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# Biostatistics BIL 311

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This 70 page Class Notes was uploaded by Quinten Beatty on Thursday September 17, 2015. The Class Notes belongs to BIL 311 at University of Miami taught by Staff in Fall. Since its upload, it has received 73 views. For similar materials see /class/205750/bil-311-university-of-miami in Biology at University of Miami.

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Date Created: 09/17/15

Excel Clarification Correct Usage of statistical functions The book has a mistake on page 232 it should say Use TINV0 199 in Excel Proper Use of Excel formulas gt t Distribution 0 To obtain tdvu when u205 use TINV2gtlt lud I Example t1070975 TINV2gtlt 1097510 TINV005102228 0 To obtain tdvu tdya td1a when u0clt05 use TINV2gtltud I Example tloyoms TINV2gtlt 002510 TINV005102228 To obtain a 2sided Prtdgt x with x20 use TDISTxd2 To obtain a 2sided Prtdlt x with xlt0 use TDISTxd2 To obtain a lsided Prtdgt x with x20 use TDISTxdl To obtain a lsided Prtdlt x with xlt0 use TDISTxdl 0000 0 Table 5 in the Appendix only shows 1 sided values for ugt05 The Excel function TDIST returns probabilities for areas to the RIGHT of positive x values right tail areas I Example To obtain the 2 tailed probability for Prt50lt 200 TDIST2602 005 2gtltTDIST2601 When using the tables look for Prt50gt 200 and do lPrt50gt 200l0975005 Example To obtain the lsided value for t50005 nd t507095 in table 5 and because ult05 use l67l In Excel use TINV016016706 gt 98 Distribution 0 To obtain the twosided value for xzdyu use CHINVlud 0 To obtain Prx2dlt x use lCHIDISTxd 0 Table 6 in the Appendix only shows values for probability areas to the LEFT left tail area of x201 The Excel function CHINV returns values for probability areas to the RIGHT of xzdyu right tail areas I Example To obtain the value for 98970025 in table 6 27 and 98970975 in table 6 1902 In Excel use CHINV0975927 for x2990 and use CHINV002591902 for 98970975 To obtain Prx29lt 2103 use lCHIDIST21039001 gt F Distribution The Excel function FINV returns values for probability areas to the RIGHT of Fnydyu right tail areas Biostatistics Lecture 5 BIL 311 Lecturer Dr Patricia Buendia Lecture 5 Outline I Chapter 4 Discrete Probability Distributions The Expected Value of a Discrete Random Variable The analog of the arithmetic mean is called the expected value of a random variable or the population mean and is denoted by 1335 u ixi PrX xi i1 Example PrXx 0008 0076 0265 0411 0240 x 0 1 l l 2 3 4 Book page 85 The Variance of a Discrete Random Variable The variance of a random variable is also called the population variance Varx 02 ZR xi 02 PIquotX 2 xi 39 or I R Varx 0392 EX u2 x12 PrX x1 2 Book page 8 7 The CumulativeDistribution Function of a Discrete Random Variable The cumulativedistribution function cdf of a random variable X is denoted by FX and for a specific value x of X is defined by PrXS x Example Let s draw the cdf for table 42 PrXX 0008 0076 0265 0411 0240 x o 1 2 3 4 Book page 89 Example of a CDF PrXS xi PrXXi Xi Book page 89 Permutations and Combinations Examgle How to estimate the distribution of a type of white blood cell in the blood What is the distribution of k neutrophils out of 100 white blood cells The number of neutrophils follows a binomial distribution Use the binomial distribution to calculate the I robabilit of k out of n events occurrinv 39iA39 Book page 90 Permutations and Combinations To understand the binomial distribution learn about permutations and combinations first The number of permutations of n things taken k at a time is nPk nn 1gtltgtltn k1 The number of ways of selecting k out of n where the order of selection is imgortant Bock page 91 Permutations We have 3 mental patients and need to assign a control for eacn by choosing from 6 available controls In this case the order matters How many ways are there for selecting 3 controls I 6P36gtlt5gtlt412 quot5A39 n factorial n or n factorial is defined as 39 ngtltn1gtltn2gtlt gtlt2gtlt1 It gives the number of ways of selecting n objes v f Alternative formula for permutations n nPk n k abc and bca are two CO m b n a O n S permutations but one combination The number of ways of selecting k out of n objects withour respect to order is referred to as the number of combinations