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Unformatted text preview: Chapter 8 — Conﬁdence Interval Estimation Example 6: Vocabulary Practice
For each of the following, identify the sample, the estimator (statistic), population, and parameter involved.
a) Of 20 fish randomly taken from a local lake, 4 were found to be poisoned. sample I 20 €r‘5k populaiﬁ‘onl all #lﬁhr‘n facet! [aka ﬁ'é‘t'é i‘a'l'il‘c. ‘ ’8 = o . 2. mecm'ld id ‘. +feie om know/l papa/419M?)
Poiboll r46: poison reuse ). ,0
b) The average temperature of 15 randomly selected doctors at a local hospital was 99.8.}
5am‘OlC ‘. 1F; aloc+or5 Pqulqtl‘on: a,” avalvrs 45/064 hospi¥4/ ﬂair/.651. Y 7: q‘l % Parame'éey‘ 1 if”? unknown PoPuL‘z’b'l ‘ eves temp . Na temp ) A.
c) In a poll of 100 students at a large university, it was found that 55 of the students polled were in favor of having a reading day before the first day of final exams.
sample ‘ [up binden‘bs P0P“ la‘ér‘m I 4:” sing/«$5 af'ﬂrv‘s u""v€rﬁl47 é'éa'lr'j'llil‘C‘, [3 > 056 Parana(ref: ﬁrm: “"knm’l papa/4151294
,ppromlm‘éc mppmw‘) r4 62‘ P, d) Thirty “1 liter” bottles of Pepsi were randomly sampled from a shipment at the
distribution center. The average was found to be .98 liters.
squle ’. 30 ”llfiter'” POPMIQ'EFOKV, nil ”//.‘6er"bo‘H'lar o‘p Pepsi battle 6  Prom Hwé 54k :‘pmcnL afﬁni—
di‘dﬁ earlier. ' 6194 $3521 7( c 048 new ParaMetrrz {rue ml known FOP‘A'f‘ "an
Example 7: Survey on“). volume ) M.
A poll of 18 randomly sampled people at a grocery store was taken to rate the preference for a new cleaning product intended for use in kitchens. The results
(1 = awful to 5 = wonderful) were as follows: 4 3 4
3 4 4 4501 5 2 3 1 . 4
4 2 5 4 3 a) What is the mean preference score of the 18 people in the poll?
3? 1' 3 .' '5
b) What is the range of preference scores in the poll? 5 .— 1 : Ll  c) 'What is the sample SD? d) What parameter‘coLIld the' mean preference score of the poll estimate? 'ﬂx. 'Hm Mean prevﬂere‘nce' Score aids”
Casiorder: emf “2514’s sﬁrc (whaml‘ hfuse,
e) What parameter could the sample SD estimate? {—14.3 [OpOC/q'cqzc) 61‘ {he populqi‘l‘dn 5D 07: {Jam .
Pfe‘ieef‘cnc; ﬁussf‘c 5 o‘FCYCPyDYlC ﬂ‘éi—A13
ﬁﬁre WAD mfgA19 use fA‘SIOfoc/Hclﬁ, ii Chapter 8 — Conﬁdence Interval Estimation 8.2 The LargeSample Distributions of X and ,5 The following central limit theorem tells us that even if the population distribution is not normal, the theoretical probability distribution of X will stillrbe approximately normal if
the sample siZe n is large enough. The Central Limit Theorem (CLT) for X (with replacement) Suppose that we draw n times at random with replacement from a box of numbered
balls. Assume either that the distribution of the numbers inthe box is approximately
normal orxthat the nUmber of draws is reasonably large. Then the theoretical probability histogram for the sample'm‘ean X will approximately follOw a normal curve
with mean ' ' #7=E(X):Iubox and standard deviation. _
07=$D(X)=ab0X/x/F. ' i In particular, 7 is likely to be around pm give or take ohm/J}?— or so. Note: The larger n is, the closer X is to being normal. If the distribution of the numbers in the box is not too skewed and does not have extreme values, a good rule of
thumb for “reasonably large” is n 2 20. Example 8: Casino Dice _ .
