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# Advanced Statistical Methods ISQS 5347

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This 22 page Study Guide was uploaded by Agustina Batz on Thursday October 22, 2015. The Study Guide belongs to ISQS 5347 at Texas Tech University taught by Westfall in Fall. Since its upload, it has received 119 views. For similar materials see /class/226413/isqs-5347-texas-tech-university in Informational Systems at Texas Tech University.

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Date Created: 10/22/15

Review Notes for ISQS 5347 Andrew J Patterson Spring 2002 CONTENTS PROBABILITY RULES DISCRETE PROBABILITY DISTRIBUTION FUNCTIONS Examp e I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I CONTINUOUS PROBABILITY DISTRIBUTION FUNCTIONS Examp e I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I CONDITIONAL PROBABILITY INDEPENDENCE BINOMIAL DISTRIBUTION Assumptions I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Example I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I HYPERGEOMETRIC DISTRIBUTION Assumptions I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Example I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I EXPECTATION Example I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Properties of Expectation I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Example I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I VARIANCE Properties of Variance I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I CHEBYCHEV S THEOREM SKEWNESS AND KURTOSIS EXCEL COMMANDS NORMAL DISTRIBUTION Example 1 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Example 2 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I PERCENTILESQUANTILES Continuity Correction I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Example I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I CENTRAL LIMIT THEOREMCLT Garza NNOJG N4 MAXIMUM LIKELIHOOD ESTIMATION Basics Steps For MLE I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I xample I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I LOG LIKELIHOOD FUNCTION Example I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I UNBIASEDNESS Example I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I EFFICIENCY OF ESTIMATORS Example I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I CONFIDENCE INTERVALS Con dence Interval For M When a2 Is Known I I I I I I I I I I I I I I I I I I SAMPLE SIZE DETERMINATION CONFIDENCE INTERVAL FOR p CONFIDENCE INTERVAL FOR 02 CONFIDENCE INTERVAL FOR M WHEN 02 IS NOT KNOWN CONFIDENCE INTERVAL FOR CONFIDENCE INTERVAL FOR ME 7 My HYPOTHESIS TESTS 2 Testing M M0 From A Distribution With Known a I I I I I I I I I I I I I I Testing For p p0 For Binomial Data Or Bernoulli I I I I I I I I I I I I I I I Testing 02 a I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Testing M M0 From a Distribution With 02 Unknown I I I I I I I I I I I I I TwoSample Tests I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Testing MX My TwoSample tTest I I I I I I I I I I I I I I I I I I TESTING 0 a The F TEST I I I I I I I I I I I I I I I I I I I I I I I I DISTRIBUTIONS ChiSquaredx2 Distribution I I I I I I I I I I I I I I I I I I I I I I I I I I I F Distribution I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I The Student7st Distribution I I I I I I I I I I I I I I I I I I I I I I I I I I I I MATHEMATICS PROPERTIES of LOGARITHMS Natural Logarithm I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I ESTIMATE of 02 MULTINOMIAL DISTRIBUTION Assumptions for the Multinomial Distribution I I I I I I I I I I I I I I I I I I Example I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I GOODNESS of FIT TESTS Connection of Multinomial with X2 I I I I I I I I I I I I I I I I I I I I I I I I ApplicationGooclness of Fit Tests Parameters Known I I I I I I I I I I I I I Goodness of Fit Tests Parameters Unknown