and denoted by n n an k kn k This is expressed as n choose k Combinations property of combinatorials k Calculatewmmm Results k L1 6 6 6 6 1 6 o J 2 66x5x4x3 Using the symmetry n n k l 4 4x3x2x1 59 The Binomial Distribution The binomial distribution can be applied to calculate tne probability of events with this structure a sample of independent trials each of which can have only 2 possible outcomes which are denoted as success or failure The Binomial Distribution What is the probability that 2 out of 5 white blood cells will be neutrophil given that the probability that any one cell is a neutrophil is 06 and the cell outcomes are presumed to be independent If p06 and qO4 no neutrophil then 062O43 M this is not the probability we are looking for This is the probabii that 2 WWW WIN W fcuw in a particular arrangementlorder eg xxooo the rst 2 are neutrophis 1 20 Combinations for 2 neutrophils 239 order 12345 12345 3 5 L J A O O O O 4 25 order 12345 123 5 The Binomial Distribution There are 10 orderings for 2 out of 5 events 5 5x4 10 2 2x1 xxoooxoxooxooxoxoooxoxxoooxoxo ooxxooxooxooxoxoooxx The probability of any one ordering is O62O43 Thus the searched probability is 10 062O43 O23 16 The Binomial Distribution Let s consider the question What is the probability of k success in n statistically independent trials n PrXk k pkq k k01n where q 1p Book page95 The Binomial Distribution There are three approaches to obtain the answer to the question What is the probability of k success in n statistically independent trials 1 Use the binomial distribution formula 2 Use the binomial tables 3 Use BINOMDIST in Excel Using Binomial iaoies What is the probability of 2 lymphocytes out of 10 white blood cells if p02 Mp n k 005 01 015 02 025 0 0903 081 0723 064 0563 1 0095 018 0255 032 0375 2 0003 001 0023 004 0063 10 0 0599 0349 0197 0107 0056 10 1 0315 0387 0347 m8 0188 10 2 0075 0194 0276 0302 0282 10 3 001 0057 013 Fi 025 Book page817 39i 7 Using the formula What is the probability of 2 boys out of 5 children if the I robabilit of a bo is 051 at each birth and the sexes of successive children are considered independent random variables s 1 05120492 j 5 l 239 305130492 3 O515gtlt4gtlt32gtlt1 5 c 4 2J05120493 PrX k ijkqn k Using Binomial names I Example Children develop chronic bronchitis in 3 out of 18 household where both parents suffer from chronic bronchitis I How likely are infants in at leaSt 6 om or 15 households to develop chronic bronchitis if the national incidence s c u I We want to know PrX23 21 Using Binomial IaDIes I We want to know PrX 2 3 I PrX 2 3 1 PrX s 2 I PrX S x is the cumulative probability that there are at most x successes 18 PrX 2 3 Z k 005k09518k k3 Use BINOMDIST in Excel a e aim39twmf ya quotF L 1 1 113005quot 0931 PrX 2 3 1 03972 03763 01683 0082 Book page97 99 F Use the binomial table to answer the question How likely are infants in exactly 1 out of 10 households to develop chronic bronchitis if the national incidence is 5 05987 03763 O5987O3151 03151 Biostatistics Lecture 17 BIL 311 Lecturer Dr Patricia Buendia Lecture 17 Outline l Chapter 10 Hypothesis Iesting Categorical Data Mc Nemar s ie5t I The McNemar s Test is a twosample Test for binomial proportions for matchedpairs data I Example Patients are assigned t pairs matched on age and clinical condition and assigned to two treatments The 5year survival is p10847 for treatment A p20829 for treatment B I The Yatescorrected chisquare statistic returns a pvalue of 059 which is not significant but that test can be used only if the samples are independent I Use the Nemar s test for matched pairs Survival Treatment gt5 ys s5 ys Total A 526 95 621 B 515 106 621 Book page4 08 Total 1041 201 1242 3 Mc Nemar s ie5t I Modify the observed table Survival Treatment gt5 ys s5 ys A 526 95 B 515 106 Total 1041 201 I To a matched pairs table survival or ts patients Survival of gt5 ys s 5 ys A patients gt5 ys 510 16 s 5 ys 5 90 Total 515 106 Total 621 621 1242 1 concordant pairs Total 16 type A gt5 ys discordant pairs 526 5 type B gt5 ys 95 discordant pairs 621 quot Survival of 39 quotA patients survival or ts patients gt5 ys s5ys 510 15 5 90 515 105 How many type A gt5 ys discordant pairs Total 525 95 620 Mc Nemar s Test Normal Theory Test Let pthe probability that a discordant pair is of type A If the treatments are equally