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Unformatted text preview: Chapter 8 Interval Estimation Outline:
0 Interval Estimation of a Population Mean —Large Sample Case
—Small Sample Case 0 Determining the Sample Size
0 Interval Estimation of Population Proportion Interval Estimation of a Population Mean: LargeSample Case (1) Probability Statement about the Sampling Error (11) Constructing an Interval Estimate:
LargeSample Case with 0' Known (IV) Calculating an Interval Estimate: Large—Sample Case with 0' Unknown Margin of Error and the Interval Estimate: A point estimator cannot be expected to provide the
exact value of the population parameter. An interval estimate can be computed by adding and subtracting a margin of error to the point estimate.
 Point Estimate + Margin of Error
 Motivation for Interval Estimation:
The purpose of an interval estimate is to provide
information about how close the point estimate is to the value of the parameter (which is unknown).
Slide
The upper limit on the sampling error. interval Estimation:
An interval estimate of a population parameter is
constructed by subtracting and adding a value, called the margin of error, to a point estimate, Le,
a? iMargin of Error. From last chapter, we know how to compute
P(—ks X — p: k) lntervals and Level of Confidence: Sampling
Distribution 101, of all HI 1' "fake" .2 .n «91‘. mil:34 (1OL)%Of als contain u. Interval extend from
3? —~— zmo; to
J? + zmo; t'fEﬁﬁ‘ﬁﬁfi’ﬁzfﬂf We are 100(1d)% confident that the interval constructed
from i—ZWO'E to 3—C+zmaiwill include the population mean u. 1. This interval is established at the 100(1— 00% confidence level.
2. The value (1— a) is referred to as the conﬁdence coefﬁcient.
3. The interval estimate fizmaf is called a 100(1 a)% confidence interval (CI). What are the factors affecting intervals? Example:
A simple random sample of 50 items resulted in a sample mean of 32 and a sample standard deviation of 6.
a. Provide a 90% confidence interval for the population
mean.
b. Provide a 95% confidence interval for the population mean. (ll). Precision StatementProbability Statement about
Sampling Error: There is a (1 a) probability that the value of a sample
mean will provide a sampling error of 2mg} or less. e.g. SuppoSe a sample size n=100 and a population
standard deviation 0:20, there is a 0.95 probability that
the value of a sample mean will provide a sampling error
of or less. (lll & IV). Interval Estimate of a Population Mean: Large
Sample Case (n 3 30) o If 0 is known, use it.
o If G is unknown, use 3 to estimate 6. The confidence interval for the large sample case is given 0' # S
—— xiza ~— by 26312 J; or /2 J; a. What is this interval? Area=od2 0 Known: Interval Estimate of a Invoke Central Limit Theorem I
' C ' LargeSample (n 230) « i  _+ 0'
x — Za/g J; where: x is the sample mean 1 a is the confidence coefficient
'Za/2 is the Z value providing an area of {1/2 in the upper
tail of the standard normal probability distribution (gels the population standard deviation n is the sample size Example: National Discount Inc.
National Discount has 260 retail outlets throughout the United States. National evaluates each potential location
for a new retail outlet in part on the mean annual income
of the individuals in the area. A sample of n=36 was taken
and the sample mean income is $31,000. Develop the
confidence interval for the mean with 95% confidence
level and write down the precision statement when: (1) The population standard deviation is known as $4,500. 9“. ﬁnd”! Sample size determination: have 0'
E = z —
05/2 {n
Therefore, for a given E, Z 0' 2
n=[ 05/2 J
E If 0 is unknown, we may 1. Use the sample standard deviation from a previous
sample of the same or similar units. 2. Use a pilot study to select a preliminary sample, then
compute this sample standard deviation. 3. Use judgment or a “best guess” for the values in the
population. For example, the range divided by four is
often suggested as a rough approximation of the '
standard deviation. Example: National Discount Inc.
