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Lecture8

# Lecture8 - sample from a normal population with 3.7 0.73...

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March 1, 2010 Chapter 6. Interval Estimation Example. Let 1 2 , , , n X X X L be a random sample from a normal population 2 ( , ) N μ σ 1. Find the MLE of μ 2. Please construct a 95% confidence interval for . Solution. 1. ˆ X = 2. ( ) 0.95 P a b = How to construct a CI for 1. Start with the point estimator for 2 ~ ( , ) X N n σ 2. Biased on X , we’ll construct a new R.V. that is called a pivotal quantity for . Definition. Pivotal Quantity It is a function of the sample and the parameter of interest ( ). Furthermore, we know its distribution entirely. ~ (0,1) X Z N n - = Scenario 1 : If 2 is known (e.g. 2 3.7 = ), then Z is a P.Q. for 1

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Scenario 2 : If 2 σ is unknown, then Z is NOT a P.Q. for μ 3. Scenario 1 : Draw the pdf of your PQ. e.g. Let α = 0.05 in the above figure, we have: 0.025 0.025 ( ) 0.95 P Z Z Z - = 2
0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 ( ) 0.95 ( ) 0.95 ( ) 0.95 ( ) 0.95 ( ) 0.95 X P Z Z n P Z X Z n n P X Z X Z n n P X Z X Z n n P X Z X Z n n μ σ - - = - - = - - ≤ - ≤ - + = + - = - + = The 95% confidence interval for (when 2 is known and the population is normal) is 0.025 0.025 , X Z X Z n n - + Example . Please calculate the 95% CI for the population mean based on a random

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Unformatted text preview: sample from a normal population with 3.7, 0.73, 100 x n = = = . Solution. From the Z-table we know that 0.025 1.96 Z = , Therefore the 95% CI for μ is: 3.7 3.7 0.73 1.96 ,0.73 1.96 100 100 -+ Now we present the derivation of the general formula. 3 Let α be any small positive value less than 1 (*usually less than 0.5), in the above figure, we have: / 2 /2 ( ) 1 P Z Z Z-≤ ≤ = -/2 /2 /2 / 2 / 2 /2 ( ) 1 ( ) 1 ( ) 1 X P Z Z n P Z X Z n n P X Z X Z n n μ σ--≤ ≤ = --≤-≤ = --≤ ≤ + = -∴ The 100(1-α)% confidence interval for (when 2 is known and the population is normal) is / 2 / 2 , X Z X Z n n -+ 4...
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Lecture8 - sample from a normal population with 3.7 0.73...

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