# 11_midterm2 - Q-l A population has a probability...

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Unformatted text preview: Q-l. A population has a probability distribution given by the following density function : f(\$.e):{ ﬂe—“egﬁ if 0<m<6 0 elsewhere where 9 is an unknown parameter. A sample of size n = 1 is taken ; i.e. there is a single sample random variable X1. a) Show that U = £81 can be a pivotal random variable for the parameter 9. (Hint : First find the cumulative distribution function FU(u) of U). (10 pts) b) Using the pivotal random variable U in a) find a 90% upper—sided confidence interval for 9 when X1 was measured as :54: 2.1. (10 pts) . O 2540 a) X é 'L "5 1 at} _. (9+4; )é+——« 9L~_.) Wage): 999 93 z 3 aéleg ‘ 9f?(9 A5 K13 dJL-l-rélomleé like Pwpmla—Hw ./ 0 L440 hallwapfuéu) 2PM 4149) = '2. 9—[l9fuz'l9)_u3 Q3 2_ 3 0.4“41 1' Sal—2.x; l 74%) ﬁghtbéfcuuz) oéuéi 0‘. _ ( 3 (a.z)2-— ’Z./o.2_)3 S— 0. lo) Q-2. The following are the burning times (in seconds) of flares of two types : Type I Type II 65 64 81 71 57 83 66 59 82 65 .ﬂ ”" . ﬁn? 347’ 2 Supposing that the burning times are normally distributed find a 98% two—sided confidence interval for the ratio 312. (22 pts) 2 n, “L 96 7C. .. HI 6,13 40.1; 25096 5551 ’30 4] e 2 49 6 889 43 56 9 48) 67M 4 215’ 44 8‘9 ___,___,_____ 1 I 10604 9—373 2 1' 92‘ 26604 (“fl/é 2483393,.94 £67 ‘ 5 5' ‘ 1 10, , (347—)/" 55" WM , =LL3 1:248 A 4 97' l 612' st \ 3v?— < "q < '3” » Ma Lil's/:1; #5104111?) 57’ 1.2. 049' 7’ g t 7’ 7’ z '5’“."‘ m1»! L 7’ ,,_.z,, ;)n"‘llnzfl 0(5- ot' a I: 002, Euﬂﬂl de‘);q:l§.5‘, fowhlns ’L 3 gr“. 4.41! < 91‘ 4 4.44044) 1:, 143"; 67/” a. 0.7%; 4 93—» 4 W996 W 7’ Q-3. Suppose we have a normal population with the variance 02 = 25 and unknown mean y. We want to test the null hypothesis H0 : p = 20 against the alternative hypothesis H1 : ,u = 28. We decide to adopt the following rejection rule (RR) : Reject H0 if X > 25. a) We take a sample of size 6 and apply the above rejection rule. Calculate or = P(Type I Error). (10 pts) b) Now suppose we want to repeat the test on another sample not necessarily of the same size. If we adopt the same rejection rule but would like to have ﬂ = P(Type II Error) be at most 0.10 , what should be our new sample size ? (10 pts) 4) d: PC§>2§lyz¢o)<P(Z> aim )éP(Z>2.A§) M Q 5 wgooé- (ablccz. 43‘) \$0.5\¢o¢90—(9.492q 20.007] v/in Q-4. Suppose that two normal populations have unknown means . The variances are also unknown, but are assumed to be equal, i.e. of = 0%. Consider the test for the following null and alternative hypotheses : H01#1—#2=5 agalT‘St Hlim—Maéts; two independent samples of sizes 711 and 712 are taken from the two populations ; a) Show that apart from the standard test procedure we can also use a confidence interval approach ; i.e. construct a 100(1—a)% two-sided confidence interval for M1 —,u2 and show that the rejection rule (RR) of the standard test is equivalent to : “‘Accept H0 if 6 is in the confidence interval (10 pts) b) Verify your finding in a) in the following problem 1 A study is conducted to estimate the differences in the mean exposure to radioactivity in utility workers in the years 2002 and 2007. The following data (in terms of proper units ) were obtained based on two indpendent samples from the two years : 2002 2007 m = 12 “2 = 10 f1 = 0.94 (32 = 0.62 2121051; — \$02 = 0-60 231:1(1'2J — i2? = 0-42 Test the hypotheses ; Ho : ,ul — M2 = 0.35 against H1 : M1 — H2 75 0.35 1) by the standard method with a = 0.05; (7 pts) 2) by the confidence interval method using a 95% confidence interval. Have you arrived at the same decision ? (6 pts) 4) TN 5W‘dcwcl test is ; Rejed- H Hf. t7 0‘ J“ ‘—Zm,+v)2-1 aw alcth Ho RT ~+axq+aﬂ<ﬁi4£¢+¢ 27/ __,_ ~ 6; H’) ,2. Solw 5. 3 2. I 7. might-£2 15” x-i— _ -— F— , 2 50,44)!“ F/ [+73% < 5 4 ag—xz+l\§;n1*nb2 *4 W n1. LC l/ UC L we Wu M»: 4m uwts as M‘ Lchq UCL— 0} Women? ‘48”)! MFI‘fzIMMCc/epf Ho 1'1 S 6 QM€,:M}~CWO’ L) h 32- 23.95.25?) ryéra)“ z ’3 Q3446 ‘l‘oiﬂ/le 0 l\$+lo~rz 2.0 “o'ogl §p~0~216 ‘94-— a 3 t: in ~ _;_‘Q_.. n -3 3033 .L "T - ‘ V=10 c0.’1.4.6 1171+“? 0.0968 1 if = 2a 96 . M “940894 «-c9\%c799 (N786 Awe/put 14‘) 0-07-5394; 2 H 97 5’51 mid" Pr 7,: MM.” ‘3: 7096 (041%!in £0352 {BO/Rpm l0 Q-5. Many consumers believe that automobiles built on Mondays are more likely to have serious defects than those built on any other day of the week. To support this theory a random sample of 100 cars built on Mondays were inspected, of them 8 were found to have serious defects. A random sample of 200 cars produced on other week days revealed 12 with serious defects . a) State proper hypotheses H0 and H1; (4 pts) b) Test at the a = 0.05 significance level ; (10 pts) C) Repeat the test by the P-value method. Do you come to the same conclusion ? (6 pts) zoom 9 ’ ) l r A 1..!: »v—+zv”) \ +%)) Ig’ l{(loo a 0.6944 A\.9é /¥Ccep*lh Z. "'l—qé I ‘ :0.'LS7\$Z l” Pvan WZZoég)?WWW—tablefo,05)305000’0Z422 A} anlmg l1 YNﬁ’l’ a gym-All l’wlphkilib/MCW‘P’Y Ha ...
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11_midterm2 - Q-l A population has a probability...

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