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R を用いた検定(補講) (2) — カイ二乗検定 / Other Tests Using R...
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Kenji Saito
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January 21, 2024
Business
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R を用いた検定(補講) (2) — カイ二乗検定 / Other Tests Using R (1) - Chi-Square Test
早稲田大学大学院経営管理研究科「企業データ分析」2023 冬のオンデマンド教材 第10回で使用したスライドです。
Kenji Saito
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January 21, 2024
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Transcript
generated by Stable Diffusion XL v1.0 2023 10 R (
) (2) — χ2 (WBS) 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.1/14
https://speakerdeck.com/ks91/collections/corporate-data-analysis-2023-winter 2023 10 R ( ) (2) — χ 2
— 2024-01 – p.2/14
( 20 ) 1 • 2 R • 3 •
4 • 5 • 6 ( ) • 7 (1) • 8 (2) • 9 R ( ) (1) — Welch • 10 R ( ) (2) — χ2 • 11 R ( ) (1) — 12 R ( ) (2) — 13 GPT-4 14 GPT-4 15 ( ) LaTeX Overleaf 8 (12/21 ) / (2 ) OK / 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.3/14
χ2 : : ( ) 3 ( ) 2023 10
R ( ) (2) — χ 2 — 2024-01 – p.4/14
χ2 ( ) ( ) ( ) χ2 = n
i=1 (fi − ˆ fi )2 ˆ fi ( f (frequency) ˆ f ( )) χ2 df df = n − 1 χ2 = nj j=1 ni i=1 (fji − ˆ fji )2 ˆ fji (ni × nj ) df = (ni − 1) × (nj − 1) 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.5/14
R 10 4 14 2 4 6 12 8 20
= × ( 12:8 14 ) 8.4 5.6 14 3.6 2.4 6 12 8 20 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.6/14
χ2 ( ) 1 6 (H0 ) # ( H_0
) > chisq.test(c(28, 33, 25, 35, 27, 32), p=c(1/6, 1/6, 1/6, 1/6, 1/6, 1/6)) # ( H_0 ) > chisq.test(c(42, 33, 25, 35, 27, 18), p=c(1/6, 1/6, 1/6, 1/6, 1/6, 1/6)) ‘chisq.test(. . . )’ ( ) ( ) ( ) ← R χ2 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.7/14
( ) 0 5 10 15 20 0.00 0.05 0.10
0.15 χ2(5) x dchisq(x, df) χ2 0.05 (5) = 11.1 χ2 = 2.53 “ .R” source chisqdistg(5, chisq=2.5333) 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.8/14
( ) 0 5 10 15 20 0.00 0.05 0.10
0.15 χ2(5) x dchisq(x, df) χ2 0.05 (5) = 11.1 χ2 = 11.9 chisqdistg(5, chisq=11.867) 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.9/14
χ2 (1) R A 16 4 20 12 8 20
28 12 40 options(digits=7) # A <- matrix(c(16, 12, 4, 8), 2, 2) chisq.test(A, correct=F) 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.10/14
0 5 10 15 20 0.0 0.2 0.4 0.6 0.8
χ2(1) x dchisq(x, df) χ2 0.05 (1) = 3.84 χ2 = 1.9 options(digits=3) chisqdistg(1, chisq=1.9048) 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.11/14
χ2 (2) B 160 40 200 120 80 200 280
120 400 options(digits=7) # B <- matrix(c(160, 120, 40, 80), 2, 2) chisq.test(B, correct=F) 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.12/14
0 5 10 15 20 0.0 0.2 0.4 0.6 0.8
χ2(1) x dchisq(x, df) χ2 0.05 (1) = 3.84 χ2 = 19 options(digits=3) chisqdistg(1, chisq=19.048) 2023 10 R ( ) (2) — χ 2 — 2024-01 – p.13/14
2023 10 R ( ) (2) — χ 2 —
2024-01 – p.14/14