effective then about an equal number of type A and type B discordant pairs would be expected and p should be 12 OthenNise we would expect plt12 or pgt12 Thus we wish to test HO p12 versus H1p 12 Mc Nemar s Test Normal Theory Test To test HO p12 versus H1zp 12 1 Create a 2x2 table of matched pairs as described reviousl 2 Count number of discordant pairs nD and the number of type AB discordant pairs n A nB respectively 3 Compute X2nAnB12nAnB 4 For a twosided level 0c test if X2gtx211a then reject H0 5 The exact pvalue is given by pvaluePrx21 2 X2 5 Use this test only if nqu 2 5 or nD1212 2 5 gt n 2 20 A up 1 2 ND lm grail Aquot distribution I Frequency X2 SXfJ 1 Acceptance region 2 tlon39region Mc Nemar s Test Normal Theory Test Example If we define an event as Surviving for 5 years there are nA 16 type A discordant events and nB 5 type B discordant events X251612516467 Because 384381395 lt X2 467 ltx210975 502 it follows that 0025ltplt005 and the results are statistically significant The treatment A members are significantly more likely to survive for 5 years than treatment B members Survival of B patients Survival of gt5 ys s 5 ys Total A patients gt5 ys 510 16 526 s 5 ys 5 90 95 Book page 411 Total 515 106 621 8 Mc Nemar s Test Exact e5t McNemar s Test for Correlated Proportions Exact Test 1 Follow the procedure in step 1 in Equation 1012 2 Follow the procedure Ln step 2 In Equation 1012 1 quotD k 15 if PIA ltan2 7 H quotu cgained 5 192 f 1171 if quotA gtnU2 MW knA Computer 7 101 H114 nZ demos 4 This test is valid for any number of discordant pairs nu but is particularly 39 i 3 a 7 useful for no lt 20 when he normallt heory test tn Equation 1012 cannotf 1 9 hemch 7 Example Com are BP measurements of 20 eo Ie done b a Computer and by a trained observer There are 9312 concordant pairs and nD8 discordant pairs We have nA7 We have and b usin39 Table 1 for n8v O5 and obtain p2gtltPrX2720031300039007 22 Book page41 24 4 Flowchart Categorical Data m 5 pm No Use McNemar s m 95 mdmndm pages 410 412 Use Fisher39s exact tat page 406 R x C contingency table R gt 2 c gt 2 Use chisquare test for R x C tables plug 428 Use msgglglg test for binnmig W W o oo ding is present or the aemzel test If unclng is t pages 387 33996 Use chlsquare test for trend if no confounding u confounding ls present page 431 5e chisquare test or heterogeneity for 2 x k tables page 428 Power Analysis for Binomial Tests Power Analysis can be used to I Calculate the power for an independent samples binomial test for Hozp1p2 versus H1zp1 p2 I Estimate sample size for 2 binomial proportions independent samples I Calculate the power for a McNemar s test for H0p12 vs H1p 12 uses a modified version of the 1 sample power formula I Estimate sample size for 2 binomial proportions paired samples Book page41 6 RgtltC Contingency Tables I So far our studies had variables with only 2 categories I Use an RgtltC contingency table with R rows and C columns to display relationships between two variables where the variable in the rows has R categories and the variable in the columns has C categories RgtltC Contingency Tables r l Example The parasite Euhaporchis californiensis uses three hosts during it s life cycle birds snails and fishes It has been observed that infected killfish spend excessive time near the water surface where they may be more vulnerable to bird predation l Using a large outdoor tank investigators owe veu t at 1 out of 50 uninfected fishes 10 out of 45 lightly infected and 37 of 46 highly infected fishes were eaten by birds iiig Create a Contingency table What was their hypothesis RgtltC Contingency Tables l Example The parasite Euhaporchis californiensis uses 39 three hosts during it s life cycle birds snails and fishes It has been observed that infected killifish spend excessive time near the water surface where they may be more vulnerable to bird predation The table below lists the numbers of killifish eaten acco 0 t e levels of parasitism Uninfected Lightly Highly Total Infected Infected Eaten 1 1 0 37 48 Not eaten 49 35 9 93 Total 50 45 46 141 Not in Book RgtltC Contingency Tables I The ex ected table is obtained usin39 the same procedure as before Uninfected Lightly Infected Highly Infected Total Eaten 170 153 157 48 Not eaten 33 297 303 93 Total 50 45 46 141 I As before we want to compare observed and expected