I I I I I I I I I I I I I I I I I I 10 10 10 11 11 11 11 11 11 11 TESTING for PROBABILITY DISTRIBUTIONS Method I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I X2 Test For Independence I I I I I I I I I I I I I I I I I I I I I I I I REGRESSION Linear Regression I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Exponential Regression I I I I I I I I I I I I I I I I I I I I I I I I I Con dence Interval for 61 I I I I I I I I I I I I I I I I I I I I I I I I HYPOTHESIS TESTING COVARIANCE and CORRELATION Covariance I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I Correlation Coef cient I I I I I I I I I I I I I I I I I I I I I I I I I I ANOVA Testing 1 M2 Mk When 02 IS KNOWN I I I I I I I I I I I Testing 1 M2 Mk When 02 IS UNKNOWN I I I I I I I I I PAIRWISE COMPARISONS Tukey7s Method for Multiple Comparisons I I I I I I I I I I I I I I I Simultaneous Tukey Intervals I I I I I I I I I I I I I I I I I I I I I I Paired tTest I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I PROBABILITY RULES PA 2 0 Probability of an event is 2 0i 135 1 probability of all events is i PA u B PA PB e PA n B if A and B are mutually exclusive then PA u B PA PB since PA n B 0 pg 31 DISCRETE PROBABILITY DISTRIBUTION FUNCTIONS A random variable has a discrete pdf if it can only take on a certain number of probabilities For example the number you roll on a dice can only be 12345 or 6 You can t roll a 1i5i 0 25 138 1 sum of all probabilities must be li Example Let the probability distribution functionpdf of a discrete random variable X be x Px 1 i2 0 i3 15 45 2 05 This is a valid pdf since all the probabilities add up to 1 ie 2 3 45 05 L To nd PX lt 2 we would add up the probabilities for 7101i5 PX lt 2 i2 3 45 i95 CONTINUOUS PROBABILITY DISTRIBUTION FUNC TIONS Continuous pdfls have an in nite number of values For example in the stoplight example the pg 42 light stays on for 2 minutes Therefore the values range between 0 and 2 minutes but you can have any number in between 0 and 2 For example we can have 1 minute 15 minutes 153234 and etc For a variable to have a valid pdf it must satisfy 00 fYy dy 1 700 meaning the total area under the curve must li Example 1 fz7 z1 for0ltzlt2 This is a valid pdf since 2 271 71 fzdz izldzlt712zgt lg712701 0 0 2 4 To nd PX gt 1 2 2 2 PXgt11fzdz1z1dzltz2zgt17127lt1gti CONDITIONAL PROBABILITY o This is used When you are nding the probability of an event given the fact that a different event already happened o PAlB 1155 read as Probability of A happening given B already happened 0 Bayesls Theorem says that PBlAj 39 PAj PAle Where Ai is a set of n eventsi INDEPENDENCE 0 events A and B are independent if PAlB PAand alsoPBlA PB So the probability that event A happens does not depend on the event that B happened 0 also Written as PA NB PA PB BINOIVIIAL DISTRIBUTION o it is a discrete distribution 0 given by PX I 21010 10 Where Assumptions X has a binomial pdf if there are n trials each classi ed as a success or failure the trials are independent the probability of success 135 is constant for all trials F905 X is the number of successes out of the n trials Example The probability of hitting a submarine With a torpedo is p 6 Assuming that each trial is independent What is the probability we a submarine exactly 5 out of 10 shots Solution Pz 5 1 oi65045 201 What is the probability that we hit the submarine at least 5 times Solution 1 7 5 PI 5 Pz 6Pz 7Pz 8Pz 9Pz 10 15 oi65045 1 065044Mig0610040 201251215i121i040i006 834 pg 51 pg 66 pg 70 pg 135 HYPERGEOMETRIC DISTRIBUTION 0 given by PX z pg 128 7 N is the population size 7 n is the size of the sample taken from the population 7 T is the of successes 7 w is the of failures Assumptions X has a hypergeometric pdf if i i i If there is a population of N items T of Which are a success and w N 7 T are a failure 2 n items are randomly selected from the population N 3 X the number of successes