Statistics 300: Elementary Statistics

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Statistics 300: Elemetary Statistics Sectios 7-, 7-3, 7-4, 7-5 Parameter Estimatio Poit Estimate Best sigle value to use Questio What is the probability this estimate is the correct value? Parameter Estimatio Questio What is the probability this estimate is the correct value? Aswer zero : assumig x is a cotiuous radom variable Example for Uiform Distributio 1

If X ~ U[100,500] the P(x 300) (300-300)/(500-100) 0 100 300 400 500 Parameter Estimatio Pop. mea µ Sample mea x Pop. proportio p Sample proportio pˆ Pop. stadard deviatio Sample stadard deviatio s Problem with Poit Estimates The ukow parameter (m, p, etc.) is ot exactly equal to our sample-based poit estimate. So, how far away might it be? A iterval estimate aswers this questio.

Cofidece Iterval A rage of values that cotais the true value of the populatio parameter with a... Specified level of cofidece. [L(ower limit),u(pper limit)] Termiology Cofidece Level (a.k.a. Degree of Cofidece) expressed as a percet (%) Critical Values (a.k.a. Cofidece Coefficiets) Termiology alpha a 1-Cofidece more about a i Chapter 7 Critical values express the cofidece level 3

Cofidece Iterval for m lf s is kow (this is a rare situatio) E x ± z E α Cofidece Iterval for m lf s is kow (this is a rare situatio) if x ~N(?,s) x ± z α Why does the Cofidece Iterval for m look like this? x ± z α 4

x ~ N( µ, ) make a x value ito a z - score. The geeralz - score expressiois ( µ) z x for x, µ x ad x is µ : uchaged is 5

0. 4 5 0. 4 0 0. 3 5 0. 3 0 0. 5 0. 0 0. 1 5 0. 1 0 0. 0 5 0. 0 0 so a z -score basedo x is x µ z Usig the Empirical Rule Make a probability statemet ( x µ ) P < < 95% : Normal Distributio α Relative likelihood α -3 - -1 0 1 3 Value of Observatio 6

Check out the Cofidece z-scores o the WEB page. (I pdf format.) Use basic rules of algebra to rearrage the parts of this z-score. Maipulate the probabilit y statemet : P < ( x µ ) < 0. 95 7

Maipulate the probabilit y statemet: P x < µ < x + Cofidece 95% a 1-95% 5% a/.5% 0.05 0.95 Maipulate the probabilit y statemet: multiply through by (-1)ad chage the order of the terms P x < µ < x + 0.95 Cofidece 95% a 1-95% 5% a/.5% 0.05 Cofidece Iterval for m lf s is ot kow (usual situatio) x ± t α s 8

Sample Size Needed to Estimate m withi E, with Cofidece 1-a Z α E ˆ Compoets of Sample Size Formula whe Estimatig m Z a/ reflects cofidece level stadard ormal distributio ˆ is a estimate of, the stadard deviatio of the pop. E is the acceptable margi of error whe estimatig m Cofidece Iterval for p The Biomial Distributio gives us a startig poit for determiig the distributio of the sample proportio : pˆ x p ˆ successes trials 9

For Biomial x µ p pq For the Sample Proportio x 1 pˆ ( x) x is a radom variable is a costat Time Out for a Priciple: If µ is the mea of X ad a is a costat, what is the mea of ax? Aswer: a. µ 10

Apply that Priciple! Let a be equal to 1/ so 1 p ˆ ax X X ad µ ˆ µ p a x a( p) 1 p p Time Out for aother Priciple: If x is the variace of X ad a is a costat, what is the variace of ax? Aswer:. ax a x Apply that Priciple! Let x be the biomial x Its variace is pq p(1-p), which is the square of is stadard deviatio 11

Apply that Priciple! Let a be equal to 1/ so ad 1 p ˆ ax X ˆ a p X X ( 1/ ) ( pq) Apply that Priciple! 1 ad pˆ pq pq pq pˆ Whe is Large, pˆ ~ N µ p, pq 1

What is a Large i this situatio? Large eough so p > 5 Large eough so (1-p) > 5 Examples: (100)(0.04) 4 (too small) (1000)(0.01) 10 (big eough) Now make a z-score p p z ˆ pq Ad rearrage for a CI(p) Usig the Empirical Rule Make a probability statemet: pˆ p P 1.96< < 1.96 95% pq 13

0. 4 5 0. 4 0 0. 3 5 0. 3 0 0. 5 0. 0 0. 1 5 0. 1 0 0. 0 5 0. 0 0 Normal Distributio α Relative likelihood α -3 - -1 0 1 3 Value of Observatio Use basic rules of algebra to rearrage the parts of this z-score. Maipulate the probability statemet: pq Step 1: Multiply through by : pq pq P 1.96 ( pˆ < p) < 1.96 0.95 14

Maipulate the probability statemet: Step : Subract pˆ from all parts of the expressio: pq pq P pˆ 1.96 < p < pˆ + 1.96 0.95 Maipulate the probability statemet: Step 3: Multiply through by -1: (remember to switch the directios of < >) pq pq P pˆ 1.96 p pˆ + > > 1.96 0.95 Maipulate the probability statemet: Step 4: Swap the left ad right sides to put i covetioal < p < form: pq pq P pˆ 1.96 p pˆ < < + 1.96 0.95 15

Cofidece Iterval for p (but the ukow p is i the formula. What ca we do?) pˆ ± z α pq Cofidece Iterval for p (substitute sample statistic for p) pˆ ± zα pq ˆ ˆ Sample Size Needed to Estimate p withi E, with Cofid.1-a Z α E pq ˆ ˆ 16

Compoets of Sample Size Formula whe Estimatig p Z a/ is based o a usig the stadard ormal distributio p ad q are estimates of the populatio proportios of successes ad failures E is the acceptable margi of error whe estimatig m Compoets of Sample Size Formula whe Estimatig p p ad q are estimates of the populatio proportios of successes ad failures Use relevat iformatio to estimate p ad q if available Otherwise, use p q 0.5, so the product pq 0.5 Cofidece Iterval for s starts with this fact if x ~ N( µ, ) the ( ) 1 s ~ χ (chi square) 17

What have we studied already that coects with Chi-square radom values? ( ) 1 s ~ χ (chi square) ( ) ( 1) s ( x µ ) ( ) 1 1 ( x µ ) ( x µ ) ( x µ ) z a sum of squared stadard ormal values 18

Cofidece Iterval for s LB UB ( 1) χ R ( 1) χ L s s 19