stat crip 2

stat crip 2 - I Consider a 2 x 2 table I Expected cfll...

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Unformatted text preview: I Consider a 2 x 2 table I Expected cfll courts (under 1-5,] are the row rota] times the col'mm' total divided hs'tlie overall total 5 F Tom [1 "5310],! “m “mum” WWW" "1”" I Standardized residuals flour each cell are Group 1 a b a _b .1" i!- rfifl—‘i. -: p -: '5— ra'g—‘i'fijl _ l— o . . . , _ -. I.- 4.- 1: observeo count -._- erepecteo count I'd‘fEfipECTefl count G: 3“? 2 C d C d I Example: binge drinking 511111;? T0011 11 _ C b _ d- [l Maj-101'}. confidamo htomrlfortr popduflon prop-0111011 ' "'- mm m Relative risk = R = Lot = a,.-'|a+h||-'Ic'lc-d ] F‘ P—Im r,— I- .I' I- P—Enfl .— '14 P1 1' 'l.' I II I 11534 [11:11 " ' ' ' ' "J" ‘ ' -. ' ' realism [—5.33] '7353 [313] [5 “I an: [71:94] ['1 fil - - 1:135 -53 95% confidence interval = RR 3: e l “" For a =31 15 ifpopulation is normal use 1‘ Known Variant-II Unknmrn Variance Patrol: Tests . - - I . test nanic -:-tcst 2 .saunplc t-tcst pnolctl t-tcs‘t paired fi-tcst Sign Test For n 15 rfpopl nnot normal do not use I ( I) H i) v“ ((1 fig — . - --.4 ; o2:w .2:% "'i's For a = 1: —-—l0 use 2' except if PDPI n str'onElE’ skewed S'd' “3mm” “T "I‘ "I ‘9 men—2 ‘6 24—1 “"' “firmw- . - . I; a fix Sir . 1; L La For :1 3 :10 r' can be used all the trrne S'E' “V wx + n, V n, + m. "w n, + n, a Mm“ . _ _ s 1 lllll‘iULgJJ. J . ,. Li. 3 i e 2 7 i u 21'} close to Z hilllj] { < M US" mnpum 11 +1:y n - - - - .. .— - 1- = test stat W"? W"? ST”’”' TILL. . . . ._. . _. . - ‘- --I I < n .3 I Mean IQ ofchildren sunrvrng a neonatal stroke mm “1"” ’ y“ Test: H0: [.1 = 100I versus Ha: pi 100I 1:5 known = 15] We expect the true [.1 to he about 90 How large ofa sample Err) do we need to detect a mean of 90 with 80% power using a 2-sided test with o = 0.05'.’ Pfireject H[,: p. = 100 .u =90}= :“r—iuu ' 'r—mo ix of in " o... I Pgreject HE: p = 100 Ill = =K c—1.96 j—P P 2- 1.95 := 11.3, solve for n J? I J \ JT—90 10 '. .I—PU 10 =P {—1.96— r —P‘ . 5-1.96— . L odd; cod; ..' 'u.‘ 5.345 tin-"Jo ,. / ~, ,z . .' 10 10 '- =P Z-i—lflo— I. +P 3}].96— . =03 Ila GLUE; .'| lg GENE; -'l Very close to 0 I From I‘:I:U_.l] table P{Z <12 2,3] = 0.8 means 2,] = 0.34. so 10 mar—1.95115 are? 10 so :1 = I: 4.2 ]: =l'.-'.|5_. 13 children are needed in the study I So in this case we want to calculate PEIEJECT H3: p. = 300 p. = T50mg] I Recall that our test statistic and rejection strategy for this test was to reject HE, if z =iI -1.|5-l5 {ct = 0.05 1-tailed test] where _r—aj _ r—aoo _r—aoo z _ I _ — _ — an“; ream—a 44.31 I So we reject HEl if 0.84 = —1.96 — If we can manipulate the data into a "success" versus “failure” structure we can use a binomial distribution WlThp = 132 for analysis In a sign test we test Hg: p = 132 versus Ha: p 95 1.52 and use the binomial X, the number of successes in a trials: as the test statistic In a sign test for matched pairs we use X. the number ofpositive differences for the pairs — ignoring pairs with difference 0 f—SDO Z = — I: -l.|545 44 11 Which is equivalent to rejecting H.) if E =::-i.64ix.j44.3ij—3430:3311 Iii-FF So power = P{reject Hg: n = 300 In = T50mg] EL. _ pfl— p] .5 _ =Ptr<t2t11 n=T50mg] :‘r—rsu _: nan—350"; '-._ 44.31 44.31 .-' =91: 41.53: =o.3us Thus the power for this test situation is 30.5% I To test H0: [.11 - !.L_-_~ = 0 versus H_,_: pl - !.L_3 at 0 use the two—sample r-test statistic A very low power - hecause .II is only 18 Has an approximate t—distribution under HE, where the degrees of freedom it are either approximated by software or are the smaller C'le — 1 and n3— 1 We want to be able to estimate the credit card debt of'L'S families to within : $100 with 95% confidence We know the standard deviation o = 51.420 How many families do we need to study?I Set 100 =1..'