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MIE364S_T_2A_09 - MtESGQ S Tutorial#2A 1 The diameters of...

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Unformatted text preview: MtESGQ S Tutorial #2A 1. The diameters of bolts are known to have a standard deviation of 0.0001 inch. A random sample of 10 bolts yields an average diameter of0.2546 inch. (a) Test the hypothesis that the true mean diameter ofbolts equals 0.255 inch, using a = 0.05. (0) Find the pwvalne for the test in part (a). (c) What size sample would be necessary to detect a true mean bolt diameter of 0.2552 inch with ' probability at least 0.90? 2. The Rockwell hardness index for steel is determined by pressing a diamond point into the steel and measuring the depth of penetration. The index is approximately normally distn'buted with a standard deviation of 8. For 50 specimens of a certain type of steel, the Rockwell hardness index averaged 62. (a) The manufacterer claims that this steel has an average hardness index of at least 64. Test this claim at the 1% significance level. Find the p value for this test. - (b) Steel is mfiiciently hard for a Certain use as long as the mean Rockwell hardness measure does not drop below 68. For the test in part (a), find the probability of type II error if the true mean hardness index is 68. (c) Find the sample size required for a: = 0.01 and film 0.05. 3. The life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated tempemnne until failure, and the following lives are obtained 25.5 - 26.8 24.2 25.0 27.3 25.1 23.2 28.4 27.8 25.7 (:1) Calculate the sample average and standard deviation (1)) Find the sample median and the lower and upper ‘quartiles. to) Construct a 90% two-sided confidence interval on mean life in the accelerated test. (d) The manufacturer wishes to be sure that’the mean battery life exceeds 25 b. What conclusions can be drawn from these data? (c) Find the pnvalue for the test in part (e). 4. The compressive strength of concrete is being tested by a civil engineer. He tests 16 specimens, and obtains the following data: 2216 2237 2249 2204 2225 2301 2281 2263 2318 2255 2275 2295 2250 2238 2300' 2217 (a) Construct a 95 percent two—sided confidence interval on the mean strength. (b) Suppose that we wanted to be 95 percent confident that the‘error in estimating the mean compressive strength is less than lSpsi. What sample size should be used? - (c) Construct a 95 percent two—sided confidence interval on 0’2. (d) Construct a 95 percent lowermnfidence interval on 0'2. / é. A company engaged in the casting of pig iron is concerned with the percentage of silicon that it contains. Let X denote the mnnber of grams of silicon per 100 grams of pig iron. Preliminary analysis indicates that X is normaily distributed. A random sample of 20 values of X produced the foilom'ng results: 1.13, 0.94, 0.73, 0.97, 0.71, 0.80, 0.72, 0.82, 0.48, 0.97, 0.85, 1.17, 0.79, 1.00, 0.89, 0.60, 0.8-7, 0.87, 0.92, 1.16 Find a 99% confidence interval for the mean percentage of silicon. How many measurements are necessary (approximately) to have the absolute error in estimating a with :17 less than or equal to 0.05 with 99% confidence? Test the hypothesis that the process from which