out of n Example There are N 50000 engineers in the US T 10000 of them are considered nerdy i What s the probability that out of the engineers you known 6Me Mike Cary Cody Jono Ryan 2 of them are nerdyi 0500 42000 1 Solution PX 2 0246 6 EXPECTATION pg 190 o The expected value of a random variable is equal to its mean 0 EX M o For a discrete probability distribution functionpdf the expected value is pg 196 1900 Mac 2161005 allk Where 107506 is the associated probability of kl Example The pdf of a discrete random variable X is the following I 101 1 3 0 2 15 45 2 05 EX 7li3 1 5i45 2i05 475 o For a continuous pdf the expected value is EC My 0 y fyydy fy represents the probability distribution functionpdfi It can look like a uniform Nor mal etcisee pdf handout Properties of Expectation X and Y are random variables7 a and b are constants C Additivity l EX Y EX EY 202 o Linearity EaX b aEX b o EX EX n o If and only if X and Y are independent7 then EXY EX Example Let EX 1 5 and Y 3X 2 EY E3X 2 E3X E2 3EX 2 31 5 2 6 5 VARIANCE onM EKXM MX57 pg 218 Properties of Variance o Additivityl VaTX Y VaTX Va7 Y7 If and only ifX and Y are independent pg 223 o VaTaX b aQVaTX o VaTX 7 Y VaTX Va7 Y7 If and only ifX and Y are independent VaTX iVaMX a o VaTX Y Var X VaTY 2CovX7 Y for all random variables 0 the above equation simpli es to Var XY Var XVaTY if X and Y are independent CHEBYCHEV S THEOREM 0 basically tells you the probability that W is Within k standard deviationsa of the mean pg 229 2 PlW7illtegtlii 62 a2 WWimzagg 1 POWinl ltkagtli Written 3 different ways here 0 Law of Large Numbers states that the more samples you take7 the closer that X get to M Mathematically it says as n A 007 X A M SKEWNESS AND KURTOSIS o skewness is a measure of symmetry No skewness symmetric 0 given by w o Kurtosis is a measure of outlierness h see notes for examples 0 kurtosis means it is a normal distribution It can also be seen as how close to normal the pdf is ENG04 01 0 given by 73 positive kurtosis negative kurtosis EXCEL COMMANDS 0 To calculate the probability of a hypergeometric distribution use HYPGEOMDISTxnrNi Use the example for engineers in the hypergeometric section we would use PX 2 HYPGEOMDIST261000050000i o Binomial Distribution in Excel BINOMDISTxnp0 or The last eld is whether to use cumulative or not Using the example in the binomial distribution section we would calculate Pz 5 BINOMDIST510i60i Getting at least 5 hits Pz 2 5 17 Pz lt 5 17 BINOMDIST410i61 NORMAL DISTRIBUTION pg 263 403 1 7 2w 2 N 7 2 AWE L M M a gt ae The standard normal is a normal distribution with zero meanM 0 and a 1 It is given by f M o 2 1 L I 7a 75 M 27f 0 When drawing the normal draw it from M 7 30 to M 30 o to calculate probabilities with the normal you must rst standardize it and then use tables 0 to standardize any normal you subtract its meanM and then divide by its standard deviationai Example 1 Let X be normally distributed with X N N10 52 Calculate PX lt 4 Solution First standardize iti PXlt4Plt lt gtPltZlt 0 0 M 10 and a 5 in shorthand 47510 PZ lt 712 now look up 712 in the table to get PX lt 4 PZ lt 7152 7 51151 Example 2 Now nd PX gt 8 Solution PX gt 8 7 PZ gt 7417 PZ lt 74 173446 7 6554 PERCENTILESQUANTILES o c is the 10 percentilequantile of a pdf if PX S c p Continuity Correction pg 263 0 used When approximating a discrete pdfespecially binomial and hypergeometric With the normal distribution 0 to approximate a binomial pdf With a normal you must have a suf ciently large enough n7 and p is not close to 0 or 1 Only valid if np7 3xnpl 7 p gt 0 and np 3xnpl 7 p lt n o subtract When it is a lower bound and add When it is an upper bound7 let X S 5 7gt X S 55 X 2 7 7gt X 2 65 Example Question 431 on pg 272 If pX k 0i7k0i310 k for k 01Hi107 is it appropriate to approximate P4 S X S 8 by computing 35 71007 S Z S 85 71007 100703 100703 Solution No The continuity correction is done correctly