tables I Hozp1p2p3 H1 at least two of the pi are different with pi proportion of eaten fish in infectious group i II Not in Book 15 RgtltC Contingency Tables I Under HO no relationships between variables for an RgtltC continvenc table the sum of OE2E over the RC cells in the table will approximately follow a chisquare distribution with R1C1 df I ChiSquare Test for an RgtltC Contingency Table Compute X2O11E112E11ORCERC2ERC 2 If X2gtX2R1XC11Oc 0r pPrX2R 1xC 1 gtX2lt0l rejeCt Ho Use this test only if no more than15 cells have expected values lt5 and no cell has expected valuelt1 1 2 33 4 If in our example X2695 what is the approximate pvalue I I 1 pvaluelt0001 2 pvaluegt005 3 pvaluelt005 1 If in our example X2695 what U conclusion can we reach 1 eject HO the probability 0 being eaten is contingent upon whether the fish was arasitized 2 Accept HO infection level and probability of being eaten are independent in killifish 3 Reject HO infection level is significantly different to the proportion of eaten killifish RgtltC Contingency Tables Uninfected Lightly Highly Total Uninfected Lightly Highly Total Infected Infected Infected Infected Eaten 1 1O 37 48 Eaten 170 153 157 48 Not eaten 49 35 9 93 Not eaten 33 297 303 93 Total 50 45 46 141 T t I 50 45 46 141 1 17 2 9 303 2 l Example X2Qm 695 17 303 I X2R XC oclt X22o9991381lt695X2 I It follows that plt1O9990001 Therefore the results are very highly significant and we conclude there is a significant relationship between trematode parasitism and the rates of predation by birds II 20 Not in Book True or False The formula for the ChiSquare statistic for binomial tests always involves comparing the observed with the expected valuesfrequencies 1 2 False 21 ChiSquare Goodnessof Fit Test I In order to use parametric methods we need to assume that the data comes from a specific underlying probability distribution Normal binomial Poisson ChiSquare etc I Now we will study a method to test for the goodness of fit of a probability model I H0popE versus H1 por pE 22 Book page 438 ChiSquare Goodnessof Fit Test I Example We would like to assume the observed measurements below came from a normal distribution Frequency distribution of mean diastolic blood pressure for adults 3069 years old in a communityewide screening program in East Boston Massachusetts 7 Observed Expected Observed Expected Group mm Hg frequency frequency Group frequency lrequency lt50 539 2779 280 lt90 4804 44735 250 lt60 330 5471 290 lt10 21 19 2 431l1 gt160 lt70 2132 21 2637 2100 lt1 6553 6841 1270 lt80 4554 42833 2110 251 1012 1 Total 14736 14733 l The assumption can be tested by computing the expected frequencies if the data came from a normal distribution 23 Book page 438 ChiSquare Goodnessof Fit Test I Example The expected frequencies within a group interval a to b are given by 14736gtltCDbp6 lt13 a pG I For the 250lt60 group we get 14736 gtltCD60806812 cp 50806812 14736 gtltCD1723 c1gt2557 Q Q 14736 x 004240005314736gtltO 71 Frequency distribution of mean diastolic blood pressure for adults 30 69 years old In a communitywide screening program in East Boston Massachusetts 2 i yr x7 V A i 2 33 Group mm Hg 152223 Group E 39 391 39 V 7 rquot j I V V lt50 57 779 280 lt90 4604 44735 7 gt21 gig gs 250 lt60 330 5471 290 lt100 2119 24311 260 lt70 2132 21267 2100 ltno ass 6841 270 lt50 4534 425323 2110 251 1072 24 Total 14736 14736 r i Frequency d istributinn of mean diastolic blood pressure for adulils 30 69 years old a communitywide screening program in East Boston Massachusetts The expected frequencies within group interval 100 to 110 are given by 1 14736 gtltCD100806812 I 110806812 2 659 gtltCD100806812 c1 110806812 3 14736 gtltCD110806812 c1 100806812 4 659 gtltCD110806812 c1 100806812 in39j Observed Expecied Observed Expecied V Group mm Hg frequency frequency Group frequency frequency lt50 57 779 280 lt90 11604 44785 7 250 lt60 330 5471 290 100 2119 24311 1 260lt70 2132 21267 I 2100lt110 270 lt80 4584 42833 2110 251 1072 I Total 14736 i4 36 ChiSquare Goodnessof Fit Test 1 Divide the raw data into groups 2 Calculate the k parameters of the probability model For a normal distribution e ompute he mean and the variance therefore k2 3 Calculate the probability p of obtaining a value within a particular group and the expected frequency n x previous Example 4 Compute X2O1E12E1 OgEg2Eg where