There are 2 types of test:
Significance Testing
P-Value Testing
Slide 2
Null Hypothesis:- the assumption about the parameter Mean Height is 1.62m H˅0: μ=1.62 Proportion is 0.47 H˅0: p=0.47 Mean number per ml is 6 H˅0: λ=6
Alternate Hypothesis:- if the previous test concludes that the null hypothesis should be rejected
H˅1: μ>1.62
One -tailed tests
H˅1: p<0.47
Two-tailed test H˅1: λ≠6
Slide 3
If dealing with population and it's Normal, use a standarised Normal Distribution:
X~N(μ,σ^2)
Z=X-μ/σ
Slide 4
The smaller the value of p, the stronger the case for the rejection of H˅0.
p≤0.01 there is very stong evidence to reject H˅0
0.01<p>0.05 there is strong evidence to reject H˅0
p>0.05 there is insufficient evidence to reject H˅0
For an one-tailed test; p-value = P(X≥x) or P(X≤x)For a two-tailed test; p-vaule = 2P(X≥x)
P-Value Testing
Slide 5
Testing the mean of a distribution
A Single Value with a Random Sample of n;
X̅~N(μ,σ^2/n)
Z=X-μ˅0/√σ^2/n
2 means with a Random Sample of n;
X̅-Y̅~N(μ˅x-μ˅y,σ^2˅x/n˅x+σ^2˅y/n˅y)
Slide 6
Testing the mean of a Poisson Distribution
Single Value
X~Po(λ)
T=X˅1 + X˅2 + ... + X˅n where X˅n are independent observations.
Then T~Po(nλ)
If λ is large and NOT in tables then X~Po(λ,λ)This is a normal approximation question and a continuity correction MUST be used;
T~Po(nλ) → T~N(nλ,nλ)
Slide 7
Testing a Proportion/Binomial Probability
X~Bin(n,p)X'~Bin(n,q) - if p is not in the tables
If X~Bin(n,p) with np>10 and nq>10 then;
X~N(np,npq) approx.
As this is now a Normal approximation, a continuity correction is NEEDED!
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