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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, 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