CHAPTER 6 Statistical Inference & Hypothesis Testing
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CHAPTER 6 Statistical Inference & Hypothesis Testing . 6.1 - One Sample Mean μ , Variance σ 2 , Proportion π 6.2 - Two Samples Means, Variances, Proportions μ 1 vs. μ 2 σ 1 2 vs. σ 2 2 π 1 vs. π 2 6.3 - Multiple Samples Means, Variances, Proportions - PowerPoint PPT Presentation
Transcript of CHAPTER 6 Statistical Inference & Hypothesis Testing
• 6.1 - One Sample
Mean μ, Variance σ 2, Proportion π
• 6.2 - Two Samples Means, Variances, Proportions μ1 vs. μ2 σ1
2 vs. σ22 π1 vs. π2
• 6.3 - Multiple Samples Means, Variances, Proportions μ1, …, μk σ1
2, …, σk2 π1, …, πk
CHAPTER 6 Statistical Inference & Hypothesis Testing
RANDOM SAMPLE size n
POPULATION X = random variable, numerical (discrete or continuous)
X ~ Dist(, ) = mean 2 = variance Parameters
Statistics
1 2 3{ , , , , }nx x x x
Parameter Estimation
1
1 n
ii
x xn
2 2
1
1 ( )1
n
ii
s x xn
variance
mean
Sampling Distributions
,X Nn
2 2 21
ˆ(Chi-squared)
nS
1 2 3{ , , , , }nx x x x
Sampling Distribution
ˆ ,X Nn
1 2 3 where each 0 or 1{ , , , , }, in yy y y y
POPULATIONPOPULATION Success Failure
RANDOM SAMPLE size n
Discrete random variableX = # Successes in sequence of n Bernoulli trials (0, …, n)
For any randomly selected individual, first define a binary random variable:
Y 1 if Success, with prob
0 if Failure, with prob 1 Parameter
Sampling Distribution
ParameterEstimate = ?
ˆ ??? X
n
POPULATIONPOPULATION Success Failure
For any randomly selected individual, first define a binary random variable:
Y 1 if Success, with prob
0 if Failure, with prob 1 Parameter
Sampling Distribution
ˆ ,X Nn
ParameterEstimate = ?
Discrete random variableX = # Successes in sequence of n Bernoulli trials (0, …, n)
If n 15 and n (1 – ) 15, then via the Normal Approximation to the Binomial…
RANDOM SAMPLE size n
n 2 (1 )n
, .X N
ˆ X
n
Sampling Distribution
We know...~ Bin( , )n
POPULATIONPOPULATION Success Failure
For any randomly selected individual, first define a binary random variable:
Y 1 if Success, with prob
0 if Failure, with prob 1 Parameter
Sampling Distribution
ˆ ,X Nn
Sampling Distribution
ParameterEstimate = ?
Discrete random variableX = # Successes in sequence of n Bernoulli trials (0, …, n)
If n 15 and n (1 – ) 15, then via the Normal Approximation to the Binomial…
RANDOM SAMPLE size n
n 2 (1 )n
ˆ X
n
(1 ),Nn
, (1 ) .X N n n
We know...~ Bin( , )n
s.e. d
oes not
depend on s.e
. DOES
depend on
Sampling Distribution
Example
0 0:H
Null Distribution
Example
0 0:H
Null Distribution
0 0:H 0: 0.20H
Example Null Hypothesis 0: 0.2H Alternative Hypothesis : 0.2AH Sample
n = 100 X = 10
(100)(0.2) 20 15n (1 ) (100)(0.8) 80 15n
(1.96)(s.e.?)
Example Null Hypothesis 0: 0.2H Alternative Hypothesis : 0.2AH
Samplen = 100 X = 10
10 0.1100
95% Margin of Error (0.1)(0.9)
100(1.96) .0588
95% Confidence Interval (for ) = (.1 .0588, .1 .0588) (.04, .16)
0: 0.20H ˆ 0.10 .04 .16
does not contain null value = 0.2 Reject at = .05Statistical significance at = .05… Evidence that < 0.2, based on study.
point estimate of true ˆ X
n
(1.96)(s.e.?)(0.1)(0.9)100(1.96) .0588(0.2)(0.8)100(1.96) .0784
(.1 .0588, .1 .0588) (.04, .16)(.2 .0784, .2 .0784) (.12, .28)
Example Null Hypothesis 0: 0.2H Alternative Hypothesis : 0.2AH
10 0.1100
95% Margin of Error
95% Acceptance Region (for H0) =
ˆ X
n
does not contain null value = 0.2 Reject at = .05Statistical significance at = .05… Evidence that < 0.2, based on study.
point estimate of true
Example Null Hypothesis 0: 0.2H Alternative Hypothesis : 0.2AH
Samplen = 100 X = 10
0: 0.20H ˆ 0.10 .04 .16
(0.2)(0.8)100(1.96) .0784(0.2)(0.8)100(1.96) .0784
(.2 .0784, .2 .0784) (.12, .28)does not contain null value = 0.2 Reject at = .05does not contain point estimate = 0.1 Reject at = .05
Example Null Hypothesis 0: 0.2H Alternative Hypothesis : 0.2AH
10 0.1100
95% Margin of Error
ˆ X
n point estimate of true
Example Null Hypothesis 0: 0.2H Alternative Hypothesis : 0.2AH
Samplen = 100 X = 10
0: 0.20H ˆ 0.10 .12 .28
Statistical significance at = .05… Evidence that < 0.2, based on study.
95% Acceptance Region (for H0) =
?
?Z 0
ˆ
s.e.Z
2 0.1ˆP
Example Null Hypothesis 0: 0.2H Alternative Hypothesis : 0.2AH
10 0.1100
ˆ X
n point estimate of true
Example Null Hypothesis 0: 0.2H Alternative Hypothesis : 0.2AH
Samplen = 100 X = 10
0: 0.20H ˆ 0.10 .12 .28
p-value =
(0.2)(0.8)100
.10 .20
2 0.1ˆP
2.5
2 2.5P Z 2(.0062) .0124 .05
Reject at = .05, etc.0H
