Chapter 8 Interval Estimation

Chapter 8 Interval Estimation - Chapter 8 Interval...

Info iconThis preview shows pages 1–20. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
Background image of page 19

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 20
This is the end of the preview. Sign up to access the rest of the document.

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: Large-Sample Case (1) Probability Statement about the Sampling Error (11) Constructing an Interval Estimate: Large-Sample 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 1-01, of all HI 1' "fake" .2 .n «91‘. mil-:34 (1-OL)%Of als contain u. Interval extend from 3? —~— zmo; to J? + zmo; t'f-Efifi‘fififi’fizfflf We are 100(1-d)% 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 Statement--Probability 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 ' Large-Sample (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 (margin-0f 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 t-distribution. What is the distribution of 37 ? 3 g 3? N MGM-4 (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 T-distribution pdf . - ' \ t-v 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 -2-1.5 — std normal distribution t—distribution d.f.=6 —t-distribution 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 t-distribution: 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 1-a is the confidence coefficient. o ta,”1 is the t—value corresponding to the 1‘- distribution with n-1 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 one-bedroom apartments within half- a-mile. 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 . 11-161.. c o .9 i. _ s in > 55° 3 1.1-? 12 lnterval Estimation of Population Proportion: When np25 and n(1-p) 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 190-19) 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; : 0-03 ' ‘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~ 0-05: 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 (fie-cram 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/ Gen-fidence Jn fem/N1"! I _ I” ' 1‘ "- Valueé (i..- Zo-azffi 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 qo-f. :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, 85-833V0Mlz) qbl :1 :52» 0h? q‘HL-Q .L/z': ‘ 2‘! =7 {61516: = 52+ £41205? :2 9°1- mqs m (amass) .-. 30:}: 5. m = (We, 56-0) 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:. .g-zOZtt'. 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‘ : [78-55, 8135)} 71:. 90/264 (5, m awf'atem inferwgfl meeafl. -* . 10-23013?- 0 ta; Y: tqz%“~ will W13)mf 5 I 2 : (3-2-50, 37-50} 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‘ wand-F- 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) = 0-02 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‘ 0-02. :' OJS'i 0-02. : (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) (0-5?) 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 ...
View Full Document

Page1 / 20

Chapter 8 Interval Estimation - Chapter 8 Interval...

This preview shows document pages 1 - 20. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online