but it does not satisfy the conditions In this problem n 107 p 7 np7 3xnpl 7 p gt 0 1007 7 3x100i7l 7 0 3 265 gt 0 this is okay np 3xnpl 7 p lt n 100i7 3x100i7l 7 0 3 1135 gt n this is NOT okay CENTRAL LIMIT THEOREMCLT pg 265 0 Sum of n random variables coming from the same distribution can be approximated using the normal distribution if n is large enough MAXIMUM LIKELIHOOD ESTIMATION pg 310 o estimates an unknown parameter7 97 from a known distribution fWw 9 o chooses 9 that maximizes the likelihood of the sample 0 the likelihood function L9 is a product of the pdf fW w 9 evaluated at n data points7 ie7 L6 H fWw 9 i1 o 9 does NOT change when you are doing the MLE Basics Steps For MLE l substitute 9 for the variable in the pdf to obtain fW w 9 2 Count the number of data pointsn Substitute each data point into fW w 9 and multiply them to obtain L9 3 Take the derivative of L9 with respect to 9 dig and set it equal to 0 4 solve for 9 to obtain the MLE of 9 Example Let A Ae for z gt 0 We observed X1 2X2 4X3 1 Find the MLE of A Solution Step 1 We notice this is an exponential distribution We want the MLE of A so substitute 9 into fX z to get fXz9 96 Step 2 We have n 3 data points so substitute them into fx 1 9 and multiply to get 3 H 667926679466791 636729749719 636779 i1 Step 3 Take derivative of L9 936 79 with respect to 9 and set it equal to 0 ogt 392679 93pm 926796 7 79 0 Solve for 9 For 926 793 7 79 07 then 926 79 0 or 3 7 79 0 The rst term92 7 07 9 07 but that would make the pdf 0this is actually a minimum of the likelihood function and would not make since so solve the second term3 7 79 0 377 0 37 so istheMLEo o LOG LIKELIHOOD FUNCTION 0 denoted lnL9 o it is the natural logdenoted ln of L9 0 follows same steps as the MLE7 except after step 2 and before step 3 you take the ln of L9 o sometimes the log likelihood function can not be used 10 Example Use previous Example 1 Solution After steps 1 and 2 take the natural log of Lt9 936 79 Step 2 extended lnLt9 lnt93 79 Slm 7 79 Step 3 Take the derivative and set it equal to 0 M g I 7 0 Step 4 Solve for 6 so istheMLEo UNBIASEDNESS pg 338 0 an estimator is said to be unbiased if the expected value of the estimator is equal to the actual parameter being estimated 0 an unbiased estimator satis es7 E 9 Example Assume we take take a random sample from a binomial distribution Let X be the number out of n that are a success Prove that g is an unbiased estimator of the true pi Solution We must show that t97 or in this case p X n E A E 7 E X 7 p n lt gt n in so this is an unbiased estimator EFFICIENCY OF ESTIMATORS pg 345 0 we want an estimator with the smallest variance 0 if we have two unbiased estimator7 we want the want with the smallest variance Example Let 91 and 92 be unbiased estimators7 ire Elt61gt Elt62gt 9 Let Va39rt91 5 and VaTt92 3 We would want to use 92 as our estimator because it has a smaller variance This means that the estimate from 92 would be closer to the true 9 than 91 most of the timer CONFIDENCE INTERVALS pg 323 Con dence Interval For M When 02 Is Known 0 Use this equation only if a is known 0 the l 7 a100 con dence interval for M is i 7 2042 lt M lt iza2i n w W 11 SAMPLE SIZE DETERMINATION Step 1 specify d maximum margin of error p g 330 Step 2 specify the CI or risk factor a usually a 05 Step 3 guesstimate the unknown parameters7 eig 0 Meta Step 4 solve the of Margin of Error d for n CONFIDENCE INTERVAL FOR p Let Xn7 then the CI for p is A Pier A Pier 1072042 0 0 lt p lt pw2 0 0 TL TL CONFIDENCE INTERVAL FOR 02 pg 415 the sample variance is given by 1 7 e 52212 i1 the con dence interval for a2 is n 7 US2 lt a2 lt n 7 052 2 2 X17n71 Xn71 CONFIDENCE INTERVAL FOR n WHEN 02 IS NOT KNOWN pg 430 the sample variance is given by 52 g Y e W the con dence interval for n is S YitenAi lt M lt Ytn i 2 w 1 w CONFIDENCE INTERVAL FOR 4 pg 512 the sample variance is given by 1 S2nilYiiY2 i1 2 u the