# 2.' -4-5 I fo. - /30 + ;3, x + G: ~ / ~ ) ~ ov. Fd'r evt.'i') cutckf' ()y\e.._o OYLt dtt:vl. t'"'i ~ _) y =.5_21/2-+. 8"'- 2.

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1 Statistics 100 Sample FINAL Instructions: I. WORK ALL PROBLEMS. Please, give details and explanations and SHOW ALL YOUR WORK so that partial credits can be given. 2. You may use four pages of notes, tables and a calculator but no other reference materials. 1. The following data were obtained from a random sample of eight oak trees on their age in years (x), and the diameter of their trunk in inches (y). x: 22 y: ( 4) a. Write down the regression model and explain its components. (10) y - /30 + ;3, x + G: ~ / ~ ) ~ ov Rrsl"'mJ~ Y-I"'tr-rUpt Slop'" &p(av1a.~e>v'j Ve-IA. VOcA.. b. Obtain the estimated least-squares regression equation, and explain the meaning of the estimated slope in the context of this problem. )( y ( ;< -x) (y-y) -)l... - ~ (X-X ( Y-Y) 22 ;to -2. -~ y 14 Ho lo IOU ~ 3'1 9 I ~" loo (X-x)( Y-'/) I ~.2. C> 2.' -4-5 I fo L... I'\ 2..'.20~ X= 1'12./ 8 == / 0./ 8S-2 2.S G, q b I _ ~ (_)(-x)( y-y )/ ~ c x -7<)"2. = T- 3 8 / 8 r:;; 2 - Y- b x - 2. fa - '8G:,~ 2.. ( 2.Lf) ~2. I I l A y =.5_21/2-+. 8"'- 2. x Fd'r evt.'i') cutckf' ()y\e.._o OYLt dtt:vl. t'"'i ~ _) OY1 C-<.VcY"aJL 17vt. t rv.. v'l It cl.ta me.. f < v w ''/I (t"! C-U.tl/l~. 8 (:, fo 2_ t'., ch t.e

2 Statistics 100 FINAL page2 (8) c. Test the null hypothesis that the correlation coefficient is equal to zero versus the alternative hypothesis that the correlation coefficient is positive at 0.01 level of significance and find the p-value f = 0 ) H,-q = p > o ; wht-u P =-- p 0 p. urv. ~ Y' = :2. { x-;t ){ y-y) - 7-?:. ~ ~, q ~ I l y ~lx-xjz.lly-'f)l yl89-) ({..,e;z.} t = Y-)/Cn-2-)/(1-rz.) ~.C}(pll (tp/cl- 'i<,tl) t OI = Wt.. Cctll'\ njrd Ht> at- 'ol ~.e.q.. 0 3,143 -}?-Vo.lt..u < OCJOS (8) d. Find the 95% prediction interval for the diameter of the trunk when the age of the tree is 30 years. Explain the meaning of this prediction interval. Th..<.,,loC>(1-~}'f. rr~d., t "'+. ~ Y +to( S \/1,_.L +()(!1-'X) 2. - A.- e ~ ~l' x.-:il.)'- o( =.(I s- ol..1 = ) df- = ~ t-. ()l..s- :: ~ ~ ~ -;-\'l./ _ ~ '+38 sscn.s1.d) = ~t.y-y)'-- ( ~c)(-3i...)ly-vh, ~ot.-j<) = t#12.- m =.<::2.1'ttos- Se- - J :SS(rHid)/(f'l-1..) = 2., q f = 'S. 7-// Z"''- L( 30) = 3 /, l'f 7-'?-. q 5' /. f rul.'e--f-, l/n i"'ftor u...i '-1 : 3 I I q q. 41- '( '2.. Cf IP so). J I +..L8 + C 3o - 2. W '\.. ' - ' y 8S~ (5) e. Find the coefficient of determination and explain its meaning. 1 q2.3t-.. o' tfv._ (/~~~ ""' +r-...,,,..1<. cltc. m..hy- Ct'VI ~ exp lai1111d J.,') r 'a; ~~~ JLJaftt?vii'1t f' wrtt. ay.

3 Statistics 100 FINAL page3 2. Researchers in an agricultural college want to test a new variety of corn seed, which they have developed. They got eight farmers to plant the new seed and another seven farmers to plant old seed. The number of bushels per acre each farmer obtained are: New Seed: 130 Old Seed: Is the new seed better than the old seed in terms of the yield? (3) a. State the null and alternative hypotheses. J-1 o ~ ~, -==- ft I-<, = h'lt.&i"', Yt'(.lcl ("" 1k.. n.t. w.l).u... d.... old v,,..,,, (7) b. Ass4me the yields for the new and old seeds are normal, test at 0.05 level of significance. 9, = 130 S :i. = S c. z., 'I-. :i a'") z. =- ~. s~ 1 g ~ '"2-11./:z 'L 4.'lll 10.l f-- + " t = (Yr-Yz_)-o f HA. t', > /!2.- Hl.- - s~ s~ -+n,..-i._ I "l 0 - I 2..3 = f:' t l.f.n10 _ '3.2.C.SY (3) c. Find the p-value. P-val 1.4. oos-< -p-v... \1.AA <. o l