Suppose National’s management team wants an estimate of the population mean within $500 (margin of error), with a probability
of 0.95. What should the sample size be? Assume the
population standard deviation is $4,200, and the population has a
normal distribution. ' c; 1.1.00
E : 50° If E is the maximum sampling error (margin0f enemiwé 54“;991 Small sample case: n < 30 _ ,
1. Can only handle if normal population. 2. If 5 unknown, use 3 to estimate 0.
3. When population is normal Y—w S/x/g follows tdistribution. What is the distribution of 37 ? 3 g
3? N MGM4 (may, f—distribution: 1. Degrees of freedom determine how accurately s
approximates o.
2. t—distribution approaches standard normal as degrees of freedom T. The mean of the t—distribution is 0. The more the degrees of freedom are, the less the
dispersion (variance) of t—distribution is. 5. The t—tables. PS” .1
ii... Shape of t—distribution: —The t—distribution looks like the normal distribution.
—The t—distribution with one degree of freedom is unique,
as is the t—distribution with two degrees of freedom, and soon. 1“”:  Iﬁﬁitgro Tdistribution pdf .  ' \ tv alue Compare normal and t—distribution:
Due to the uncertainty in the standard deviation, the
sampling distribution has more probability towards the tail. compare standard normal and t 0.450 
0.400 0.350
0.300
pdf 0.250
0.200 0.150
0.100 r 0.050 0.000 T’T1177777.":I: l . .
4 ~35 3 ~2.5 21.5 — std normal distribution t—distribution d.f.=6 —tdistribution cl.f.=2 '1 as b.1"b.'e"1_1”’1.'e 2.1‘ 2.5 3.'1 3.6 t or 2 value 10 lnterval Estimate of a Population Mean: Small Sample
Case (n < 30): Case I: Population is not normally distributed.
> Sample size must be increased to 30 and large sample
procedures should be used. Case ll: Population is normally distributed and c is known. . . . _ o
> The confidence Interval IS x i 2m T
H Case lll: Population is normally distributed and 0' is unknown.
__ s ’
> Use the tdistribution: x i tmm W More on Case III... Interval Estimate (Note assumed that probability
distribution is normal and o is unknown) ‘7 i tor/2,114 m
where o 1a is the confidence coefficient. o ta,”1 is the t—value corresponding to the 1‘
distribution with n1 degrees of freedom remaining
for the standard deviation. 0 The area in the upper tail is 0/2.
0 s is the sample standard deviation. 11 Example: Apartment Rental.
A reporter for student housing is writing an article on the cost of student housing. A sample of 10 on  edroom
units within half a mile of campus is taken. rovides a mean of $550 with a standard deviation of $60.
Provide a 95% confidence interval estimate for the mean ' rent per month of the onebedroom apartments within half amile. Assume. monthly rents are Wdistributed. 1%le
‘i : 5n
5:“
K“). c ' '9
‘1. 3 Zﬂ/LE (v.0an
0'5
'23: hr, i
and “a
5503' t.°"g"£
143.17.:on w 156
I . 11161..
c o .9 i. _
s in > 55° 3 1.1? 12 lnterval Estimation of Population Proportion: When np25 and n(1p) 2 5, the interval estimate is given 170*?) Note p is replaced Nete normal distriti'utilon by its estimate Examplerﬁolitical Science Inc.
Political SCience Inc. specializes in voter polls and surveys
to keep political office seekers aware of their popularity In a recent campaign, P81 found that 220 out of‘SOﬂvotteEsp _‘ contacted favored a candidate} 'Develop a 95%
confidence interval estimate for. the proportion of
population of registered voters thatifavprsthe candidate. up .: tiger; :3,“ '2 “ ('.,) 3 I; '4"; 3.3‘. w‘ .43: I; _ F 3 "' ts .q.., :5“
‘F W “I: a“;
T Sample Size for interval estimate for Population
proportion: Let E be the maximum sampling error or margin of error.
