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6 6.1 - One Sample  M ean μ, Variance σ 2, Proportion π 6 6.2 - Two Samples  M eans, Variances, Proportions μ1 vs. μ2 σ12 vs. σ22 π1 vs. π2 6 6.3 - Multiple.

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Presentation on theme: "6 6.1 - One Sample  M ean μ, Variance σ 2, Proportion π 6 6.2 - Two Samples  M eans, Variances, Proportions μ1 vs. μ2 σ12 vs. σ22 π1 vs. π2 6 6.3 - Multiple."— Presentation transcript:

1 6 6.1 - One Sample  M ean μ, Variance σ 2, Proportion π 6 6.2 - Two Samples  M eans, Variances, Proportions μ1 vs. μ2 σ12 vs. σ22 π1 vs. π2 6 6.3 - Multiple Samples  M eans, Variances, Proportions μ1, …, μk σ12, …, σk2 π1, …, πk CHAPTER 6 Statistical Inference & Hypothesis Testing CHAPTER 6 Statistical Inference & Hypothesis Testing

2 RANDOM SAMPLE size n POPULATION X = random variable, numerical (discrete or continuous) X ~ Dist ( ,  )  = mean  2 = variance Parameters Statistics Parameter Estimation variance mean Sampling Distributions

3 Sampling Distribution POPULATION Success Failure RANDOM SAMPLE size n Discrete random variable X = # Successes in sequence of n Bernoulli trials (0, …, n) For any randomly selected individual, first define a binary random variable: ParameterParameter Sampling Distribution Parameter Estimate = ? Parameter

4 POPULATION Success Failure For any randomly selected individual, first define a binary random variable: ParameterParameter Sampling Distribution Parameter Estimate = ? Parameter Discrete random variable X = # Successes in sequence of n Bernoulli trials (0, …, n) If n   15 and n (1 –  )  15, then via the Normal Approximation to the Binomial… If n   15 and n (1 –  )  15, then via the Normal Approximation to the Binomial… RANDOM SAMPLE size n Sampling Distribution

5 POPULATION Success Failure For any randomly selected individual, first define a binary random variable: ParameterParameter Sampling Distribution Parameter Estimate = ? Parameter Discrete random variable X = # Successes in sequence of n Bernoulli trials (0, …, n) If n   15 and n (1 –  )  15, then via the Normal Approximation to the Binomial… If n   15 and n (1 –  )  15, then via the Normal Approximation to the Binomial… RANDOM SAMPLE size n s.e. does not depend on  s.e. DOES depend on 

6 Sampling Distribution Example Null Distribution

7 Example Null Distribution

8 ExampleNull Hypothesis Alternative Hypothesis

9 ExampleNull Hypothesis Alternative Hypothesis 95% Margin of Error  95% Confidence Interval (for  ) =.04.16 null value does not contain null value  = 0.2  Reject at  =.05 Statistical significance  < 0.2 Statistical significance at  =.05… Evidence that  < 0.2, based on study. point estimate of true 

10 ExampleNull Hypothesis Alternative Hypothesis 95% Margin of Error  95% Acceptance Region (for H 0 ) = null value does not contain null value  = 0.2  Reject at  =.05 Statistical significance  < 0.2 Statistical significance at  =.05… Evidence that  < 0.2, based on study. point estimate of true  ExampleNull Hypothesis Alternative Hypothesis.04.16

11 null value does not contain null value  = 0.2  Reject at  =.05 point estimate does not contain point estimate  = 0.1  Reject at  =.05 ExampleNull Hypothesis Alternative Hypothesis 95% Margin of Error  point estimate of true  ExampleNull Hypothesis Alternative Hypothesis.12.28 Statistical significance  < 0.2 Statistical significance at  =.05… Evidence that  < 0.2, based on study. 95% Acceptance Region (for H 0 ) =

12 ExampleNull Hypothesis Alternative Hypothesis point estimate of true  ExampleNull Hypothesis Alternative Hypothesis.12.28 p-value =  Reject at  =.05, etc.


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