g is the number of groups If X2gtngk110c or pPrXZgk1 gtX2ltoc reject H0 Use this test only if no more than 15 of the expected value are lt5 no expected value is lt1 Book page440 26 What does it mean to reject the Null Hypothesis for a Goodnessof fittest 391 1 There isasignificantdifference between the means 7 2 The proportions are the same The model fits the data well 4 The model does not fit the data ChiSquare Goodnessof Fit Test UITDHWSE E alii e mete 39 quot H Wei tie H r yetiinf Examgle We have k2 98 and 8215 df X2577792779 2511O7221O7235O2 Because X235O2gt2052x250999 HO can be reected with 0001 This means that the normal model does not provide an adequate fit to the data Frequency distribution af mean diastglic blond pressure for adults 30 69 years old In a communitywide screening program in East Boston Massachusetts Obsewed Expected Observed Expected 39 Group mm Hg freqlmricy frequency Group iraquancy frequency lt50 57 779 280 lt90 4604 44785 250 lt60 330 5471 290 100 2119 24311 260 70 EN 32 21267 2100 lt110 659 6841 Z70 lt50 4584 42833 2110 251 1072 BOOkpClg634 Total 14736 14736 39 28 Quiz Name the two approaches for testing the Hypothesis H p1p2 vs H1p1 tp2 wheln the sample sizes of two independent samples with categorical data are arge I Use 2sam Ie test for binomial I ro ortions or YatesCorrected 2gtlt2 contingency method Name one approach for testing the Hypothesis H0p1p vs H1p1 tp2 when the sample sizes of two independent samples are smalE I Fisher s exact test True or False In an RxC contingency test the alternative hypothesis is H1 at least two of the pi are different I True True or False You use the McNemar s test with independent samples I False True or False A concordant pair is a matched pair in which the outcome is the same for each member of the pair I True True or False The binomial power formulas are the same for independent as for paired samples I False You use the ChiSquare Goodness of fit test to test what I To test how well the data fits a model of distribution I ll 29 l Type of Data Goal Parametric Methods Rank Score or Binomial Survival Time Measurement from Measurement Two Possible Gaussian Population from Non Outcomes Gaussian Population Describe one group Mean SD Median Proportion Kaplan Meier survival interquartile range CU rve Compare one group to a hypothetical value Onesample t test Wilcoxon test Chisquare or Binomial test Compare two Unpaired t test MannWhitney test Fisher39s test Logrank test or Mantel unpaired groups chisquare for largeHaenszel samples Compare two paired Paired t test Wilcoxon test McNemar39s test Conditional proportional groups hazards regression Compare three or more unmatched groups Oneway ANOVA KruskalWallis test Chisquare test Cox proportional hazard regression Compare three or more matched groups Repeatedmeasures ANOVA Friedman test Cochrane Q Conditional proportional hazards regression Quantify association Pearson correlation Spearman Contingency between two variables correlation coefficients Predict value from Simple linear regression Nonparametric Simple logistic Cox proportional hazard another measured or e ression regression regression variable Nonlinear regression rgt Predict value from Multiple linear regression inegmillg s ggr Multiple logistic Cox proportional hazard several measured or binomial variables or Multiple nonlinear regression ones in class regression regression Biostatistics Lecture 13 BIL 311 Lecturer Dr Patricia Buendia Lecture 13 Outline l Chapter 8 Two sample Hypothesis Testing When carrying out a paired samples t test x L we go ahead and use the 1 sample formula 39 for the differences but for a 2 independent samples t test We use the 2 ind samples t test for unequal variances We carry out an F test first to find out if variances are unknown We use the 2 ind samples t test for unknown vanances e carry out an F test firs to find out if variances are equal j Testing Decision Are Variances Equal Pi in two Independent Normal Samples l Strategy for testing for the equality of means in two independent normal r samples 1 Perform F Test for the equality of two variances a If significant variances differentl Performt quotin test assuming unequal variances b If NOTsngnIflcant variances are the same tne Perform ttest assuming equal variances Hypothesis Testing TwoSample TTest for Independent Samples with