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# STATISTICAL METHOD STA 291

UK

GPA 3.87

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This 8 page Class Notes was uploaded by Helga Torp Sr. on Friday October 23, 2015. The Class Notes belongs to STA 291 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/228280/sta-291-university-of-kentucky in Statistics at University of Kentucky.

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Date Created: 10/23/15

Review for Exam II A Outcamgr epemme mun den expemem A Sempxe Space 7 set mlmwn den msxhle outcomes e1 en expmlmml Evmt r speeme mueeem e1 outcomes results den expmlmml lhsl mee M39Am euzene EXAMPLE An expem e An hemenumhexemmSem 111 one endwm Ee the ou39mme m be e 2 s p s c eu could he E hmh menu exeevvm awe we can em he dmhles we whee m end n exe some whde numhe 1 Lhmugq 6 Female eveme mhen hmh c 7 hmh me numhes exe less men or equal to 315 11 A anpmmes vi anbabxhby Emcmam r mommy functions heve the n mnng pmeues 1 me probehxmy deny event 1 hemween 1 end 0 me sum euhe thehdmes e1 en the ou39mmes m esempxe speee 1e 1 A Camp mmtr me eemplemem e1 en evem A dsnmd A e everythmgthsz e m m evem A This means pAlt e PA A Mme mm A shaded 1e helm S A Immecmn vi Evem n e the newsan e1 evmls A end 5 1 me seme as me even A end 5quot that use mm a envouwane lhsl 1e hmh m evem A end In even 5 A meme wnh A ns sheaea 1e helm A B A Unmn vi Evmts u r the mm denems A end 5 s the same es the mm A or 5quot um 1 n comm a any outcome um 13 m mum event A or m mm 5 A Mme mm AUBshsdedxshelmv A B A We m have mae mmle mm by mmhmeum mime shove Maw 13 epmme d A n By A Mubua y Exclusvgr evm39s A end exe mutually excluswg or ramp 1 um 13 no Wigwam m mm mm 1 they have mmhmgm common Sn PA n5 o A Pmure mm mm A end 5 1 hm A B EXAMPLE Lem me even39s O E B he mm es 917m w in one am mum aflwn The even39s 0 end E exe mutually exclunve ans A whxls ous mus EUA LuAe end EnA m Faua y lea e rem that n even39s are equally My end we are there exe only a fume hhhheh exeweeheee thl e we the hhmheh ex ehhe h he eehhhhe eheee s 5 me hhmheh we eeh eempwe he wobshxmy exeh ewehw A e wltAgt 7 he hhhheh emweemee h A P 75 the 1111ka cf mwmnes m 5 chhhmhmh peheehhhey e the eehheheheh hmhehhw ex A even 5 deemed MB 5 the heehehhhw he A wuheeeh eweh thl heewehws hehhehea so P09 0 m e eeee whehe we wehe m hh e the whewv eehhhhe sheee ee 5 ghee we hhew 5 memeh we he he eheehwee callus eeee seeheeehe may he afA he 5 efhheheew 5 he he h e we heye the mwe 7 PA n 5 my W heew he emhhe d 0111 A A Indepmdmb EvEmseev Aeh em ere mdepehdem m Lh mheh MeLhmelmeuV we mu eheek heh epehaehee 0f evmls A end 5 by shme n e ewem aeee hm eneee e hsl PA x my Ammlm pmsxhxhly would he showmglhel mm PAe ehe wobshih y he muewhhghmhee a eempuung probabilnms o eoee em 57 mm exoneeehhee oh ehmge ok whehe eheeeeme en mm mm pm PA P5 e 05 n A em 5 ehe mheueny emhmwe than A Pms x pa PBA x PA h A em 5 ehe hhdehhehaehw than PA x pa The pmhehdny PA 05 he sanemmes eeuea ehe 1mm 392 EXAMPLE whh e heme e We ere gvm the fdlomnglehle 0f 1an pxdmhdmes s ethxs e etehlenfnmt thebdmes we slmP y heed the velues mm the probeth e1 the hteheeethm e1 evmts For etempxe P5 n o 3 Te compute the meme pmhebmty em we 51mph edd eteythmgxh the A mhhhh 5e pm 0 10 2 o a Sxmxlele PE o 2 0 1 0 1 e o 1 To compute eny ehththhe ptehehhtee we simply nd the thehdxty cf the mtmsecum end the Prehehxhty d the madame event 7 PCnD o 2 o 2 N P DgtWm EXAMPLE wth eeee Your evante teem he h the the pleyz s You heveessxyled e Prehehhhty cf 60 thet they WI wm the chempmhlp Pest mad mdmele thet whm teem wm the dqemmmshlpethey wm the fust Q1112 nithe smes 70 nithe tme he The m t me over eha he Fhstwewm aehheetehte Let Whe the eteht thet m teem we the ahemplmshw end F be the evmt thet they Wm the fust we so we 1 the evmt they lose the hempth end F e the eteht thet they lose the rst Qme We ere gven pa o 6 me o 7 Paww 0 25 se usmgthe Ms thet PAPAlt 1 end PowAH pom 1 we eh cmplete the fdnwmg P1d7eh1hty thee hmewewhheheeme PU ewemdynesimfmd pomp mm m PF W x pa o a x o e 0 13 We hate thet pm pow n m pwlt n m the teem etheh loses the rst we end Wm the eehes or loses the test we end km the senee end these exe mutueny 1 exclusive events We already know PW F07 so we only need to nd Pchach PW0 F0 PFchc gtlt PW0 075 gtlt 04 03 So PFc 018 03 048 Thus7 c i PW F0 i 018 i amniwi ioam 0 Expected Value the expected value is given by the formula Ex So if we had the probability table 3 the expected value would be 7303 205 502 709 11 11 0 Remember that probability corresponds to area 0 Binomial Distribution the binomial distribution results from an experiment con sisting of 71 trials where each trial results in a success with probability p7 or a failure with probability 17p usually we use 1 for success and 0 for failure The mean or expected value is mp7 and the variance is np1 7p so the standard deviation is 1np1 10 0 Normal Distribution this is the distribution with the bell shaped curve The normal distributions depends on the mean usually denoted M and the standard deviation usu ally denoted 039 We have the table for probabilities of a normal with M 0 and 039 1 in the back of the book So7 to use this table for any normal distribution7 we must compute the z