con dence 1nterval for 73 1s y 52 e lt 2 Fnrimyemeg 3y qm l SN 3 e 7Fnrlanvil lt 3y CONFIDENCE INTERVAL FOR ME My pg 510 the pooled sample variance is given by pg 486 S 7 n7 7 05 my 7 0532 p 7 TLX Jr y 7 2 Where7 52 7 Z Xi 7 Y 532 2231 K 7 37 i1 sometimes they use n nX and m ny so the Cl for Mac 7 My is7 1 1 S S 1 1 I y tnny723p 71 lt Mm l by lt I y l tgm rny723p 71 HYPOTHESIS TESTS Testing M M0 From A Distribution With Known 02 Let 2 pg 372 a To test H0 M M0 versus H1 M gt M07 reject H0 ifz 2 2a b To test H0 M M0 versus H1 M lt M07 reject H0 if 2 S 720 C To test H0 M M0 versus H1 M M07 reject H0 if 2 2 gt lt Testing For p po For Binomial Data Or Bernoulli Let 2 i where z is the number of successes out of mi pg 376 x7lz7o1717o7 a To test H0 2p p0 versus H1217 gt p0 reject H0 ifz 2 2a b To test H0 p p0 versus H1217 lt p0 reject H0 if 2 S 720 C To test H0 p p0 versus H1 p p0 reject H0 if 2 2 gt lt Testing 02 03 2 pg 418 Let x2 7 n 7 Dig pg 418 a To test H0 a2 03 versus H1 a2 gt 03 reject H0 if X2 2 iamil b To test H0 a2 03 versus H1 02 lt 03 reject H0 if X2 S xinil C To test H0 02 03 versus H1202 03 reject H0 if 12 S x nil or 22 2 xiianil Testing M M0 From a Distribution With 02 Unknown 7 pg 436 Let t pg 436 a To test H0 M M0 versus H1 M gt M07 reject H0 iftZtan1 b To test H0 M M0 versus H1 M lt M07 reject H0 if t S 770671 versus H1 M M07 reject H0 if M 2 tn1 C To test H0 M M0 TwoSample Tests Testing MX My TwoSample tTest pg 484 pg 486 the pooled sample variance is given by pg 486 S2 n1 e we my 7 05 p TLX my 7 2 Where7 n 52 Z Xi 37V 532 223109 0 37V i1 a To test H0 MX My versus H1 MX gt My7 reject H0 ift 2 tanzny2 b To test H0 MX My versus H1 MX lt My7 reject H0 ift S itamzny72 C To test H0 MX My versus H1 MX My7 reject H0 if M 2 tnzny2 TESTING 0 0 The F TEST 2 pg 497 a To test H0 0 0 versus H1 0 gt 07 reject H0 if if S Fany1nz1 b To test H0 2 0 0 versus H1 0 lt 07 reject H0 if if C To test H0 2 s 2 7 2 2 2 0X 7 0 versus H1i0X 0y7 reject H0 1f 5 if S Fanyiim i 0r 1f if 2 Fligmyiim i S Fliamyilmzil Test Reject Criteria H0 0 0 vs H1 0 gt 0 12 S Fany1nz1 H0 0 0 vs H1 0 lt 0 g S F17any1nz1 H0 0 0 vs H1 0 f 0 71 S Fny1nz1 or if 71 2 F1ny1nz1 DISTRIBUTIONS Chi Squared X2 Distribution pg 410 Let Z1 Z2 m Zn be a set of n independent standard normal random variables Then 2211 Z1 has a Chisquare distribution With n degrees of freedomi Let Y1Y2 HWY be a random sample from a normal distribution With mean M and variance 02 De ne S2 as 1 7 7 5224812402 i1 Then the ratio n 7 US2 a2 has a Chisquare distribution With n 7 1 degrees of freedomi F Distribution pgi421 Let U and V be two independent Chisquare random variables With m and n degrees of freedomrespectivelyi Then the ratio iUm W has an F Distribution with m and n degrees of freedom F The Student7st Distribution pg 425 Let Z N N0 l and U N xi then the ratio L m has a studentls tdistribution With n degrees of freedomtn 72 is distributed as tn1 in case you forgot 32 Til 211Xi7 XV MATHEMATICS PROPERTIES of LOGARITHMS Natural Logarithm also can be written as loge said log base e 1nXY1nXlnY ln lnX7lnY lnXC clnX lneX X lneC c 1nXc Yb clnX b1nY ilnY loggchm ESTIMATE of 02 0 an estimate of 02 is n 1 5392 YiYQ 771le gt MULTINOMIAL DISTRIBUTION o It is an extension of the binomial distribution to kVariables pg 527 Assumptions for the Multinomial Distribution there are n trials each classi ed into only I of k categories Cl 52 ck the trials are independent the probability set 101102 pk is constant for all trials ie p1trial l p1 trial n F905 44 77 X out of n for category 2 o The pdf of a multinomial distribution is l n 061 12 Tie Illlgl1klpl p2 H pk PX117X2I27Xk1k Example Suppose we are going shing and we know that the distribution of sh in the lake are as Bass 15 Cat sh 35 Crappie 4 Other 1 What s the probabilty that if we caught three sh 2 are bass and l is a crappie Solution P2 Bassl Crappie P2 Bass0 Cat shl Crappie0 