Equal variance I Example The time it takes for the blood to clot is recorded for people in tWU groups not necessarily on the same size that are taking drugs B or G I The data in both groups are normally distributed I After running an F test we accept the Null hypothesis of I equal variances Book page3 05 Hypothesis Testing TwoSample TTest for Independent Samples with Equal variance I The first step in a 2sample t test for independent samples with equal variance test is to Calculate Lne pooed variance I The pooled estimate of the variance from two independent samples is given by S2 quot1 1S12 2 DSg quot1 712 2 Book page3 05 Hypothesis Testing TwoSample TTest for Independent Samples with Equal variance To test the hypothesis HO p1 p2 versus H1 p1 2 p2 with a signifiesw ewl of x t 371 372 We compute 511111 1n2 quot1 1512 2 1S 2 S I If tgttmn221a2 we reject HO quot1 quot2 2 I ltlstn1n22 1a2 tnjng z distribtiondistribution oftundeng I W39th3 2 x Prtnln22 s r z fts 0 p 2gtlt1 Prtnln22 St1l39ftgt 0 b a Rejection region Rejection region 6 H Acceptance region Book page 3 05 amp3 06 r I Hypothesis Testing TwoSample TTest for Independent Samples with Equal variance 37 3 6 t 1 2 W s2 quot1 1512n2 1s quot1quot2 2 I Example Blood clothing times of two independent Vrou s Vrou B has 6 I ersons Vrou G has 7 I 25O58226O818211169540211O5193 and sO72 min I t875974O7216172475 I Because t1109752201lt2475 we reject HO Book page3 06amp3 07 In the previous example we decided to reject H0 l I i What does it mean in reference to the blood clotting times and the problem at hand 5 1 Drug B and drug G have significantly different blood clotting variances 2 Drug B should be used to treat blood clots 39 3 Blood clotting times between the drug G and the drug B groups are significantly different Hypothesis Testing TwoSample TTest for Independent Samples with Equal Variance I You can obtain the exact pvalue or approximate the p value using the tables For p2gtltPt11lt2475 I t110975lt2475lt 01099 or equivalently 2201 lt 2475 lt 2718 it then follows that I 001 lt p2 lt0025 and 002 lt p lt 005 or equivalently 002 lt 2gtltPt11lt2475 lt 005 04 I The exact Pvalue 03 TDIST2475112003 1f tgt 0 p va lue l twice the right hand tail area Book page3 08 j Confidence Intervals for Independent Samples with Equal Variance I A twosided 100gtlt1oc CI for the true mean difference p1p2 between two independent samples is given by X1 X2 i tn1n2210L2 5 1n11n2 l Example A 95 CI for the true mean blood Clotting difference is given by a uw a r tm975 v7LV1617 MW r 2201gtltO4 099 min i 088 min187O11 Book page3 09 We obtained a 95 CI of 187O11 in the previous example What does it mean that O is not part of the CI Nothing he probability that the mean blood Clotting times or the two groups are the sam m1m20 is less than 5 3 Zero is not the population mean 4 The true mean blood Clotting time of each of the two groups is different to zero m1zOm2 2 O N i Q Hypothesis Testing TwoSample TTest for i I Independent Samples with Unequal Variance Reminder The ttest is quite robust for equal sizes and large sample sizes l Assume that the samples are normally distributed with the first sample from a Np1o12 distribution the second from a Nu2c522 distribution and 01272022 I We want to test H0 u1 2V8 H11H1 H2 Book page3 I 7 Hypothesis Testing TwoSample TTest for Independent Samples with Unequal Variance I Use Satterthwaite s Method and compute the test statistic t and round down the degrees of freedom d to the nearest inte er d 12 V512 1S 2 I If tgttd1a2 we reject HO V S12 n1s My Lon2 we accept HO Sfquotlynl 1S 22 2 1 I tdquotdistribtiondistributionoftunderHo it 2XPrtdvv 5 p 2 X Prl d S 11gt O a Rejection region Rejection region Does E Acceptance region look familiar Book page31 8 ag W T was i Hypothesis Testing TwoSample TTest for I Independent Samples with Unequal Variance I Example Test for the equality of the mean Cholesterol levels of the Children whose fathers have died from heart disease versus Children whose fathers do not have a history of heart disease knowin that the variances are different F test I We compute t340 and d 1514 and round cl down to d 151 the approximate degrees of freedom I The critical value method leads to t34gtt1200975198gtt1510975 therefore reject HO Using Table 5 Book page3 I 9

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