Statistics 300: Elementary Statistics Sections 7-2, 7-3, 7-4, 7-5 - - PDF document

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Statistics 300: Elementary Statistics

Sections 7-2, 7-3, 7-4, 7-5 Parameter Estimation

• Point Estimate

–Best single value to use

• Question

–What is the probability this estimate is the correct value?

Parameter Estimation

• Question

–What is the probability this estimate is the correct value?

• Answer

–zero : assuming “x” is a continuous random variable –Example for Uniform Distribution

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If X ~ U[100,500] then

• P(x = 300) = (300-300)/(500-100)
• = 0

100 300 400 500

Parameter Estimation

• Pop. mean

–Sample mean

• Pop. proportion

–Sample proportion

• Pop. standard deviation

–Sample standard deviation

x s p ˆ µ p σ

Problem with Point Estimates

• The unknown parameter (µ, p,

etc.) is not exactly equal to our sample-based point estimate.

• So, how far away might it be?
• An interval estimate answers

this question.

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Confidence Interval

• A range of values that contains

the true value of the population parameter with a ...

• Specified “level of confidence”.
• [L(ower limit),U(pper limit)]

Terminology

• Confidence Level (a.k.a. Degree of

Confidence)

– expressed as a percent (%)

• Critical Values (a.k.a. Confidence

Coefficients)

Terminology

• “alpha” “α” = 1-Confidence

–more about α in Chapter 7

• Critical values

–express the confidence level

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Confidence Interval for µ lf σ is known(this is a rare situation)

      ⋅ = ± = n z E E x σ

α 2

Confidence Interval for µ lf σ is known(this is a rare situation) if x ~N(?,σ)

      ⋅ ± = n z x σ

α 2

Why does the Confidence Interval for µ look like this ?

      ⋅ ± = n z x σ

α 2

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score.

• z

a into value an make ) , ( ~ x n N x σ µ σ µ) ( is expression score

• z

general The − = x z n unchanged x

x x

σ σ µ µ is and : is , for

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n x z x σ µ − = is

• n

based score

• z

a so

% 95 2 ) ( 2 : statement y probabilit a Make =           < − < − n x P σ µ Using the Empirical Rule

Normal Distribution

• 3
• 2
• 1

1 2 3

Value of Observation Relative likelihood       2 α       2 α

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Check out the “Confidence z-scores”

• n the WEB page.

(In pdf format.)

Use basic rules of algebra to rearrange the parts of this z-score.

( )

95 . 2 2 : statement y probabilit the Manipulate =             < − <       − n x n P σ µ σ

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95 . 2 2 : statement y probabilit the Manipulate =             + − < − <       − − n x n x P σ µ σ

Confidence = 95% α = 1 - 95% = 5% α/2 = 2.5% = 0.025

95 . 2 2 terms the

• f
• rder

the change and (-1) by hrough multiply t : statement y probabilit the Manipulate =             + < <       − n x n x P σ µ σ

Confidence = 95% α = 1 - 95% = 5% α/2 = 2.5% = 0.025

Confidence Interval for µ lf σ is not known(usual situation)

      ⋅ ± = n s t x

2 α

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Sample Size Needed to Estimate µ within E, with Confidence = 1-α

2 2

ˆ         ⋅ = E Z n σ

α Components of Sample Size Formula when Estimating µ

• Z

α/2 reflects confidence level

– standard normal distribution

• is an estimate of , the

standard deviation of the pop.

• E is the acceptable “margin of

error” when estimating µ σ ˆ σ

Confidence Interval for p

• The Binomial Distribution

gives us a starting point for determining the distribution

• f the sample proportion : p

ˆ trials successes n x p = = ˆ

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For Binomial “x”

npq = σ

np = µ For the Sample Proportion

x is a random variable n is a constant

( )

x n n x p 1 ˆ = =

Time Out for a Principle:

If is the mean of X and “a” is a constant, what is the mean of aX? Answer: .

