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Unformatted text preview: I Consider a 2 x 2 table
I Expected cﬂl courts (under 15,] 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 Maj101'}. conﬁdamo htomrlfortr popduﬂon prop0111011
' "' mm m Relative risk = R = Lot = a,.'a+h'Ic'lcd ] 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 VariantII Unknmrn Variance Patrol: Tests
.   I . test nanic :tcst 2 .saunplc ttcst pnolctl ttcs‘t paired ﬁtcst 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 2sided 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‘ . 51.96— .
L odd; cod; ..' 'u.‘ 5.345 tin"Jo ,.
/ ~, ,z .
.' 10 10 '
=P Zi—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 1.5l5 {ct = 0.05 1tailed 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 IiiFF
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 rtest 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 IIcI 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} IfPt'alue =11 (ti = 0.05]. reject HI:I in favor of HEl
{13] IfPvalue '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 II3 HI:I is false] = Pﬁmake a correct rejection of 1ij T.li"ould like to minimize Type I and Type II errors.
but there is a tradeoff 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 31131: 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 ILJ: Ill;  u] = 0' versus Hi: In]  Ill] i 0  Using the pooled 2sarrip1e 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 twowayr table is a 2x2 table andE=abcd
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 chisquare {f} test h 1 _ T [observed count  expected count]2 ’1: _ H expected count
This has a chisquare distribution with [r  1]:c  1)
degrees of freedom under HE, ifexpected counts lie 5
Reject 11,: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 lJJ'H: P 231
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 twosample t—test statistic is 3—1—2 ~. “Jr 15: nt—lsi
t=—' ' wheres;=—I‘1 3' {‘ j'
l 1 ' arm—2
' Jl1 “3 dfisn1+JI32 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 nilply] '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 IIﬂ: 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 IIcl 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 ltarled 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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This note was uploaded on 04/10/2008 for the course STAT 578 taught by Professor Kirnan during the Spring '07 term at Harvard.
 Spring '07
 Kirnan

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