ch6 - Chapter 6 SAMPLING DISTRIBUTIONS 6.1 Only if the...

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Unformatted text preview: Chapter 6 SAMPLING DISTRIBUTIONS 6.1 Only if the pieces are put on the assembly line such that every 20—th piece is random with respect to the characteristic being measured. If twenty molds are dumping their contents, in sequential order, onto the assembly line then the sample would consist of output from a single mold . It would not be random. 6.2 The subscribers to Literary Digest were not representative of the people voting in the 1932 Presidential election. At that time, people with telephones were also likely to have higher incomes than the typical voter. Instead, the poll over-represented upper income voters who were likely to vote Republican. 6.3 (a) A typical member of the population does not vacation on a luxury cruise. The sample taken would be biased. (b) This sample will very likely be biased. Those will high incomes will tend to respond while those with low incomes will tend to not respond. (0) Everyone feels that unfair things should be stopped. The way the question is phrased biases the responses. 179 180 Chapter 6 SAMPLING DISTRIBUTIONS 6.4 Number the states from 1 to 50 according to their alphabetical order. Using the first two columns of Table 7 to select 8 numbers between 1 and 50 by discarding numbers outside of this range as well as any previously drawn number gives 13, 4, 29, 14, 36, 48, 22, 35. These correspond to Illinois, Arkansas, New Hampshire, Indiana, Oklahoma, West Virginia, Michigan, and Ohio. 6.5 (a) The number of samples ( given that order does not matter) is 6 6-5 (”fit (b) The number of samples (given that order does not matter) is 25 25-24 (2177—300- 6.6 For a random sample, each outcome must have equal probability. Thus, the prob- abilities are: (a) fi = .0667. (b) 3% = .00333. 6.7 (a) The probability of each of the numbers is given in the table: — Probability 1/6 2/15 1/10 1/15 1/15 1/15 1/10 2/15 The mean of the distribution 18 1 2 2 1 6(“)+1—5(—3)+"+1—5(3)+6() 1 $14 4>+%(3‘3>+r15<2‘2)+1—15(1‘1)+1—5(O>=0 181 The variance is 8.667. (0) = i 15 i 15 (b) The 50 samples are shown in Table 6.1 along with their means. Table 6.1. 50 Samples of Size 10 Taken Without Replacement 182 6.8 Chapter 6 SAMPLING DISTRIBUTIONS Table 6.]. (continued) .50 fiwbeJHOD—JOJONr-h (c) The mean of the 50 sample means is .034. The sample variance of the 50 sample means is .8096. (d) According to Theorem 6.1, the distribution of the means has mean 0 and variance 02 N—n__ 8.667 20 71—‘N—1 —10 -E=.5977. The sample values in part (C) compare well with these theoretical values. (a) The mean and variance of the distribution are the same as in part (a) of Exercise 6.7. (b) The 50 samples, taken with replacement, are shown in Table 6.2 along with 183 their means. Table 6.2. 50 Samples of Size 10 Taken with Replacement. 184 Chapter 6 SAMPLING DISTRIBUTIONS Table 6. 2. (continued) 3 4 4 0 3 0 4 4 3 4 O 4 1 (c) The mean of the means is —.048. The sample variance of the means is .8621. (d) According to Theorem 6.1, the distribution of the means has mean 0 and variance 02/71 = 8.667/10 = .8667. The sample values in part (c) compare well with these theoretical values. 