other 3 m1523504110 31524 027 16 GOODNESS 0f FIT TESTS pg 532 0 used to test whether data comes from a speci c distribution Connection of Multinomial with X2 pg 533 Let X17X27 i i i 7X116 beamultinomialrandomvariable7iiei7 X17X27 i i i 7X116 N MNnp1iH 71016 then a the random variable k C Z Xi npi2 i1 npi has approximately a X2 distribution with kil degrees of freedomi For the approximation to be adequate7 the k classes should de ned so that npi 2 5 for all it ApplicationzGoodness of Fit Tests Parameters Known H0 p17iii7pk7r1iu7rk H1 at least one pi m when H0 is true k Ii 711002 2 E I N X1671 i1 npl so reject H0 when C Iimi2 gt 2 7 11 mm 7 Xlia il Goodness of Fit Tests Parameters Unknown pg 540 if we know what distribution a data might come from7 but not the parameters of the distri bution we use this metho i the test statistic becomes k 7 A 2 Cl 212 71 i1 n i Here the sample estimate of p1 that is is used instead of the null hypothesis value m Testing suppose that f0 is a pdf having 7 unknown parameters To test H0 fy f0 versus H1 1 fYy ye foy7 We would reject H0 if k A Cl Z i 711002 gt X2 i 1 n i 7 1704167177 TESTING for PROBABILITY DISTRIBUTIONS Method 1 categorize the observations such that there are 2 5 in each category E0 estimate the parameters of the probability distribution M 02 A 7etc 9 Calculate the probabilities for the multinomial categories using estimated parameters these are the null probabilities 9 Perform the XQTesti Subtract 1 degree of freedomdf for every parameter you estimated in step 2 X2 Test For Independence pg 548 Let Ii yi i l 2 i i n be a set of categorical data reduced to a contingency table with 7 rows and 5 columns Let the random variable Xij denote the number of data points in the ith row of X and the jth column of Y Let and ij be the observation proportions belonging the ith row of X and the jth column of Y respectively Denote 7 c A A X n V 2 c2 lt m H F1 10qu To test H0 X and Y are independent versus H1 that X and Y are dependent we would reject 2 H0 1f 52 2 X17047 71071 REGRESSION pg 559 Linear Regression Given n point zly1i i Imyn the straight line that best ts the points is given by y 50 511 where the slope 61 is given by 61 7 21 Iiyi 21 MEL 1 EL Ii and the yintercept 60 is Bo n n 211 yz39 51 211 Ii y 7 Bli Another way to do this is with the calculator So let s say we have the data points 01 1 lt5 1 3 2 5and we want to nd the linear regression for it ie y 60 611 This is how we do it 1 First make sure there are no variables storedi so hit F6 and then the clear azlli 2 now we want to store the zy valuesi Goto APPSHDataMatrix EditorHNew Hit the APPS77 button and then 6this is the DataMatrix Editor Got to New This will bring up a new menu We want a List so push right and then go down to list Give the new list a name I use a single variable name so then I can clear it as in step 1 3 Now put the 17s in column 1 and the y7s in column 2 Be sure they are in the same r er 4 push F5 Calci ln calculation type go to LinReg and press enter 5 For the I enter in cl for the rst column where we put the 1 values 6 For the y enter in C for the rst column where we put the y values 7 press enter and it will give you the correct values of 60 b 12 and 61 a 185 So our line to estimate these data points is y 60 511 12 1851 8 If you want to see or or Eziyi etci hit the F5 button againCalci and then select Tonar instead of LinRegW This will give you all of the info you need in order to show you work on a test 9 don7t forget to repeat step 1 to clear out all of your variables when you are donell Exponential Regression used when we want to estimate data with a function of the form 9 506mg we can make it a linear relationship between I and y by taking the natural log of both sides giving 1119 11150 511 so we can nd the solution to the problem by taking the natural log of our observed data and then performing linear regression or use the equations 