4 Statistics 100 FINAL page4 (7) d. Suppose the distribution of the yields for the new and old seeds are not normal. State the null and alternative hypotheses and test at 0.05 level. Ho! 7.A.e """,...,~ ""'bt-t.s/.u../s./o..cn a.,._._ -tlv. /:>OVY\t.{-,,.,- b.,tt, ~~tt,., H,q, ~ '/ of!- f'"\lw ~J '"" T # < #< ty\gvl ft.e..,. 0 l.l /l-u c.i ~J7~,~~ f I "2.Gj I 2...~ S" l<,+k'l.= hi" hz I I '1.5 I 2. '& I 2. '-I 0? I 3. o 1'2..S I 2. ( q 1..(. 1 : SS". S 0 t<'"=,s S"~.S"-+ ~ :8(~) S" c;,, =S Co ~ U.s = ss-.s- V) '-I 3. LAI~ Cc.i"' r; J et t \-\ 0 aj o s- ~~ A digital instrument is used to test the acid level of a particular drug under three different temperatures. Six replicates of each experiment are run in the laboratory and the results are shown below: Temperature Replicate The between sum of square is 10,474 and the within sum of squares is 12,670. ( 4) a. Write down the analysis of variance model and explain its components. y,. p + c. + f, - l l) I ~ ~ ) I 1. 1-# obj. ff-tit VYI &a.,,.tl c( t[,.,...,.,,y...{ -rt. +Y"t'll. ~ rt\.tll"\ -(- H-. t-...-l'r.t.

5 Statistics 100 FINAL page5 (10) b. Construct the ANOV A table. Sot.( 'r"'-4.. d~ SS fv1.s F - -- Te.h'\ p. 2 l(j41-4 :)L 3 "f ~./ q '1'.. y-y-" V' 1'5" l2g-,7(j 84s I TeitA I I? 2.3 l'+lf I (5) c. Perform a global F test at 0.05 level and explain the result. Ms ( bdwq,,,.., F = t--15 (Witt.. I... ) - 5 L... '.l? 8 <+ s- d (.. == 2. Cl...,t.( I S- F. (Js = 3., ~ (3) d. Find the p... value for the F.. test. 0 (8) e. Construct a 95% confidence interval for the difference between means in temperatures 50 and 80. Explain the meaning of this confidence interval. ;Ju. looc1...q.)'/ ~ cacnt~, ""tt.v-v~\ t..., Y.f - Y_,,! +"'h ys;<~-+~"",\ +-, 6 2..r ==- 2.. I 3 I /Ju_ ct.r '/. ~ cj..~u. (1'-14.~3s:s-8(:.) t""li<'vvej Vi ± 2.l.?1{~4s-(1;-+{-~ 5" 2 ' g :? ~ '1 'i: 3 s ' 7 G. C/- y ~ ( 2 3 ' 0 l,&'1) '14 '5 '111.f) IA)(. CUR q S- /. ~ t-l~t- fatt-f 1/,-r_ di_//. in ~a"" Ct C.c'C.I ~ (rr -+e~pe.~~&ar-11 S-o a... d So (..., b<twu.n z:s.ol.&t\ e>i... J t;4.!s'i7<-t.

6 Statistics 100 FINAL page6 4. A study was conducted to observe the relationship between the emotional stability and talent. One huncl~d individuals were selected randomly and the following cross-classification was obtained. Emotional Stability Stable Unstable To-\--. I Talent ( ~ Artist 10.c.) 16 (IOolt\ 2. <.. Musical genius 20 '2.0. 4) 14 {I J, C.J 34 Math 30 ('2.1..() 10 (IC.) 4-o,,,..,.,,.., G:.o I 40 oo (2) a. State the null and alternative hypotheses in words. I-lo? G. mofi V"'l,J /l4,,,.l:til1+j ~ L',,,cle~~Jc...J- of -J.Qle.-if /-1,q : J-1 0 "" Yl uf tv-1,.4...t (10) b. Use the chi-squares procedure to test the hypothesis stated in (a) at 0.01 level. d. F; ( r -1 )l IL -I ) r- ( '3 - I ){ 2-1) = 2 \. X.01 - Cf.2. I W t. Ca n ne-'t Y- ~)! d \-\ u Cl/{. o I ~el_. (3) c. What is the p-value?. al< t-ve..lt,a..1 ( O 2. A Brief solution of the regression example given in Lecture 24: x-bar - 2, y-bar = 14 I: (x - i)2 = 8, I:(y-y) 2..; 144, I(x - x}(y-y) = 32. b1"= 4, ho == 6, Y" = 6 + 4x., SS(resid) = 16, Se= 1.633, SbJ =.5774, t = , r =.9428.

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