§|6i and solve for H flanged 1'100 So a 4‘0 I; = 23.33 and u = tars I. .II = T'T-‘5 famihes {always round up] 1] State I-IcI and Hi and specify the significance level ti 2] Collect the data 3) Calculate the value of the test statistic. measuring the evidence against HE, 4] Compute the P—value 5) Compare the P—value to the significance level a {e.g.. a = 0.05] and draw a conclusion: {a} IfP-t'alue =11 (ti = 0.05]. reject HI:I in favor of HEl {13] IfP-value '13 [ti = 0.05], data do not provide sufficient evidence to reject HI:I Statistical power is calculated before gathering data Power is 1— |3 where |3 = Pqupe II error] Power = Pl:1'E_|E'CT I-I3 HI:I is false] = Pfimake a correct rejection of 1-ij T-.li"ould like to minimize Type I and Type II errors. but there is a trade-off for fixed a: _-'-‘is ti decreases 0 increases and vice versa Clrdinaiily, we fix PEType I error) = a = 0.05 and choose a sample size (it) so the experiment has Pnype II error) = |3= 0.2 and thus a power of 3113-1: The reliability of a “do not reject H3" conclusion is given to us b1: the power ofthe test "'_ CT _ o I“. l1 ' 3.2;: E1 +1930 3, - We want to test I-LJ: Ill; - u] = 0' versus Hi: In] - Ill] i 0 - Using the pooled 2-sarrip1e r—test because sl s'ery close to s_-_. r r (2 Lul —1:|_'.'1 +|:r«rl —1:|.,1 wherei: " nl—JII—l' “L H: . 2 , _ _I, 2 := 98.10—(9839) where S?— = {55—1]I~0.595E) +Ifi2I—l)[0. .43) 1 1 " 63+65—2 5.. —+—_ " I55 6) —0.29 _ —0.29 _ 1 —_-.29 So t=— — 0121 3555 0.1255" - From t—tables: to :1)- cntical "value for 123 df' is 1.93 - Since t 'i* 1.93 1'E_|E'CT H3 at the a = 0.05 signif. level - Conclude: women have higher average temperature 54 When the two-wayr table is a 2x2 table andE=a-b-c-d then the test (Yates version} is 1 _ {lad—be — “rs-"213}: _ unwarranted Need all 4 expected cell counts Z== 3 For confidence intervals use relative risk _ h _ _ Relative risk = RR = jam-"d: = [al..-'|'la+b:|'Jy'.:'e’.-'|:c+d}] _ . - -:].95 -55 91:91:. confidence interval = RR x e 3 W" To test HE: no association between the row and column yariables, use the chi-square {f} test h 1 _ T [observed count - expected count]2 ’1: _ H expected count This has a chi-square distribution with [r - 1]|:c - 1) degrees of freedom under HE, ifexpected counts lie 5 Reject 1-1,:I if}:3 2::- ffifir — lj|:c— 1) d.f.] If an}r expected count =12 5 use Fisher's exact test {uses sample space and simple probabilities concept] Parameter groupfs} lJ l-JJ'H: P 23-1 CDUHT CUlulIS- —— I“ I” 1 Fisher:- ‘ or dfcomuutecl by formula Key implication from independent random samples , ._ _. 5.2. J: . _. ‘s'arlx—jfil: ‘ —" andif'o'; =5]; =15! then a H ' | I If samples from 3 normal populations (equal o3‘s} pooled two-sample t—test statistic is 3—1—2 ~. “Jr -15-:- nt—lsi t=—' ' wheres;=—I‘1 3' {‘ j' l 1 ' arm—2 ' Jl1 “3 dfisn1+JI3-2 Ifsamples from 2 normal populations (unequal oj'sj then two—sample r—test statistic is P. = JLl1— x] 1 With dfappro'rt. by software or minimum of”; - 1 and“I -1 If samples from 3 normal populations. then the test of equal variances is the F—statistic with larger s1 on top and delel —1 and H] -1 F: 2 '32 To test HE: p =39: compute the 2 statistic .._ F — P's pa [1 — p5] H Has a 37(01): distribution under Hg. when H1130: 10 and nil-ply] '1= 10 The plus four (ll-'ilsonj estimate of the population proportion and it's standard error is , Elf—2 . Ilo'fl— n“) = and SE . = ; P n+4 ‘5 iii—4 r-‘i leyel (1—de 'i'eh confidence interval is p' i :3 2BEF. This works well when n 5 To test HI]: Ill = In:l versus I-Ifl: Ill # Int, ifpopulation is normal {or n 11= 30] and o is known. use a 2 statistic Z = r — its, so; Two types of errors in hypothesis testing P{Type I error) = P{reject HI:l I-Icl is true} = u P{Type III error} = anot reject HE. H.) is false) = [3 Power = Pfreject HEI HE, is false) = 1- |3- We can calculate the power to detect a specific H‘5L 1When we design experiments we usually set :1 = 0.05. I3 = 0.2 {so Power = 0.80] and then calculate n l-tarled hypothesis tests and lEZ‘Is are directly related The 93% confidence interval for a proportion. based on a sample ofsrze .II. is Ila—1.95 IM s+r9s I y: ‘ I? .1 ...
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stat crip 2 - I Consider a 2 x 2 table I Expected cfll...

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