51 7 7121 Ii lnyi 21 1119i 7 7421 9512 EL 2 ln o 2111119239 251 211 Ii g 7 Bli once we nd ln o we take the exponential of it to get the true 60 item7 Bo elquot 30 now we can nd y Beef This can also be done with the calculator except putting in lny instead of y and then nding 50 eln 30 Con dence Interval for 61 Con dence interval for the slope 61 is A 2 A 2 1 tn721 lt 51 lt 51tgmi2 where 32 is the variance of the error7 82 1 n 2 1 n A A 2 n7 Hel n72 120275076110 HYPOTHESIS TESTING Let t where z is the number of successes out of mi pg 591 a To test H0 61 6 versus H1 61 gt 5 reject H0 ift 2 teakg b To test H0 61 6 versus H1261 lt 5 reject H0 ift S 770672 C To test H0 51 6 versus H1 61 f 5 reject H0 if M 2 ta2n2 2 Prediction Interval pg 601 used to give a range of numbers of a predicted value De nition Let 117Y1l l l In7Yn be a set of n points that satisfy the assumptions of a linear model A prediction interval for Y at the xed value I is given y Where 9 80 811 Con dence Interval for EYx pg 599 De nition Let 117Y17 i i i 71n7Yn be a set of n points that satisfy the assumptions of a linear model A con dence interval for 60 511 Q i tn72 8 Where 9 30 311 COVARIANCE and CORRELATION pg 607 Covariance The covariance of two random variables X and Y is de ned as Comm EX 7 XY 7 m1 My 7 WW note CovX7 X VaTX le and Y are independent then CovX7 Y 0 This is because ifX and Y are independent then EXY EX EY7 then COUCH EXY MXW EXEY MXMY MXMY 7 MXMY 0 The inverse is not true however If the covariance is 0 this does not necessarily mean that the variables are independent VaTX Y Var X VaTY 200vX7 Y for all random variables the above equation simpli es to Var XY Var XVaTY if X and Y are independent estimate of the covariance is 1 7 7 SXy n71 XiiXYZ7Y Correlation Coe icient pg 611 the correlation coef cient tells how Y tends to increase as X increases It is a measure of the linear relationship between X and Yi the correlation coef cient7 denoted pX7 Y between random variables X and Y is given as C01 X7 Y X7Y O39XO39Y WXJN S 1 pX7 Y 1 if and only if Y aX b for constants a bi estimate of the correlation coef cient is T SXY sty where 5X and Sy are estimates of 0X and 0y respectively ANOVA pg 636 Testing 1 M2 Mk When 02 IS KNOWN to test H0 M1 M2 Mk vsl H1 not all the Mjls are equal we would reject H0 if 513R 2 X1ak1 we can get SSTR from the equation on page 636 or from Excel see Case Study 1221 on pg 641 for an example An ANOVA table generated by Excel will look like this Source of Variation SS f MS Pvalue F Crit Between Groups SSTR k1 MSTR PFk1nk 2 F Within Groups SSE nk MSE Total SSTOT n1 we would lookup the value for SSTR from the table to do our test Testing 1 M2 Mk When 02 IS UNKNOWN to test H0 M1 M2 Mk vsl H1 not all the Mjls are equal the test statistic is F 7 SSTRk 71 T SSEn 7 K we would reject H0 if F 2 Fkualwl kk PAIRWISE COMPARISONS to test H0 2 M Mj versus H11 M y Mj the tstat is 7 Y1 Y1 tij MSE n7 717 where MSE n1 71s n2 7 1s 1 H nk 7 Us n1n2ulnk7k Reject H0 if ltijl gt tnk we can then get the MSE from the table in order to compute our tstatisticl although this can be used to test multiple variables7 it is really only used to test 1 2 because the error becomes too big This is because for every test we perform we get a PType 1 error a for every comparisonl we would rather use Tukey s methodl Tukey7s Method for Multiple Comparisons 647 pg we use this for multiple comparison to avoid error discussed above Decision Rule Reject H0 M M if ltijl gt c Qkn7kolt k is the number of groups where c Q refers to the studentized range distribution 21 Simultaneous Tukey Intervals EffjiCRMSE 1 1 i i ni nj Palred tTest pg 682 used to test if data values are paired e correlated Let MD MX MYyDiXi YYi totest H0MD OVS H12MD 7amp0 the tstatistic is if D SDx Where 1 n D ED1 and 52 1 imp Dy 2211 2211 i1 nn 7 1 reject H0 if M 2 tn1

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