Sampling and Sampling Distributions

Chapter 7 Sampling and Sampling DistributionsLearning Objectives1. Understand the importance of sampling and how results from samples can be used to provide estimates of population characteristics such as the population mean, the population standard deviation and / or the population proportion.2. Know what simple random sampling is and how simple random samples are selected.3. Understand the concept of a sampling distribution.4. Understand the central limit theorem and the important role it plays in sampling.5. Specifically know the characteristics of the sampling distribution of the sample mean ( x ) and the sampling distribution of the sample proportion ( p ).6. Learn about a variety of sampling methods including stratified random sampling, cluster sampling, systematic sampling, convenience sampling and judgment sampling.7. Know the definition of the following terms: parameter sampling distribution sample statistic finite population correction factor simple random sampling standard error sampling without replacement central limit theorem sampling with replacement unbiased point estimator relative efficiency point estimate consistency7 - 1 Chapter 7Solutions:1. a. AB, AC, AD, AE, BC, BD, BE, CD, CE, DE b. With 10 samples, each has a 1/10 probability. c. E and C because 8 and 0 do not apply.; 5 identifies E; 7 does not apply; 5 is skipped since E is already in the sample; 3 identifies C; 2 is not needed since the sample of size 2 is complete.2. Using the last 3-digits of each 5-digit grouping provides the random numbers:601, 022, 448, 147, 229, 553, 147, 289, 209Numbers greater than 350 do not apply and the 147 can only be used once. Thus, the simple random sample of four includes 22, 147, 229, and 289.3. 459, 147, 385, 113, 340, 401, 215, 2, 33, 3484. a. 6, 8, 5, 4, 1 Nasdaq 100, Oracle, Microsoft, Lucent, Applied MaterialsN ! 10! 3,628,500 b.    252 n!( N n )! 5!(10  5)! (120)(120)5. 283, 610, 39, 254, 568, 353, 602, 421, 638, 1646. 2782, 493, 825, 1807, 2897. 108, 290, 201, 292, 322, 9, 244, 249, 226, 125, (continuing at the top of column 9) 147, and 113.8. 13, 8, 23, 25, 18, 5The second occurrences of random numbers 13 and 25 are ignored.Maryland, Iowa, Florida State, Virginia, Pittsburgh, Oklahoma9. 102, 115, 122, 290, 447, 351, 157, 498, 55, 165, 528, 2510. finite, infinite, infinite, infinite, finite54 11. a. x  x / n   9 i 6(x  x) 2 b. s  i n 12 2 2 2 2 2 2 (xi  x) = (-4) + (-1) + 1 (-2) + 1 + 5 = 4848 s =  3.1 6 112. a. p = 75/150 = .507 - 2 Sampling and Sampling Distributions b. p = 55/150 = .3667465 13. a. x  x / n   93 i 5 b. 2 xi (xi  x) (xi  x) 94 +1 1 100 +7 49 85 -8 64 94 +1 1 92 -1 1 Totals 465 0 116(x  x) 2 116 s  i   5.39 n 1 414. a. 149/784 = .19 b. 251/784 = .32 c. Total receiving cash = 149 + 219 + 251 = 619619/784 = .7970 15. a. x  x/ n   7 years i 10(x  x )2 20.2 b. s i   1.5 years n 1 10  116. p = 1117/1400 = .8017. a. 595/1008 = .59 b. 332/1008 = .33 c. 81/1008 = .0818. a. E(x)    200 b.  x   / n  50 / 100  5 c. Normal with E ( x ) = 200 and  x = 5 d. It shows the probability distribution of all possible sample means that can be observed with random samples of size 100. This distribution can be used to compute the probability that x is within a specified  from 19. a. The sampling distribution is normal with7 - 3 Chapter 7E ( x ) =  = 200 x   / n  50 / 100  5For 5, ( x -  ) = 5 x   5 z    1 Area = .3413  x 52(.3413) = .6826 b. For  10, ( x -  ) = 10 x   10 z    2 Area = .4772  x 52(.4772) = .954420.  x   / n x 25/ 50  3.54 x 25/ 100  2.50 x 25/ 150  2.04 x 25/ 200  1.77The standard error of the mean decreases as the sample size increases.21. a.  x   / n  10 / 50  1.41 b. n / N = 50 / 50,000 = .001Use  x   / n  10 / 50  1.41 c. n / N = 50 / 5000 = .01Use  x   / n  10 / 50  1.41 d. n / N = 50 / 500 = .10N  n  500  50 10 Use  x    1.34 N 1 n 500 1 50Note: Only case (d) where n /N = .10 requires the use of the finite population correction factor. 