We have 19019) 1’! E : Zea/2 Or in other words, n : [Zen/2] E
<Note> 5
(1) If p is unknown, use [*9 instead of p if pilot study is
done. ‘ (2) If no information is available about p, it is assumed to
be 0.5 since the sample size required is maximum for
this proportion. Example: Political Science lnc. (p. .12) _
SupposePSl would ike'*0.‘99 probabilityathat theisample
proportion is,withjn $0.03 of the population‘proportion, what shoi1ld the sample size he? I; : 003 ' ‘i m. ' _
2' on: ‘4"...7‘5” ‘9." 4° t ‘
Zil—é. 1., l .‘l z
E ‘ '7" 9.3.63
' ' 14 Chapter 8: Practice Problems Q1. We know: a. If a: = .05 , i.e. 5%, what is the value of the conﬁdence coefﬁcient? (1‘05 =I~ 005: 035‘ ’1‘
Lsz at 9%“?an b. If a = .05, what is the value of 2,2 '? 2 21/2 5 Z4.9.025 ’7 1““ c. In equation (1), what is the random variable? .— X d. In equation (1), how do you characterize p ? l/Lfis #7:: fafulab‘dn mm,
If is animawn beefs; ﬁred auméer (ﬁecram afar). c. How do you interpret equation (1) for a = _05 ?
‘Prefar'h'an a}? Jim e: )1! ' ,b‘e: wr'ﬁ’v'n jibe Cmﬁ/pafec/ Genﬁdence Jn fem/N1"! I _ I” ' 1‘ " Valueé
(i.. Zoazfﬁ I zy,g¢5ﬁ)ﬁ&7frlm x [Z [29" ?5‘/0. Q2. The average monthly electric bill of a random sample of 64 residents of Lafayette is
$80 with a standard deviation of $24. a. Construct a 95% conﬁdence interval for the population mean of monthly electric
bills of Lafayette residents. ———‘—"“"—““’7
qof. :3» 242: £695 ‘m 2v
 .s ,_. 301 1.135 .__....
./Z a 95%“ 3 xi 24/ T‘— J??? at —..==}Z a .960
l 1 n= C13“! !2, 85833V0Mlz) qbl :1 :52»
0h? q‘HLQ .L/z': ‘
2‘!
=7 {61516: = 52+ £41205? :2 9°1 mqs m (amass) .. 30:}: 5. m = (We, 560) a/
b. Construct a 99% conﬁdence interval for the population mean. a. = .1".
/‘Z 3 20:12,?? = 301‘ 2,9095% 2 80i(:_9'53)(39 so 7274* == 6% 96, 8744)
______=_—_.——I:—:
801: (2.656) C3) = 8011358 #4:. .gzOZtt'. 9—1::
tr” X ﬁat/2m 9 ooosm«Mﬂl c. What effect do you observe as the conﬁdence level is increased? u
as
0 H
93
m
as The wi'cl'th 6f #ul Con {alarm JMHggJeal. d. What happens to the 9S % conﬁdence interval for the population mean if the
sample size was 100 with the same sample mean and same sample standard deviation as above? 9 _
— s .. +1.? J—zso: 4409
ﬁgXi’Za/g‘ﬁ“go“ 6m}; ,1 (as: 2%,. WELD?)
  ' 'feru‘aél Wed
Note "fth m Maﬁa cal 3W cmﬁdwm m 80:436 — — 0 ﬂ =
t a, x i tot/2.5 = 502‘; I"??? 12%, ‘ 3011 (“wow =C?s.2‘r;3‘t?€il e. What effect wi you observe to the 95 "/5 conﬁdence interval for the populatio
mean if the sample standard deviation were $30 instead of $24? ﬂ 9X1}: Zd/zm  30— did/g; 80i:¥35‘ : [7855, 8135)} 71:. 90/264 (5, m awf'atem inferwgﬂ meeaﬂ. * . 1023013? 0
ta; Y: tqz%“~ will W13)mf 5 I 2
: (3250, 3750} f. Assume that a sample mean of $75 and a sample standard deviation of $22 were
obtained for a random sample of 25 residents of Lafayette. Construct a 95%
conﬁdence interval for the population mean of monthly electric bills. What assumption did you make? = 2. 069 [ frr «if = 2‘19 5!“ if! a)" m‘ I.
