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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% twosided 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, 0.87, 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 _ 02546"0255 = —12.6491 < 30“ 2 “3.96 Reject Ho. 2 w W ,. w
° 6N; 0.0001/Ji6
pvaluew 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
“Kwpa 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) pvalue 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" ‘ tan1 = 130.9052“? 7“ 3236!
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 , 043005 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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 Spring '10
 Makis

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