6.9 A table of each outcome and the mean follows: 185 Outcome Mean Outcome Mean 1, 1 1.0 3, 1 2.0 1, 2 1.5 3, 2 2.5 1, 3 2.0 3, 3 3.0 1, 4 2.5 3, 4 3.5 2, 1 1.5 4, 1 2.5 2, 2 2.0 4, 2 3.0 2, 3 2.5 4, 3 3.5 2, 4 3.0 4, 4 4.0 Thus , the distribution of these values is: Value No. of Ways Obtained Probability 1.0 1 .0625 1.5 2 .1250 2.0 3 .1875 2.5 4 .2500 3.0 3 .1875 3.5 2 .1250 4.0 1 .0625 Where the probability is ( no. ways )X(.25)2. Consequently, the distribution of X has mean #2? = 1(.0625) + 1.5(.1250) + - - - + 3.5(.1250) + 4(.0625) = 2.5 which may be obtained directly from the symmetry of the distribution. The dis— tribution of X has variance 032 = (1 - 2-5)2(.0625) + (1.5 — 2.5)2(.1250) + - -- 186 Chapter 6 SAMPLING DISTRIBUTIONS + (3.5 ~ 2.5)2(.1250) + (4 ~ 2.5)2(.0625) = .625. Now the mean of the original distribution is also 2.5 and the variance is 1.25. Thus, Theorem 6.1 yields 2.5 and 1.25/2 = .625 as the mean and the variance of the distribution of the sample mean of two observations. These agree exactly as they must. 6.10 The 25 means are 3.80, 4.25, 4.25, 5.05, 4.85, 4.20, 4.05, 4.15, 4.85, 4.20, 3.00, 5.20, 4.30, 4.45, 3.75, 5.40, 5.60, 5.70, 4.00, 5.10, 3.15, 4.50, 3.40, 5.00, 4.50. The sample mean of these 25 means is 4.428. The sample standard deviation of these 25 means is .714. The population mean is: 1 1 1 — — —=4. . 010+110+ +910 5 The population variance is: 1 1 1 2 2 2 2 — 1— 9—— . =. 010+ 10+ + 10 (45) 825, so the population standard deviation is 2.87228. Theorem 6.1 says that the mean of the distribution of X is 4.5 and the variance is 8.25/20 = .4125 Thus the theoretical standard deviation is .642. These compare well to the sample results. 6.11 The variance of the sample mean X, based on a sample of size n, is a2/n. Thus 187 the standard deviation, or standard error of the mean is 0/ \/7—z. (a) The standard deviation for a sample of size 50 is 0/\/5—0. The standard deviation for a' sample is size 200 is U/\/ 200. That is, the ratio of standard errors is a/m_ x/SE_1 a/x/E _ 2 E7 so the standard error is halved. El (b) The ratio of standard errors is 0/ 900 _ M400 _ g 0/ 400 «900 3’ so the standard error for a sample size 900 is 2/ 3rd’s that for sample size 400. (c) The standard error for a sample size 25 is W 2553 x/fi times as large as that for sample size 225. (d) The standard error for a sample size 40 is :1 __6__=4 m times as large as that for sample size 640. 6.12 The finite population correction factor is given by (N — n)/ (N — 1). Thus, when (a) n = 5 and N = 200, the finite population correction factor is 200—5 200-1 = .9799. 188 Chapter 6 SAMPLING DISTRIBUTIONS (b) n = 10 and N = 400, the finite population correction factor is 400 —- 10 400 _ 1 = .9774. (c) n = 100 and N = 5, 000, the finite population correction factor is 5,000— 100 = . 2. 5,000 —- 1 980 6.13 We need to find P([X — pl < .6745 - a/fi). Since the standard deviation of the mean is (IA/E, the standardized variable (X — u)/(0/\/fi) is approximately a normal random variable for large n (central limit theorem). Thus , we need to find: X- M P(| < .6745). a/f‘ Now, interpolating in Table 3 gives X— < .6745 = .75. HUM > Thus, _ _ X — M X—— P g —.6745 P( ”> 6745 * .25 (0N5 )= U/x/_ ) so _ X e u P(| a/fil < .6745) —— .75 —— .25 -— .50. The probability that the mean of a random sample of size n , from a popula— _ tion with standard deviation 0, will differ from ,u by less than (.6745)(a/\/fi) is approximately .5 for sufficiently large n. 6.14 (a) Chebyshev’s theorem for X ( which has standard deviation (IA/H ) states that: ka 1 P(lX— Ml<fi)2 1_E‘ 189 In this case, a = 2.4 and n = 25 so we take k = (1.2)5/2.4 = 2.5. Then _ — 2.50 1 (b) Under the central limit theorem Pu)? — l < 3) ~ F<k> F(—k) M W N 7 