Imagine we toss a fair casino die‘and let X denote the average of the score on n tosses of the fair die. Why can we use CLT to approximate P(3.3 S X s 3.7)? Find
mm and Ubox and then use the CLT to approximate the probability for the following values of n. Bquuse {Jug 3"‘2314419‘6‘4 (‘5 71146 (Jani‘l’lj n. 190165
a—branoiom w/ re plenum/lent prom the box El 23: 'l 5 4:]
what {a no‘é skewed. ””335 , 0’1“le.708 P(sEAYé 3.7) = P(% ,1 Z 512:3; 5 own(l 03““ : P(’OL€5é 25 (an—{5) ‘— 0 u73e— 0 324,7 = i
b)n=150 . 7=Mb0x=35 3 0’§=6‘m/\{ﬁ = '7°8/\/77.=>=0H'll P(5.%9.X—f;g,7) : PC2237??? 4 2,; 3.74.5 . r éI‘l‘ : P(‘lri'b s as 1.93) ‘: 0.9254: ~ 0.074%! a
,— 87 Chapter 8 — Conﬁdence Interval Estimation C) n = 1000 M2: 3.5 3 5‘; = l‘708/Vlooo : 0.05%
._ __ 3'5 35 33735
P(§.?>£XS'57>' PCméZé c.0511 :P(~'5.7 5 2s 3.7) : 1.~ o 2&3
d) What do ydu notice when comparing paIts (a) through (5)?
A5 n cacti/‘6 b‘\3ﬁer)x I!) —\f<\f“yIf13 166$ (6D is cleavessins)‘ ‘50 X If: Sei‘élﬂa c/oser‘
+0 ,(4: 9,. 5 more esvaz‘en, The CLT for the Sample Mean X Computed from a Simple Random Sample
Drawn from a Large Real Population (without replacement) Consider a simple random sample of size n drawn without replacement from a large real
population of size N with mean upop and standard deviation opop. Assume either that (i) the population distribution is approximately normal, or that
(ii) both n and N  n are reasonably large and the population distribution is not too
skewed and does not have outliers. Then the sampling distribution of’X will be approximately normal with parameters and #YZIUPOP
0' _0pap'_ N—n
'X J3 Iv—1' Example 9: CLT Without Replacement Consider the mean 7 of n = 10 numbers drawnwithout replacement from the
population {1, 2, 3, ...,. 23, 24, 25}. a) Determine ,u)7 and a)? , using the facts that upqp = 13 and opop = J33 .
1M. :" ﬂpcp = ’5 N__j:_’l _ ___V$Z ZSID _.
0== Q\PoP/‘m' ’ I/TZ ' illsI. ‘ /.g
b) Use the CLT to estimate P(10N 5 < X < 15.5). Pélo BLXé l5 5).: P(Io.‘>—l3 eff. ls.’5 ﬁrs)
P(— (3‘14. as I '53) "  0.9177 , 0~osz3 :— 88 i
I Chapter 8 — Conﬁdence Interval Estimation CLT for 7 computed from a Simple Random Sample Drawn from a Many
Times Larger Population Consider a simple random sample drawn from a population at least 20 times larger than the sample. Let upop and amp denote the population mean and population standard
deviation. Assume either that (i) the population distribution is approximately normal, or that
(ii) the population distribution is not too skewed and does not have outliers, and the
sample size n is reasonably large (n _>. 20, say). ‘ Then the sample distribution of the sample mean X is close to the'normal distribution. Example 10: Household Size
Suppose that the average household size in a small city, Le. a city in the country, is p = 4.60 persons per household with a standard deviation of o = 1.75 persons. Let 3?
denote the average size of n = 500 randomly chosen households. a) Compute 5Q?) and SD(/\_’). Assume more +£1qu [0,008 house Lola’s
#3: racer$1.90 6% z Shop/f7) : l75/lsoa = 0.078>
b) Use the CLT to estimate P(4.4O s 7 s 4.80).
POM—l0 s YéL—l.80> : PCiﬂf _4_. a: z ”'5‘“? °~°78 0.075
Z .<. 2.64,) ll
.U
’7 l“ 01 9
lb The CLT for [7' from a Random Sample with Replacement Suppose that we draw n times at random with replacement from a box of Os and Is.
Let p denote the proportion of 1—balls in the box and ,5 the proportion of 1balls in the  sample. If n is so large that hp 210 and n(1  p) 2 10, then the theoretical histogram
of ,5 will approximately follow a normal curve with mean ' l\ #,~,=E(p)=p 6e=50(ﬁ)=\/p(1P)/ﬁ Example 11: Flipping a Coin
Let ,6 denote the proportion of heads in n tosses of a fair coin. Use the CLT to estimate P(0.45 s ,5 s 0.55) for the following values of n.
a) n= 100 np: “>0, néip) > 52) => CLT‘. ”’13 :' P = 0‘5 5576‘ = l<P>0~P3/V77_ ‘ l»5(l—.¢s)/V}Zs = 0.05
P(O.L{5’;$ 50,55) . Pd °"“”°5 4 1:: é 0.550.5 0.05 " 0.05 89 =P(—zezsl) and standard deviation ...
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 Spring '09
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