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Toetsende Statistiek - stappenplan H0 en Ha

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Dit is een stappenplan voor H0 en Ha waarmee je een ruime voldoende kunt halen voor het vak Toetsende Statistiek gegeven op de Radboud Universiteit. Makkelijk om bij het tentamen te houden! Dan weet je precies welke je wanneer moet kiezen/gebruiken. Vak is gegeven door Liesbeth Veenstra.

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Uploaded on
October 30, 2021
Number of pages
2
Written in
2020/2021
Type
Class notes
Professor(s)
Liesbeth veenstra
Contains
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H0 en Ha
Toetsende Statistiek

Eenzijdig Tweezijdig

Analyse In woorden In symbolen In woorden In symbolen df

One sample t-test H0: μ = 27,1 H0: μ = 27,1 n–1
Ha: μ > 27,1 Ha: μ ≠ 27,1
Ha: μ < 27,1
Onafhankelijke groepen H0: In populatie hebben H0: μ1 = μ2 // μd = 0 //μ1 H0: In populatie hebben H0: μ1 = μ2 // μ1 – μ2 = 0 n1 + n 2 – 2
Independent t-test twee groepen hetzelfde – μ2 = 0 twee groepen hetzelfde Ha: μ1 ≠ μ2 // μ1 – μ2 ≠ 0
groepsgemiddelde Ha: μ1 < μ2 // μd < 0 // μ1 groepsgemiddelde
Ha: In populatie is – μ2 < 0 Ha: In populatie
gemiddelde groep 1 Ha: μ1 > μ2 // μd > 0 // μ1 verschillen twee
groter (/kleiner) dan – μ2 > 0 groepsgemiddelden
gemiddelde groep 2
Afhankelijke groepen Zie onafhankelijke H0: μ1 = μ2 // μd = 0 //μ1 Zie onafhankelijke H0: μ1 = μ2 // μ1 – μ2 = 0 nd – 1
Two paired t-test groepen – μ2 = 0 groepen Ha: μ1 ≠ μ2 // μ1 – μ2 ≠ 0
Ha: μ1 < μ2 // μd < 0 // μ1
– μ2 < 0
Ha: μ1 > μ2 // μd > 0 // μ1
– μ2 > 0
2 2
Levene H0: Homoscedasticiteit in H0: σ 1=σ 2
Alleen tweezijdig de populatie = gelijke 2 2
Ha: σ 1 ≠ σ 2
varianties in de populatie
Ha: Heteroscedasticiteit in
de populatie = ongelijke
varianties in de populatie
ANOVA H0: Geen invloed X op Y in H0: μ1 = μ2 = μ3 dfM = k – 1
Altijd tweezijdig de populatie Ha: Niet H0 // μ1 = μ2 ≠ dfR = N – k
Ha: Minstens 1 μ wijkt μ3
af // tenminste 2
gemiddelden verschillen
van elkaar
Chi-kwadraat Test proporties gelijk H0: geen samenhang Test proporties gelijk k–1

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