where F is the standard normal distribution in Table 3. Thus, lX—ul N _ _ 0N7? <2.5)~F(2.5) F( 2.5) .9938 — .0062) = .9876. P(|X—1L|<1.2) = P( 6.15 Here U/fi = 16/10 = 1.6 and X —— 176 P1 <X<178=P~1<X—17<2=P—.2< (75_ _ ) ( _ 6_) (65_ 16/10 3 1.25). Since n = 100 is large , the central limit theorem yields the approximation X—176 16/10 22 P(——.625 g g 1.25) F(1.25) ~ F(—.625) .8944 — .266) = .628. H 6.16 Since the population is normal, the distribution of X is exactly normal and we need to find X — 7.75 — 10 X — 10 P( ” < —3.00) = F(—3.00) = .0013. a/fi < 1.5/\/Z )=P( 1.5/2 Thus, if the process is deemed “out of control” by the decision rule ( i.e. X < 7.75 means that the process is termed “out of control”), then it is very unlikely that 190 Chapter 6 SAMPLING DISTRIBUTIONS the process is actually in control. However, the process may actually be out of control with relatively high probability of not being detected. 6.17 We need to find 36 P(Z X.- > 6,000) P(X > 166.67) = P(X — 163 > 3.67) i=1 X—163 18/6 P( > 1.222). Since n == 36 is relatively large, we use the central limit theorem to approximate this probability by 1— F(1.222) = .111. 6.18 (a) If X has density f(x) , and Y = X — [1, then HY = /°°<x—u>f(x) dx= /::vf(x) div—u/_:f($) do:- —00 _ But, /00 $f(2:) dz = IL, and /00 f(:r) da: =1, so, HY = u — u = 0. (b) 012/ = expected value of (Y — py)2 =1WX*M”W2 : /m(m*#)2f($) d-‘E = 0'2- 6.19 Let 11); = E(X) be the expected value of X. First we will show that E(X + Y) = 191 E(X) + E(Y). Let f(a:, y) be the joint density function of X and Y. Then, if X takes on discrete values $1- and Y takes on discrete values yj, i<$1+yj>f<mwfl 1:0 || M8 E(X + Y) F o = 293 Zf (rm) +29 Zflmuyfl i=0 jO: =0 i=0 But, 22:0 f (xi, yj) = f1($), the marginal density of X. Similarly for Y. Thus E(X +Y) = 2mif1($i) + Zyjf2(yj) = E(X) + E(Y) i=0 j=0 Thus, i; X.- MX = E( ”I ) 1 n l n = 5m: X1) = — ZEOQ) i=1 n i=1 But7 each E(X,-) = ,u so u)? = p. 6.20 Since the sample is from a normal population 93—“ _ 47.5—42.1 s/fi _ 8.4/5 t = = 3.214 is the value of a t random variable with 24 degrees of freedom. From Table 4, X— SN— P( > 2.797) = .005. Before sampling, the probability of getting a value of t > 2.797 is .005. The 192 6.21 6.22 6.23 Chapter 6 SAMPLING DISTRIBUTIONS probability of being as large or larger than the observed t = 3.214 is even smaller. Consequently, the data tends to refute the claim that the population mean is 42.1 . It is likely that the true mean is greater then 42.1. The mean of the data is 11‘: = 23 and the sample standard deviation is 6.39. Thus, if the data is from a normal population with ,u = 20, the statistic is the value of a t random variable with 5 degrees of freedom. The entry in Table 4 for a = 10 and 1/ = 5 is 1.476. Before the data are observed, we know that X—u “SM > 1.15) > .10. Thus, the data does not give strong evidence against the ambulance service’s claim. If the sample is from a normal population With ,u = .5000 # :2 - ,1 _ .5060 — .5000 SN?) .0040/@ t = 4.74 is the value of a t random variable With 9 degrees of freedom. Horn Table 4, X—# P‘ S/x/fi > 3.25) = .005. Before observing the data, the probability of getting a value of t > 3.25 is .005. The probability of being as large or larger than the observed t = 4.74 is even smaller. Thus it is very unlikely that the process is under control. The data suggest that the true mean is greater then .5000 cm. We need to find —1 S2 —1 39.74 P(S2 > 39.74) = P(£32—) > (71—)2——— U U ) 193 2 P((n— 1)S2 > (14)39.74 ) 1452 a2 21.32 0-2 =p( > 26.12). Since the data are from a normal population, the statistic (n — 1)S2/a2 has a X2 distribution with n — 1 degrees of freedom. From Table 5, with 1/ = 14 degrees of freedom, we see that 145'2 0-2 P( > 26.12) = .025. Thus, the probability that the claim will be rejected even though 02 = 21.3 is .025 or 2.5 percent. 