STAT$401$–$Homework$3$–$Solutions$$!1. Questions!about!sample!standard!deviation!s!,!standard!error!of!sample!average,!and!half8width!of!confidence!intervals!!(a) Since!we!are!concerned!with!the!variation!in!the!population!distribution!of!Y,!the!rat!weight,!the!10!g!should!represent#s,!the!sample!standard!deviation!measures!variation!among!16!rats!in!sample.!(b) If!the!SE!of!the!sample!average!10=ns!and!n!=!16,!then!1016=s,!so!s!=!40.!(c) A!CI!with!95%!confidence!has!the!form:!!nstyn)2/1(1α−−±.!!Since!16,05.0 == nα,!131.2975.015211==−−ttnα.!!!The!half8width!is!1016131.216)21(1==−−sstnα,!so!77.18131.21610==s.!(d) Given!SE(!)!=!10,!df!=!n#8!1!=!15,!and!α!=!1!8!0.90!=!0.10,!the!t8value!for!a!95%!confidence!interval!would!be!t15(0.95)!=!1.753.!!Thus!the!confidence!interval!half8width!is!1.753(10)!=!17.53!~!18,!and!the!confidence!interval!is!150!±!18,!or!(132,!168)!g.!!!!!!(e) The!value!for!µ0!=!160!is!included!in!the!interval,!and!so!a!p8value!for!the!g 160:0=µHwould!be!larger!than!0.10.!!That!is,!there!is!no!evidence!that!the!mean!weight!of!the!rats!is!different!from!160!g.!!!2. Freezing!temperature!of!milk!!Let!!µ!=!mean!freezing!temperature!of!milk!from!the!milk!supplier!!(a) CH545.0:0−=µ!!(assume!milk!from!supplier!is!“ordinary”!and!has!usual!mean!freezing!! !! ! ! ! temperature)!CHa545.0: −>µ!!(want!to!show!mean!freezing!temperature!is!higher!than!usual)!!(b) The!test!statistic!is!39.3002066.0007.015/008.0)545.0(538.0/0==−−−=−=nsYtµ!!(c) Since!CHa545.0: −>µ,!our!p8value!should!be!the!fraction!of!t!values!larger!than!our!observed!t!=!3.39!for!a!t8distribution!with!141 =−= ndf.!!Note!t!=!3.39!is!between!3.326!and!3.787!in!Table!A.2,!so!we!need!to!express!the!p8value!as!an!interval.!!That!is,!the!fraction!of!t!values!larger!than!3.39!(area!under!t8curve!to!right!of!3.39)!is!between!the!fraction!of!t#values!larger!than!3.326!(area!right!of!3.326)!and!the!fraction!of!t#values!larger!than!3.787!(area!right!of!3.787).!!So!p8value!is!between!(1!8!0.9975),!or!(1!–!area!left!of!3.326),!and!(1!8!0.999),!or!(1!–!area!left!of!3.787).!!This!means!that!the!p8value!is!between!0.001!and!0.0025.!!(d) We!have!convincing!evidence!that!the!milk!supplier!is!adding!water.!!!The!p8value!is!very!small,!implying!that!if!CH545.0:0−=µ!were!true,!our!observed!t8ratio!would!be!unusually!high!(only!between!0.1%!and!0.25%!of!t8ratios!are!higher!with!df!=!14).!!!!!2 3. Weight!gain!under!dietary!supplements!!The!two!observations!on!a!hog!are!paired.!!If!we!were!doing!this!by!hand,!we!would!begin!by!calculating!the!difference!in!weight!gain!for!each!pair.!!If!we!use!Y!=!weight!gain!of!hog!on!treatment!diet!–!weight!gain!of!hog!on!placebo,!then!we!are!interested!in!making!inferences!about!!µ!=!mean!difference!in!weight!gain!for!the!population!of!hogs.!!!!The!hypotheses!are!H0#:!µ!=!0!vs.!Ha#:!µ!>!0!!(one8sided!because!we!are!interested!in!increased!weight!gain!from!the!treatment!diet)!!!The!test!statistic!is!t!=!nsY/0−!=!(!1.0333!–!0!)!/![!0.516!/!sqrt!(6)!]!=!49.02!!The!p8value!is!p!<!0.0005!(this!is!calculated!by!taking!half!of!the!28sided!p8value!in!the!output!in!SAS,!or!!<!0.0001!divided!by!2)!!There!is!convincing!evidence!that!the!diet!supplement!causes!increases!in!weight!gain.!!Note!that!we!can!use!the!“causal”!statement!because!this!is!a!randomized!experiment.!!SAS!code!for!problem!3.!!title 'HW 3 Q3 Hog siblings and weight gain data' ; * Part (a) suggests a paired sample, so we need to calculate the difference and conduct a paired t-test i.e., 1-sample t-test on differences) ; data one; input sibling treatment placebo ; diff = treatment - placebo ; datalines; 1 70.4 69.3 2 38.2 37.2 3 59.7 58.7 4 40.3 39.3 5 93.5 92.5 6 50.7 49.6 ; run; * Method 1: proc ttest on the difference ; proc ttest data=one H0=0 ; var diff ; title2 'part a - paired t-test of difference in weight gain (treatment - placebo)' ; run; * Method 2: proc ttest using the paired statement ; proc ttest data=one H0=0 ; paired treatment*placebo ; title2 'part a - paired t-test of difference in weight gain (treatment - placebo)' ;
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