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# Ch 5 L20 - Mother A” Rich boyfriend A Poor father Rich...

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Unformatted text preview: Mother A” Rich boyfriend A . Poor father Rich boyfriend B .. If only I have a :1 . :1 ﬂ§¥4§g§§15i§§~§ .ﬂanavﬂarﬂaaﬁvnﬁﬂdgﬁaﬂﬁWnuﬁﬂﬂh. , poor father tests some hypotheses IS answer th To Daughter, A‘ Test statistic: f e . XA = (Time boyfriend A chats with daughter A) ' ' - (Time he chats with other women) "' N(9,1) p-value, pA =p(:xA>1.955 |6=O)I = 1 - @(1.955) = 0.02529 Test statistic: " A . xB = (Time boyfriend B chats with daughter 3)» . - (Time he chats with other women) "' N(7\,1) p-value, pB I R = P (=xB >-1.955 |_ A: 0‘) W _ = 1 - (I)(-1.955) ' = 0.97471. -law m having In ificant... 02529) giﬂﬁnggﬁﬁagﬁbﬂuaﬁghuaﬂaﬂg ign O ( pretty 5 Looks like I’ hson One p-value a ric Poor- father used figure ' 0.02529: min(pA, p3) ‘ to assess his evidence for atleast one daughter to marry ’ a rich guy. His significance testiwas actually done for: Small min(pA,_pB) :> evidence against H0 or, equivalently, Large 1 - min(pA, p3). :> evidence against HO Poor father was actually using test statistic Poor father obsérved a 575—. ~ 4:- -. .rze: n-:- e ,—.—2 v: .A ~ 7:», r.-:-'-: .—:-‘: :14". 1:5": w.“ e 173‘: e.-:- -. z—te: ---- . ~ ' If he employd'his‘tb’. is ' p-value, = P(T > 0.97471| e = A = 0) . = P(min(pA, p3) < 0.02529 | e = A = 0) = 1 -- P( pA 20.02529, pB 2 0.02529 | e = A = 0) f Ass’umeXA, XB independent _ (observations from the two daughters are independent). ‘ p-value ’ . . = .1 — P(pA 2 0.02529 | 0 =0) P(pB 2 0.02529 IA = 0) I t __ ‘ = 1 — P(XA s (1) -1(1 -0.02529)|0 .= 0) ' - ” P(x-B s (13'1(1 - 0.02529) IA = 0) I ' == 1 — (I) ((1) - 0.02529” (I) ((1) ‘1(1 -'0.02529)) _ .-= 1 - (1 - 0.02529)2 ' 0.;049943 ———> correct p-valae is bigger. . . although still marginally significant .J Po‘or father How does he look? Could you send me a photo? sooaooo charming! m .. Po’or father What e same gu ? Th \$000000 ! tible If I“ 8515 5 Suppose: rich guy is dating only two girls at the moment—L. To daughter A, ' ‘ the “other” Woman 1'5 daughter B. To da 11gb tar B, i the ‘btbéf” woman 1'5 daughterA. " xA =v-xB ~ man [9 =' -7\] (two-sided test) . Obviously; XA, XB'(_= -XA) n0 longer independent. p—value ‘ = 1- P( pA 2 0.02529, pg 2 0.02529 | 0 = A ="-0)' ‘ = 1 — P(XA s :1) -1(1 -. 0.02529), ‘ xB s (I) -1(1 - 0.02529) | 0 =0) = 1— P( lels (Ia-H1 - 0.02529) I 0 = 0) : P(.|XA|\$ (I)'1(1 - 0.02529) | 0 = 0) ' _==2 P(XA>CI)'1(1-O.02529) |‘ 0:0) : 2 {1— (I) (<1) -1(1 - 0.02529)) } = 2(0.02529) ‘ ’ A ~ r by PWFMy d = —'i Name aner PmPeW ‘ ,. cmﬁﬁansg u ...
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