δt ) = log 10 T (1 ± δt T )= log 10 T log 10 (1 ± δt T ) = A ± δa ±δa = log 10 (1 ± δt T )= (log 10 e) ln(1 ± δt T ) (1) if δt/t << 1 δa = (log 10 e) δt T 75 (2) if δt T δa = (log 10 e) ln 2 +δa = (log 10 e) ln 0 + x = e y log 10 x = y log 10 e ln x = y ln e = y log 10 x = ln x log 10 e 76

20 c / M T ±δt c / M T ±δt c / M A ± A

21 wi fx xi = w i i x = i w ix i = σ 2 = = i w i i w ix i i w i(x i x) 2 i w ix 2 i x 2 1 = x = σ 2 = = x dxf(x) dxxf(x) dx(x x) 2 f(x) dxx 2 f(x) x random walk random walk 83 84

22 random walk random walk random walk p q (= 1-p) B,p (r) = r=0 C r p r q r =(p + q) =1 r=0 r rrr B,p (r) =! = 15! : rr! r!( r)! pr q r = C r p r q r 87 r r r = rb,p (r) = C r p r q r r = p r=0 r=0 σr 2 = r r 2 = r 2 r 2 = pq (p + q) = C r p r q r r=0 p d dp (p + q) = p(p + q) 1 = C r rp r q r r=0 p d dp p(p + q) 1 = p + ( 1)p 2 = C r r 2 p r q r r=0 88

23 Probability - 2r 2r =( 2r) = (1 2p) =0 (p =1/2) ( 2r) ( 2r) 2 = ( 2r) 2 ( 2r) 2 =[ 2 4r +4r 2 2 (1 2p) 2 ] 2 =[ p + 4(pq + 2 p 2 ) 2 (1 4p +4p 2 )] 2 =4pq 2 = 2 (p =1/2) lim + 1) t = r/ (Law of large numbers) 2) r : p = λ = const (Poisson distribution) 3) t = r r (Gauss distribution) σ r lim + t = r/ law of large numbers P (t) = dr dt B,p(r) =B,p (t) r = r = p r r 2 = r2 2 r2 2 p + ( 1)p2 = 2 2 p 2 2 = pq lim + F ( t ) t = r r σ r = r p pq F (t) = dr dt B,p(r) = pqb,p (r) G 0,1 (t) t t lim F (t) = G 0,1 (t) + G µ,σ (x) 1 2πσ exp[ (x µ) 2 /(2σ 2 )] 91 92

24 G µ,σ = 1 e (x µ)2 2σ 2 2πσ lim + p = λ = const. B,p P λ (r) = λr e λ r! p = 1 r 93 94!t t = t The probability that an event occurs in!t is equal to "!t, where " is the rate of the event. The probability that the events occur r times between t = 0 and t = t is given by binomial distribution P λ (r) = =! r!( r)! (λ t)r (1 λ t) r r! λ t 1 λ t r r!( r)! = (λt)r r! = (λt)r r! = (λt)r e λt r! ( 1)...( r + 1) r (1 λt/) r (1 λt/) (/λt)( λt) 1(1 1 )(1 2 )...(1 r 1 ) (1 λt/) r [(1 λt ) λt ] λt 1 1 e Poisson distribution 95 u t 96

25 data1 data2 data3 data4 data5 (xi - <x>) 2 xi <x> u = [#i(xi - <x>) 2 / (-1)] 1/2 t-1(95%)u/# ± ±

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