the data were obtained has the mean percentage of silicon equal to 0.8. Use a z 0.05. Estimate the p—value. For the hypothesis in 50), find the value ofthe power fimcfion by assuming that the true mean is 0.9 and 0‘2 a s2. How many measurements are necessary to have this value greater than or equal to 0.9? ‘ Q; (a) (b) (a) (a) (b) (a) (a) (‘3) Tutorial #ASqution n = 10 3? = 0.2546 ‘ cr= 0.0001 Test Ho :y=0.255 vs HA {##0255 X "flo _ 0-2546"0-255 = —-12.6491 < -30“ 2 “3.96 Reject Ho. 2 w W ,. w ° 6N; 0.0001/Ji6 p-valuew 2 P (Z < z“) = 2 P (z < 42.6491) = 0 .5045) "‘3 P(_zo.025 < Z» (zonal a“ m£11)S 0‘! 1 '2 w n 2 [mm — m) 0*] g [(—L285w196) 0.0001] z 163 E 3 #6 —,u, 410002 X~ Nut, 0‘ m 8) :1 =50 f = 62 Test Haws“ vs HA:;1>64 Z zngszi: ° 6/6”; 3‘86 There is no evidence to support manufacturer’s claim potatoes P (2 > 20) a P (Z > 4.7678) = 0.9616 ~176’78 < 2“, 2: 233 Faii to reject H“. X. .. ... mo: Par. <zo.0l#=#,) =P[ ”° +i‘°—f~‘— <z.o.+”7°f‘~lp=m] o“ n 0/ n o- n 64 ._ 68 = 19%: (BMW! p = M] = Hz1 <~1.21) = 0.1131 fl{#:) = P [Zn (30.01 'i' ii??? I!” = #1] 2 0‘05 :5 30.6: +£9'ffi = ”2006 2 m n = [HM—2°“ “3““) 0'] == [WELMSJBN]: : 63.2025 :5 64 #0 "" #1 64 " 68 n = 10 X~N0rmal ix. .Zm—f): 55 2 M a 2.5.9. = 25 S = ___W, 2 ’W23'76 = 1.6248 n 10 n ~1 9 position of median wig—31 = 55 m = W = Eatgg—l = 25.9 3 X —X _ position of lower quartile =23}- : 2.75 q, 2 X +_[_E)4_‘1_31 = 242 +flg—Egj—"2—l «1 24.8 (1) 3m+g position of upper quartile = m 825 X -X , H. q, 2 Km 4— Lm—l‘” 4 ‘ ’ = 273 +—-~.----.~~-m-(27'8 4 373) = 27.425 (c) 90% CI 011;; : [YMWJ ill) = [261L833 5%) = (26:1:0.9418) = (25.0582 ,26.9418) r: (d) Test HomSzS vs Hd1p>25 “Kw-pa 26w25 2'; —~ W w- W m 1.945) rm”, 3 L333 3. Reject H; There is strong evidence mat the mean battery life exceeds 25 heurs. (c) p-value w P(’I‘ > 1;) m P(T > 1.946) 2 0.0434 (Use interpoiau'on) i X. 941 n=16 ‘=L=3—5i}631=225735 n .. S 345147 a 959/ CI on .' [X it ] : [2257.752t2J31 ] 3 22393623 ,2276137 ( J O I“ 9.61335 :5: J13“ { 7) 5' tom 4 15 r 515 2»; ' "' 5 20.4346 (b) “2‘“ 3;, J; 345147 f whenn = 22, m” = M2 2 04435 J- 22 22 tom 31 2.074 6.27 =(550.017o,2349.9151) (c) 95%ciong2: (12%).?’(n—1)S*]fl[15(34.5I47)2 15(34.5147)’] z gaunt: I «1975,15 (n-‘-1)S1 ; 1564.514?)z 1:05.15 25 (d) 95% L01 on 0'2 : s 714.758? Q5) X’VNGRMRL. mam , T= Era-”"5 0339; , Sian m: 0.03:2,» w .3) 963% CI 50" ,0" ‘ tan-1 = 130.9052“? 7-“ 32-36! fit teamfiih (0.86% 1: 2.3eiI°-°3%9) = (0.7565, 0.98241; .b) mfg; s: ELM a» >[Zé3 I =a~ n 7,; [3.5750f2éne'1) "‘ «#9: n» 83.9»! G: 83/ z?“ 15“; = 2.573 C) ‘Ho’ #:08 «$0.01 .f "t9! .. z “t .. Ha = [1,L4=O.Q ‘ “n: (mania! 2.0?3 - ’56 44“.; ‘33.!“ r; 0.959.: ~04; _ Sum, ‘5 4 to‘oxd‘z , 'FTR Ho 13 W Fagin/29 " "75% . p—wm 2 afltiq >12”) = fiPétmr.7§¢e) -.~— ileWT‘i) = o.c-‘i’.t'8/, p~va£ua 130.va €0.05 amt 040,; d) AWE OJ: Sm , 04-30-05 r13"“?fitfilfiififlt ”tub I", = 2am)»: $1.95 Pomu = PLRnd'uL Ha 1 H034 3w“) :PCZo—i: -Z|§'.‘lu.fl20fi) +P£Za>zzoilufl=o.¢j) ~PLZé~Ie.¢?)+ PLz=p ~03?) : o + oms‘g =0fnc1/f, TE! BJWSE Yb, _P(Z€"Z¢.og+%fi&iflfl=0q3 m —% w ... m 2.- m , 29.? z z ”47 Zack-:4” "##‘1‘4420. 6?: «- {g a» 20 36"] 6-: >8 - i~‘?5)co.f‘tb‘0 1 “71:61”, #13 (me—9.?) ”33.79 0- 33/ ...
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