Reduce ME
1. Lower confidence
2. reduce σ
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3. increase n
HT for pop mean
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z=x-μσ√n
Errors
I=P(reject H0/H0 true)=α
II=P(accept H0/H0 wrong)
Power=1-β
Increase pow
1. increase α
2. consider alternative µ that is farther than hypothesized µ
3.increase n
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4.decrease σ
One sample t CI x±t*s√n One sample t-test
T=x-µ0s√n
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Df=n-1
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P is 2 closest
Two pop means (before and after) xD±t*sD√nD xD=x1-x2 sD=x2-(x)2nn-1 Hypothesis test
H0:µa=µb
H1: µa≠µb
T=xd-D0Sd√n
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Df=n-1
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P is 2 closest
Two pop means (2 samples one condition) σ known
CI:
xd±z*σ12n1+σ22n2
Hypothesis test:
H0:µa=µb
H1: µa≠µb
Z=x1-x2-Dσ12n1+σ22n2
s-known
CI:
xd±t*s12n1+s22n2
Hypothesis test: t=x1-x2-Ds12n1+s22n2 df= lower of (ni-1)
σ unknown (σ1=σ2)
CI:
xd±t*sp21n1+1n2 sp2=sampled pool variance df=n1+n2-2 Hypothesis test:
T=x1-x2sp21n1+1n2
------------------------------------------------- sp2=n1-1s12+n2-1s22n1+n2-2 -------------------------------------------------
P is 2 closest
Compare 2 pop variances
Assumptions:
1. 2 pops are normally dist
2. samples are random and indep
H0:σ12=σ12
H1: σ12≠σ12
F=s22s12
(or f=s12s22 when H1:σ12>σ22)
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f: numerator degrees of freedom and denominator degrees of freedom (n-1)
Single proportion p=sample prop=X/n
(X= number of success) p0=pop prop= S/n
CI:
p±z*p(1-p)n
Hypothesis test
H0:p=p0
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Z=p-p0p(1-p)n
Two pop proportions
CI:
p1-p2±z*p1(1-p1)n1+p2(1-p2)n2
Hypothesis test:
H0:p1=p2
Z=p1-p2p(1-p)(1n1+1n2)