score o Zscore the z score for a particular value of our random variable X is given by the formula 2 z T M Whenever using the table and z scores to compute probabilities DRAW A PlCTUl TE so you can see how the probability from the table corresponds to what you are calculating Most of the time you will have to use the fact that the area on one side of the mean is 057 so you will subtract a value from 05 or add 05 to your value EXAMPLE The weight of whatsits are normally distributed with a mean of 22 stones and a standard deviation of 4 stones What is the probability that a whatsit weights less than 32 stones Let X be the weight of the whatsit The z score that corresponds to z 32 is 25 So PX lt 32 PZ lt 25 The picture looks like this 0 22 30 So we see that we need to add 0 5 to the value lrorn our table The table gves 0 4932 as the area between 0 and 25 under the standard norrnal curve So PX lt 32 PZ lt 25 0504932 09938 EXAMPLE You should be able to work bukwalds For mstance uslng the prevlous exarnple n u r n n a wlue say or such that PX lt o 0 33 The proture would look hke ths So we need to nd 0 5 a 0 33 017m our table the table only gwes the area between 0 and a zrscore We look ths up m the table to get a ewlue ol 0 44 However the 6 actual value is 7044 since we are to the left of the mean7 0 We now use the formula z 20 M or solve the z formula for z to nd the value of y y 70444 22 7176 22 2024 So the 33 d percentile whatsit weight is 2024 stones 0 Population Distribution vs Sample Distribution vs Sampling Distribution for a nice summary of this see slide 8 of Lecture 18 0 Sampling Distribution of the Sample Mean this is also referred to at least by me in the labs as the distribution of X When we want to compute the z score for the Z This comes from 2 mean of a sample of size n we use the formula 2 a 0 Central Limit Theorem the distribution of the sample mean7 X of a random sample of size n from a population with mean M and standard deviation 039 is approximately normal for large n with mean M the same as the original population and standard deviation U EXAMPLE Back to our whatsit example What is the probability that a sample of size 9 will have a mean weight greater than 24 stones 24 corresponds to a z score of 247 15 Using our table and a picture that YOU made we can see that our probability is PX gt 24 PZ gt 15 05 7 04332 00668 0 Point Estimators for this class point estimators will basically be the sample equiv alent So the point estimator of the population mean7 M7 will be the sample mean7 i the point estimator of the population variance7 02 will be the sample variance7 52 and so on For a quick discussion about estimators unbiased7 consistent7 ef cient7 etc see slides 6 11 of Lecture 19 0 Con dence Interval a con dence interval is an interval with an associated level of con dence some of these de nitions are tricky The level of con dence refers to the probability of the method containing the true population parameter for us either the mean7 M7 or the proportion7 p Once an interval is calculated there is no probability7 the true parameter is either in the interval or it isn7t The formulas for the con dence intervals of M and p are iiz oriiz piz 1 Realize how changes affect the interval width7 such as increasing n or decreasing the level of con dence decreases the interval width EXAMPLE Compute the 85 con dence interval for the population mean when a sample of 121 observations had a mean of 46 and a standard deviation of 52 Here i 467 s 527 and n 121 So we still need to nd 2 Remember7 2 corresponds to the level of con dence We want a 2 such that the area from 72 to z is 085 So on our table we need to look for g 0425 We look for 0425 in our table to nd that z 144 So our con dence interval is 46 i144 46 i 0681 V121 M XAMZPLE Suppose a department store wants to estrmate the average age ol rts cus mars correct to wrthrn 2 years wrth a 95 level ol con dence The standard devratron rs belreve to be 2 years at sample 5122 should they t 27 Here we must use one ol the lormulas lor the sample 5122 ll we let B be the allowable range then the lormulas are 2 2 n o2 lor the mean and 39n pm 75 lor the proportron Here we are loolnng lor the mean so we use the rst lormula Usrng the table we nd the e that corresponds to an area of g 0 475 s 196 2 22 351 9604 34 5744 So srnce we almys round up when computrng sample slzes we need a sample ol 35 Hypothesis Test 2 there are two hypotheses ln a hypothesrs test the null hypothesls HE and the alternatrve hypothesls H Remember the null hypothesrs cannot be proven to be true we can only show that the alternatrve hypothesrs rs true thrnk ol quotrnnocent L rnnoc c rs the null hypo hesrs Pevalue 2 the Prwlue s the probabrlrty ol seerng somethrng at least as contradrctory to the null hypothesrs So we generally want somethrng small thrs means that s unlrkely that we got a wlue contradrctory to the null hypothesrs by chance EXAMPL Perhaps we are consrdenng the mean ol a populatron So we wlll observe 100 and our alternatrve n at 100 Suppose we take a random sample and obtarn asample mean ol 30 The prcture ol the Prwlue rs below 2 v 3 5 a g 3 3 2 n The total area ol the shaded regrons grves the Prwlue

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