I f ﬂff QQIWJ 0) ﬂ g. If we want to estimate the population mean of monthly electric bills of Lafayette
residents within a margin of error of $2 with 95% conﬁdence, what sample size
should we use? Assume a population standard deviation of $20. _, 2' "r1_ 2. 73031 1.?67'020)‘ = seq/e v; 335“ [Am sat/.9 M1074 in 1371”) h. If the desired margin of error is ﬁxed at $2, what happens to the sample size as the
conﬁdence level is increased to 99% in part (g)? ' 7' z 2 7. 2. 2..
.. Za 0” ,._ (“2.7. .)[20) = £9.52) (an) Z 7, 5556? ~=’ 55‘ wandF Q3. An American Express retail survey found that 180 out of a total of 1200 US.
consumers had used the Internet to buy giﬁs during the 1999 holiday season. a. What is the point estimate of the population proportion of consumers using the
Internet to buy giﬁs? (200 b. Using 95% conﬁdence, what is the margin of error associated with the estimated proportion?
a.“ 15’5"?) .._ J 75 0 ’gmgs) = 002 6T 2‘10 ____._.———I vd’ S’Wed I j. i l E I. I ) I I"
c1190. :5) 5 ? t E g"; i ‘b?
0 W Uffgl wig}. a i— z,/_ Magi!“ ? ﬁrm. i
/’ (a 120
Search. A 50'“? ‘J‘ .a..—:
c. Construct the 95% conﬁdence interval estimate of the population proportion of customers using the Internet to buy giﬁs. Era/2W: Fig : Pi‘ 002.
:' OJS'i 002.
: (0.23) 0.1:?) d. How large a sample should be taken if the desired margin of error is 3%? Use
95% conﬁdence andp = 0.16. [Ex/JPUWD t, (i’WJZ(GM) (05?) 7] 5' __....___————""”" (0.03); El
 . 5?
r 5% 63 a 9253‘ t DISTRIBUTION Entries in the table give 1 values for an area or probability in the upper tail of Area or the 1 distribution. For example, with 10 probability degrees of freedom and a .05 area in the
upper tail, {as = 1.812. 0 I
Area in Upper Taii
Degrees ' 2
of Freedom .10 .05 .025 .01 .005
1 3.078 6.314 12.706 31.821 63.657
2 1.886 2.920 4.303 6.965 9.925
3 1.638 2.353 3.182 4.541 5.841
4 1.533 2.132 2.776 3.747 4.604
5 1.476 2.015 2.571 3.365 4.032
6 1.440 1.943 2.447 3.143 3.707
7 1.415 1.895 2.365 2.998 3.499
8 1.397 1.860 2.306 2.896 3.355
9 1.383 1.833 2.262 2.821 3.250
10 1.372 1.312 . 2.228 2.764 3.169
11 1.363 1.796 2.201 2.718 3.106
12 1.356 1.782 2.179 2.681 3.055
13 1.350 1.771 2.160 2.650 3.012
14 1.345 1.761 2.145 2.624 2.977
15 1.341 1.753 2.131 2.602 2.947
16 1.337 1.746 2.120 2.583 2.921
17 1.333 1.740 2.110 2.567 2.898
18 1.330 1.734 2.101 2.552 2.373
19 1.328 1.729 2.093 2.539 2.861
20 1.325 1.725 2.036 2.528 2.845
21 1.323 1.721 2.080 2.518 2.831
22 1.321 1.717 2.074 2.508 2.819
23 1.319 1.714 2.069 2.500 2.807
24 1.318 1.711 2.064 2.492 2.797
25 1.316 1.708 2.060 2.485 2.787
26 1.315 1.706 2.056 2.479 2.779
27 1.314 1.703 2.052 2.473 2.771
28 1.313 1.701 2.048 2.467 2.763
29 1.311 1.699 2.045 2.462 2.756
30 1.310 1.697 2.042 2.457 2.750
40 1.303 1.684 2.021 2.423 2.704
60 1.296 1.671 2.000 2.390 2.660
120 1.289 1.658 1.980 2.358 2.617
on 1.282 1.645 1.960 2.326 2.576 22 STANDARD NORMAL DISTRIBUTION Entries in the tab1e give the area under Area Or the curve between the mean and z stan
PFObabiliW dard deviations above the mean. For
example, for z = 1.25 the area under
the curve between the mean and z is .3944.