µ µ ⋅ a

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Apply that Principle!

• Let “a” be equal to “1/n”
• so
• and

n X X n aX p =       = = 1 ˆ p np n np a a

x p

= ⋅       = = = 1 ) (

ˆ

µ µ

Time Out for another Principle:

If is the variance of X and “a” is a constant, what is the variance

• f aX?

Answer: .

2 x

σ

2 2 2 x aX

a σ σ =

Apply that Principle!

• Let x be the binomial “x”
• Its variance is npq = np(1-p),

which is the square of is standard deviation

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Apply that Principle!

• Let “a” be equal to “1/n”
• so
• and

n X X n aX p =       = = 1 ˆ

( )

) ( / 1

2 2 2 2 ˆ

npq n a

X p

= = σ σ

Apply that Principle! n pq and n pq npq n

p p

= = = ⋅      

ˆ 2 ˆ 2

1 σ σ When n is Large,

        = = n pq p N p σ µ , ~ ˆ

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What is a Large “n” in this situation?

• Large enough so np > 5
• Large enough so n(1-p) > 5
• Examples:

– (100)(0.04) = 4 (too small) –(1000)(0.01) = 10 (big enough)

Now make a z-score

n pq p p z − = ˆ

And rearrange for a CI(p)

Make a probability statement: ˆ 1.96 1.96 95% p p P pq n     −   − < < =       Using the Empirical Rule

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Normal Distribution

• 3
• 2
• 1

1 2 3

Value of Observation Relative likelihood       2 α       2 α

Use basic rules of algebra to rearrange the parts of this z-score.

( )

Manipulate the probability statement: Step 1: Multiply through by : ˆ 1.96 1.96 0.95 pq n pq pq P p p n n   − < − < =      

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Manipulate the probability statement: ˆ Step 2: Subract from all parts of the expression: ˆ ˆ 1.96 1.96 0.95 p pq pq P p p p n n   − − < − < − + =      

Manipulate the probability statement: Step 3: Multiply through by -1: (remember to switch the directions of < >) ˆ ˆ 1.96 1.96 0.95 pq pq P p p p n n   + > > − =       Manipulate the probability statement: Step 4: Swap the left and right sides to put in conventional < < form: ˆ ˆ 1.96 1.96 0.95 p pq pq P p p p n n   − < < + =      

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Confidence Interval for p

(but the unknown p is in the

• formula. What can we do?)

n pq z p ⋅ ± =

2

ˆ

α

Confidence Interval for p

(substitute sample statistic for p)

n q p z p ˆ ˆ ˆ

2 ⋅

± =

α

Sample Size Needed to Estimate “p” within E, with Confid.=1-α

q p E Z n ˆ ˆ

2 2 2 ⋅

          =

α

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Components of Sample Size Formula when Estimating “p”

• Zα/2 is based on α using the

standard normal distribution

• p and q are estimates of the

population proportions of “successes” and “failures”

• E is the acceptable “margin
• f error” when estimating µ

Components of Sample Size Formula when Estimating “p”

• p and q are estimates of the

population proportions of “successes” and “failures”

• Use relevant information to

estimate p and q if available

• Otherwise, use p = q = 0.5, so

the product pq = 0.25

Confidence Interval for σ starts with this fact then

( )

square) (chi ~ 1

2 2 2

χ σ s n −

) , ( ~ if σ µ N x

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What have we studied already that connects with Chi-square random values?

( )

square) (chi ~ 1

2 2 2

χ σ s n −

( ) ( ) ( ) ( ) ( )

2 2 2 2 2 2

1 1 1 x n n s n x µ σ σ µ σ − − − − = − =

∑ ∑

( ) ( )

2 2 2 2 a sum of squared

standard normal values x x z µ µ σ σ     − − =           =

∑ ∑ ∑

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Confidence Interval for σ

( ) ( )

2 2 2 2

1 1

L R

s n UB s n LB χ χ − = − =