6.24 We need to find 9(3.14)2 (n — 1)S2 9(8.94)2 2 ._ _ P(3.14 < S < 8.94) — P( 42.5 < 02 < 42.5 ). Since the data are from a normal population, the statistic (n — 1)S2/c72 has a X2 distribution with I/ = n — 1 degrees of freedom. From Table 5, with V = 9 degrees of freedom, we see that — 1 S2 — 1 2 P(3.14 < 82 < 8.94) = P(%— < 16.925) — p(£7}_;éls_ < 2.0879) = 95 — .01 = 94 6.25 We need to find 1 — P(1 < 512. < 7) 7 _ $2 _ ' Since the samples are independent and from normal populations, the statistic Sf/Sg has an F distribution with m — 1 = 122 — 1 = 7 degrees of freedom for both the numerator and the denominator. From Table 6(b), we see that 194 Chapter 6 SAMPLING DISTRIBUTIONS Using the relation 1 F- 1/ ,1/ 2 ——~—, 1 oz( 1 2) Fa(V2,V1) we know that 52 1 P —i — = .01. (s; < 6.99) Thus, approximately 82 1 82 P —1 < — —i > 7 (s; 7 or s; l S2 1 S2 =P—1<— P—1>7 2.01 .01 :02. 6.26 (a) F (1215)— 1 —- 1 — 382 ‘95 ’ _ F05(15,12) ’ 2.62 _' ' (b) 1 1 F 6 2 =—=~——=.135. ‘9“ ’ 0) F.01(20, 6) 7.40 6.27 We need to find 2 _ (n—1)S2 4-180 __ (n-1)S2 P(S > 180) — P(——————-a2 > 144 ) — P(———U2 > 5). Since the sample is from a normal population, the statistic ((n— 1)S2) / (02) has a X2 distribution with n — 1 = 4 degrees of freedom. The density of the X2 distribution with 4 degrees of freedom is Thus, 195 —:l:/2 Integrating by parts with u = a: and d1) = 6 gives _e—x/2 16—5/2 5 (3—5/27 P(S2 > 180): (2:1: + 1) loo (2 + 1): i = .2873. 6.28 The value for t_05 for 1/ = 1 in Table 4 is 6.314. Thus we need to show that 0°11 Pt>t =Pt>6.314=/ _ dt=.05_ ( .05) ( ) 6.314 7r 1 + t2 But, 00 1 1 1 00 1 71' A314 ;1 + t2 dt _ ; arctan (0 16.314 — ;(§ ~ 1.4137) _ .05 as required. 6.29 The probability that the ratio of the larger to the smaller sample variance exceeds 3 is 1 512 3 6x 1—[email protected]< §—_ <3) 1~A%(1+xydw Letu=x+1orx=u—1. Then 1 Sf 4au—n —3 2 ”fag?” = L/gTd“=‘J'4/3+ us '4/3 24 52 ._E—a_mm Thus, 5'12 1—m13<8;_<m=1<%%=m%. 2 6.30 (a) Number the states from 1 to 50 according to their alphabetical order. Using the last two columns of the second page of Table 7, starting at row 11, to select 10 numbers between 1 and 50 by discarding numbers outside of this range as well as any previously drawn number gives 17,2,12,1,44,18,33,39,40,41 196 Chapter 6 SAMPLING DISTRIBUTIONS These correspond to Kentucky, Alaska, Idaho, Alabama, Utah, Louisiana, North Carolina , Rhode Island, South Carolina, and South Dakota. (b) No. High population states like California would have many participants but could send only two. A student from a small state would have a better chance than a student in California. 6.31 (a) The number of samples ( given that order does not matter) is 8 8-7 =—-=28. (2) 2.1 (b) The number of samples (given that order does not matter) is 20 20-19 <2)“§T_190‘ 6.32 For a random sample, each outcome must have equal probability. The probabilities are : (a) 5% = .0357. (b) 1—355 = .00526. 6.33 The finite population correction factor is (N — n)/(N — 1). Thus (a) N—n 8—2 N_1 _§:—1_.857. (b) — —2 N " 29——=.947. 