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Spiekbrief Experimenteel en Correlationeel onderzoek (ECO) (samenvatting)

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Spiekbrief voor het tentamen van Experimenteel en Correlationeel onderzoek (ECO) (Leiden). Alle nodige informatie met extra lege ruimte voor als je zelf er nog meer bij wil. Overzichtelijk per onderwerp. Getypt. Hiermee heb ik een 9,5 voor de oefentoets gehaald.

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April 1, 2023
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Eenweg ANOVA: Populatiegemiddeldes vergelijken Meervoudige ANOVA

Eenweg ANOVA model: 𝑦𝑖𝑗𝑘 = 𝜇𝑖𝑗 + 𝜖𝑖𝑗𝑘 = 𝜇 + 𝛼𝑖 + 𝛽𝑗 + 𝛼𝛽𝑖𝑗 + 𝜖𝑖𝑗𝑘

𝑦𝑖𝑗 = 𝜇𝑖 + 𝜖𝑖𝑗 = 𝜇 + 𝛼𝑖 + 𝜖𝑖𝑗 → i = conditie A, j = conditie B, k = participant

→ i = conditie, j = participant, αi = effectparameter (µi - µ) ↓

↓ 𝑦𝑖𝑗𝑘 = 𝑦̅𝑖𝑗 + 𝜖𝑖𝑗𝑘 = 𝑦̅ + 𝛼̂𝑖 + 𝛽̂𝑗 + 𝛼𝛽
̂ 𝑖𝑗 + 𝜖𝑖𝑗𝑘

Schatten met: • α̂i = y̅i − y̅
𝑦𝑖𝑗 = 𝑦̅𝑖 + 𝜖𝑖𝑗 = 𝑦̅ + 𝛼̂𝑖 + 𝜖𝑖𝑗 • β̂j = y̅j − y̅

• α̂i = y̅i − y̅ ̂ = y̅ij − (y̅ + α̂i + β̂j)
• 𝑎𝛽𝑖𝑗

• σ → sp (gepoolde st.dev.) • σ → sp


Aannames controleren
1. Homogeniteit varianties: ANOVA tabel
a. Vuistregel: (grootste sd / kleinste sd) < 2 , of H0: alle αβij = 0; Ha : niet alle αβij = 0
b. Levene’s test: niet significant
2. Normaliteit residuen: Histogram of P-P/Q-Q-plot residuen
3. Onafhankelijkheid: Geen check
Source SS df MS F (dfeffect,
dfe)
A ∑𝑛𝑖 𝑎̂𝑖2 I–1 SSA / dfA MSA / MSe

ANOVA-tabel B ∑𝑛𝑗 𝛽̂𝑗2 J–1 SSB / dfB MSB / MSe
Hypotheses: H0: µ1 = µ2 = … = µi Ha: NIET µ1 = µ2 = … = µi A*B ̂ 𝑖𝑗
2 (I – 1)*
∑𝑛𝑖𝑗 𝑎𝛽 SSAB / dfAB MSAB / MSe
(J – 1)
Source SS df MS F (dfG, dfe)
Error ∑(𝑛𝑖 − 1)𝑠𝑖2 N–I SSe / dfe -
Groups (G) ∑𝑛𝑖 𝑎̂𝑖2 I–1 SSG / dfG MSG / MSe (e) of ∑(𝑦𝑖𝑗𝑘 − 𝑦̅𝑖 )2
Error (e) ∑(𝑛𝑖 − 1)𝑠𝑖2 N–I SSe / dfe - Total ∑(𝑦𝑖𝑗𝑘 − 𝑦̅)2 N–1 SST / dfT -
of ∑(𝑦𝑖𝑗 − (T)
2
𝑦̅𝑖 )
2
Total (T) ∑(𝑦𝑖𝑗 − 𝑦̅) N–1 SST / dfT - Effectmaten
𝑆𝑆𝑒𝑓𝑓𝑒𝑐𝑡
VAF effect in steekproef: 𝜂2 = 𝑆𝑆𝑇

Effectmaten 2
𝜂𝑝𝑎𝑟𝑡𝑖𝑎𝑙 = 𝑆𝑆
𝑆𝑆𝑒𝑓𝑓𝑒𝑐𝑡

2 𝑆𝑆𝐺 𝑒𝑓𝑓𝑒𝑐𝑡 + 𝑆𝑆𝑒
VAF in steekproef: 𝜂 = 𝑆𝑆
𝑇
𝑆𝑆𝑒𝑓𝑓𝑒𝑐𝑡 −(ⅆ𝑓𝑒𝑓𝑓𝑒𝑐𝑡 ∗𝑀𝑆𝑒)
𝑆𝑆𝐺−(ⅆ𝑓𝐺∗𝑀𝑆𝑒) ̂2 =
VAF effect in populatie: 𝜔
̂2 =
VAF in populatie: 𝜔 𝑆𝑆𝑇 +𝑀𝑆𝑒
𝑆𝑆𝑇 +𝑀𝑆𝑒

Vuistregels: 0.01 klein, 0.06 medium, 0.14 groot
A priori contrasten Post-hoc toetsen Familiegewijze foutenkans

Hypotheses: H0: µ1 = (µ2 + µ3)/2 Ha: µ1 ≠ (µ2 + µ3)/2 Bonferroni correctie / Tukey 𝛼𝑓𝑎𝑚 = 1 − (1 − 𝛼)𝑐

H0: µ2 = µ3 Ha: µ2 ≠ µ3 𝛼′ = 𝛼 ∕ 𝑐

→ omschrijven naar lineair contrast → c = k(k -1)/2

Lineair contrast: gemiddeldes met elkaar vergelijken → met αfam controleren of α’
𝜓 = 𝑎1 𝜇1 + 𝑎2 𝜇2 + ⋯ + 𝑎𝑘 𝜇𝑘 idd ongeveer 0.05 is
→ ai = contrastcoëfficiënt

Hypotheses: H0: ψ = 0 Ha: ψ ≠ 0



T-toets voor contrasten
𝑐
𝑡=
𝑆𝐸𝑐

→ df = N – I
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