0 z
1
1‘
1
z .00 .01 .02 .03 .04 .05 .05 .07 .03 .09 1
1
.0 .0000 ‘ .0040 .0030 .0120 .0150 .0199 .0239 .0279 .0319 .0359 .1 .0393 .0433 .0473 .0517 .0557 .0595 .0535 .0575 .0714 .0753 1
.2 .0793 .0332 .0371 .0910 .0943 .0937 .1025 .1054 .1103 .1141 .3 .1179 .1217 .1255 .1293 .1331 .1353 .1405 .1443 .1430 .1517 .4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879 .5 .1915 .1950 .1935 .2019 .2054 .2033 .2123 .2157 .2190 .2224 .5 .2257 .2291 .2324 .2357 .2339 .2422 .2454 .2435 .2513 .2549 .7 .2530 .2512 .2542 .2573 .2704 .2734 .2754 .2794 .2323 .2352 j
.3 .2331 .2910 .2939 .2957 .2995 .3023 .3051 .3073 .3105 .3133 .9 .3159 .3135 .3212 .3233 .3254 .3239 .3315 .3340 .3355 .3339 i
1
1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621 1.1 .3543 .3555 .3535 .3703 .3729 .3749 .3770 .3790 .3310 .3330 g
1.2 .3349 .3359 .3333 .3907 .3925 .3944 .3952 .3930 .3997 .4015 g
1.3 .4032 .4049 .4055 .4032 .4099 .4115 .4131 .4147 .4152 .4177 1.4 .4192 .4207 .4222 .4235 .4251 .4255 .4279 .4292 .4305 .4319 1.5 .4332 .4345 .4357 .4370 .4332 .4394 .4405 .4413 .4429 .4441 f
1.5 .4452 .4453 .4474 .4434 .4495 .4505 .4515 .4525 .4535 .4545 1.7 .4554 .4554 .4573 .4532 .4591 .4599 .4503 0 .4515 .4525 .4533 1.3 .4541 .4549 .4555 .4554 .4571 .4573 .4535 .4593 .4599 1 .4705 1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 1
2.0 .4772 .4773 .4733 .4733 .4793 .4793 .4303 .4303 4312 .4317 1
2.1 .4321 .4325 .4330 .4334 .4333 .4342 .4345 .4350 .4354 .4357 2.2 .4351 .4354 .4353 .4371 .4375 .4373 .4331 .4334 .4337 .4390 2.3 .4393 .4395 .4393 .4901 .4904 .4905 .4909 .4911 .4913 .4915
2.4 .4913 .4920 .4922 .4925 .4927 .4929 .4931 .4932 .4934 .4935 I
2.5 .4933 .4940 .4941 .4943 .4945 .4945 .4943 .4949 .4951 .4952 2.5 .4953 .4955 .4955 .4957 .4959 .4950 .4951 .4952 .4953 .4954 
2.7 .4955 .4955 .4957 .4953 .4959 .4970 .4971 .4972 .4973 .4974 2.3 .4974 .4975 .4975 .4977 .4977 .4973 .4979 .4979 .4930 .4931 2.9 .4931 .4932 .4932 .4933 .4934 .4934 .4935 .4935 .4935 .4935
3.0 .4935 .4937 .4937 .4933 .4933 .4939 .4939 .4939 .4990 .4990 21 ...
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This note was uploaded on 03/03/2012 for the course MGMT 305 taught by Professor Priya during the Spring '08 term at Purdue.
 Spring '08
 Priya

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