197 6.34 (a) Here U/fl = 9/6 = 1.5 and we will bound the one—tailed probability P(X > 66.75) = P(X — 63 > 66.75 — 63). Using Chebyshev’s theorem for the two-tailed probability 66.75 — 63 P(lX—63|>66.75—63) = P(|X—63|> 9/6 9/6) 9/6 < _____ _ (66.75 — 63 )2 = .16. Thus, P(X > 66.75) S .16. (b) We need to find X — 63 66.75 — 63 X — 63 am > 9/6 kilo/«77 P( > 2.5) which by the central limit theorem is P(Z > 2.5) = 1 — F(2.5) = 1 -— .9938 = .0062. 6.35 Let X be the mean number of pieces of mail in a random sample of size 25. (a) By Chebyshev’s theorem, set — — — 1 P(X<40 or X>48)=P(|X~44| >4) 3 E Where 8 1 4 = k - = k— — = .16 0X m 01' k2 Thus, P(X'<40 or X>48) _<_ .16 198 Chapter 6 SAMPLING DISTRIBUTIONS (b) By the central limit theorem, P(X<40 or X>48) =1—P(40§X§48) 40—44 22—” 48—44 =1—P < < (8/x/2—5 — a/x/fi ‘ 8/x/fi) m1—P(—25§Z§2.5)=2P(Z>25) 2(1— .9938) = .0124 6.36 Since the population is normal, the distribution of X is exactly normal and we need to find P<1X—uls.o2>P(1X———0;/5I<_ #235 .9938 ~ .0062)— — .9876. H )= F(2 5)— F(—2.5) H 6.37 By the central limit theorem, lX—Ml U/fi > .16l/x/3E) P(|Z| > 2.25) = 2(1 — .9878) = .0244 H P(|X — p] > .06) P( 22 6.38 We need to find Since the samples are independent and from normal populations, the statistic Sf/Sg has an F distribution with n1 — 1 = 8 and n2 — 1 = 15 degrees of freedom for the numerator and the denominator, respectively . From Table 6(b), we see that so the probability is .01 or 1 percent. 6.39 By Theorem 6.5, F = 83/312 has F distribution with 7 and 25 degrees of freedom. 199 Hence , from Table 6(a), we have P(— > 2.4) = P(F > 2.4) = .05 6.40 The variance of the sample mean X, based on a sample of size n, is 02/72. Thus the standard deviation, or standard error, of the mean is a/fi. (a) The standard deviation for a sample of size 100 is 0/V100. The standard deviation for a sample is size 200 is 0/\/ 200. That is, the ratio of standard errors is 0/\/200 _ «100 _ _1_ _ 707 (wan—«2’6”fi so the standard error for sample size 200 is .707 times the standard error for sample size 100. (b) The ratio of standard errors is 0/ 300_~\/200_ g_ 816 a/ 200 «300 3 ' ’ so the standard error for a sample size 300 is .816 times that for sample size 200. (c) The ratio of standard errors is a/x/9_0 _ a/m— 2, so the standard error doubles. 6.41 Under this sampling scheme, the observations will not satisfy the independence assumption required for a random sample. The longest lines are likely to occur together when the same cars must wait several light changes during the rush hour. 6.42 (a) The production process may not be stabilized when the first shaft is made 200 6.43 (a) (C) each morning. Alternatively, the first shaft could be given extra attention and may not be representative. Buses and large trucks would take longer to pass the fixed point and thus would be more likely to be included in the sample than a small car. No, we would expect the weight of the bags to vary because the weights of the individual apples vary. With X1- the weight of the 2' —~ th apple placed in 3 the bag, 2 OXi is the total weight of the bag. Then, i=1 :2 'M O 3‘1 |/\ 3 || 130?: 11/30) so the probability concerning total weight can be expressed as a probability concerning X. Further, PR 3 y/BO) = % ) X—u ONE 30. Thus, for any specified value y, we can approximate the probability 30 . . 21/30 ‘ M P( :Xi g y) by the standard normal probablllty P( Z S —a_/f—n ) i=1 By the central limit theorem is nearly standard normal when n = With 11 = .2 and a = .03 pound )= 1 — F(1.217) = .1118 ...
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