7 - 4 Sampling and Sampling Distributions22. a.  x   / n  4000 / 60  516.40 x 51,800The normal distribution is based on the Central Limit Theorem. b. For n = 120, E ( x ) remains $51,800 and the sampling distribution of x can still be approximated by a normal distribution. However,  x is reduced to 4000 / 120 = 365.15. c. As the sample size is increased, the standard error of the mean,  x , is reduced. This appears logical from the point of view that larger samples should tend to provide sample means that are closer to the population mean. Thus, the variability in the sample mean, measured in terms of  x , should decrease as the sample size is increased.23. a. x   / n  4000 / 60  516.40 x 51,300 51,800 52,300 52,300 51,800 z   .97 Area = .3340 516.402(.3340) = .6680 b.  x   / n  4000 / 120  365.1552,300 51,800 z   1.37 Area = .4147 365.152(.4147) = .829424. a. Normal distribution, E( x ) 4260x   /n  900 / 50  127.287 - 5 Chapter 7 b. Within $250P(4010  x  4510) 4510 4260 z   1.96 Area = .4750 127.282(.4750) = .95 c. Within $100P(4160  x  4360) 4360 4260 z   .79 Area = .2852 127.282(.2852) = .570425. a. E( x ) = 1020 x   / n  100 / 75  11.551030 1020 b. z   .87 Area =.3078 11.551010 1020 z   .87 Area =.3078 11.55 probability = .3078 + .3078 =.61561040 1020 c. z   1.73 Area = .4582 11.551000 1020 z   1.73 Area = .4582 11.55 probability = .4582 + .4582 = .9164 x  34,000 26. a. z   / nError = x - 34,000 = 2507 - 6 Sampling and Sampling Distributions250 n = 30 z   .68 2(.2517) = .5034 2000 / 30250 n = 50 z   .88 2(.3106) = .6212 2000 / 50250 n = 100 z   1.25 2(.3944) = .7888 2000 / 100250 n = 200 z   1.77 2(.4616) = .9232 2000 / 200250 n = 400 z   2.50 2(.4938) = .9876 2000 / 400 b. A larger sample increases the probability that the sample mean will be within a specified distance from the population mean. In the salary example, the probability of being within 250 of  ranges from .5036 for a sample of size 30 to .9876 for a sample of size 400.27. a. E( x ) = 982x   /n  210 / 40  33.2 x   100 z    3.01 Area = .4987  /n 210 / 402(.4987) = .9974 x   25 b. z    .75 Area = .2734  /n 210 / 402(.2734) = .5468 c. The sample with n = 40 has a very high probability (.9974) of providing a sample mean within  $100. However, the sample with n = 40 only has a .5468 probability of providing a sample mean within  $25. A larger sample size is desirable if the  $25 is needed.28. a. Normal distribution, E( x ) = 687x   /n  230 / 45  34.29 b. Within $100P(587  x  787) 7 - 7 Chapter 7787 687 z   2.92 Area = .4982 34.292(.4982) = .9964 c. Within $25P(662  x  712)712 687 z   .73 Area = .2673 34.292(.2673) = .5346 d. Recommend taking a larger sample since there is only a .5346 probability n = 45 will provide the desired result.29.  = 1.46  = .15 a. n = 30 x   .03 z    1.10  /n .15/ 30P(1.43  x  1.49) = P(-1.10  z  1.10) = 2(.3643) = .7286 b. n = 50 x   .03 z    1.41  /n .15/ 50P(1.43  x  1.49) = P(-1.41  z  1.41) = 2(.4207) = .8414 c. n = 100 x   .03 z    2.00  /n .15/ 100P(1.43  x  1.49) = P(-2  z  2) = 2(.4772) = .9544 d. A sample size of 100 is necessary.30. a. n / N = 40 / 4000 = .01 < .05; therefore, the finite population correction factor is not necessary. b. With the finite population correction factor7 - 8 Sampling and Sampling DistributionsN  n  4000  40 8.2  x    1.29 N 1 n 4000 1 40Without the finite population correction factor x   / n  1.30Including the finite population correction factor provides only a slightly different value for  x than when the correction factor is not used. x   2 c. z    1.54 Area = .4382 1.30 1.302(.4382) = .876431. a. E ( p ) = p = .40 p(1 p ) .40(.60) b.     .0490 p n 100 c. Normal distribution with E ( p ) = .40 and  p = .0490 d. It shows the probability distribution for the sample proportion p .32. a. E ( p ) = .40 p(1 p ) .40(.60)     .0346 p n 200 p p .03 z    .87 Area = .3078  p .03462(.3078) = .6156 p p .05 b. z    1.44 Area = .4251  p .03462(.4251) = .8502 p(1 p) 33.   p n(.55)(.45)    .0497 p 100 (.55)(.45)    .0352 p 2007 - 9 Chapter 7(.55)(.45)    .0222 p 500(.55)(.45)    .0157 p 1000 p decreases as n increases (.30)(.70) 34. a.    .0458 p 100 p p .04 z    .87  p .0458Area = 2(.3078) = .6156(.30)(.70) b.    .0324 p 200 p p .04 z    1.23  p .0324Area = 2(.3907) = .7814(.30)(.70) c.    .0205 p 500 p p .04 z    1.95  p .0205Area = 2(.4744) = 0.9488(.30)(.70) d.    .0145 p 1000 p p .04 z    2.76  p .0145Area = 2(.4971) = .9942 e. With a larger sample, there is a higher probability p will be within  .04 of the population proportion p.35. a.7 - 10 Sampling and Sampling Distributions p(1 p ) .30(.70)     .0458 p n 100 p .30The normal distribution is appropriate because np = 100(.30) = 30 and n(1 - p) = 100(.70) = 70 are both greater than 5. b. P (.20  p  .40) = ?.40 .30 z   2.18 Area = .4854 .04582(.4854) = .9708 c. P (.25  p  .35) = ?.35 .30 z   1.09 Area = .3621 .04582(.3621) = .724236. a. Normal distribution, E( p ) = .56 p(1 p ) .56(1  .56)     .0287 p n 300 b. Within ± .03.59 .56 z   1.05 Area = .3531 .02872(.3531) = .7062.56(1 .56) c. For n = 600,    .0203 p 600.59 .56 z   1.48 Area = .4306 .02032(.4306) = .8612.56(1 .56) For n = 1000,    .0157 p 10007 - 11 Chapter 7.59 .56 z   1.91 Area = .4719 .01572(.4719) = .943837. a. Normal distributionE ( p ) = .50 p(1 p ) (.50)(1  .50)     .0206 p n 589 p p .04 b. z    1.94  p .02062(.4738) = .9476 p p .03 c. z    1.46  p .02062(.4279) = .8558 p p .02 d. z    .97  p .02062(.3340) = .668038. a. Normal distributionE( p ) = .25 p(1 p ) (.25)(.75)     .0306 p n 200.03 b. z   .98 Area = .3365 .03062(.3365) = .6730.05 c. z   1.63 Area = .4484 .03062(.4484) = .896839. a. Normal distribution with E( p ) = p = .25 and 7 - 12 Sampling and Sampling Distributions p(1 p ) .25(1  .25)     .0137 p n 1000 p p .03 b. z    2.19  p .0137P(.22  p  .28) = P(-2.19  z  2.19) = 2(.4857) = .9714 p p .03 z    1.55 c. .25(1 .25) .0194 500P(.22  p  .28) = P(-1.55  z  1.55) = 2(.4394) = .878840. a. E ( p ) = .76 p(1 p ) .76(1  .76)     .0214 p n 400Normal distribution because np = 400(.76) = 304 and n(1 - p) = 400(.24) = 96.79 .76 b. z   1.40 Area = .4192 .0214.73 .76 z   1.40 Area = .4192 .0214 probability = .4192 + .4192 = .8384 p(1 p ) .76(1  .76) c.     .0156 p n 750.79 .76 z   1.92 Area = .4726 .0156.73 .76 z   1.92 Area = .4726 .0156 probability = .4726 + .4726 = .945241. a. E( p ) = .17 p(1 p ) (.17)(1  .17)     .0133 p n 800.19 .17 b. z   1.51 Area = .4345 .01337 - 13 Chapter 7.15 .17 z   1.51 Area = .4345 .0133 probability = .4345 + .4345 = .8690 p(1 p ) (.17)(1  .17) c.     .0094 p n 1600.19 .17 z   2.13 Area = .4834 .0094.15 .17 z   2.13 Area = .4834 .0094 probability = .4834 + .4834 = .966842. 112, 145, 73, 324, 293, 875, 318, 61843. a. Normal distributionE( x ) = 3 1.2  x    .17 n 50 x   .25 b. z    1.47 Area = .4292  /n 1.2 / 502(.4292) = .858444. a. Normal distributionE( x ) = 31.5  12  x    1.70 n 501 b. z   .59 Area = .2224 1.702(.2224) = .44483 c. z   1.77 Area = .4616 1.702(.4616) = .923245. a. E( x ) = $24.07 7 - 14 Sampling and Sampling Distributions 4.80  x    .44 n 120.50 z   1.14 Area = .3729 .442(.3729) = .74581.00 b. z   2.28 Area = .4887 .442(.4887) = .977446.  = 41,979  = 5000 a.  x 5000 / 50  707 x   0 b. z    0  x 707P( x > 41,979) = P(z > 0) = .50 x   1000 c. z    1.41  x 707P(40,979  x  42,979) = P(-1.41  z  1.41) = 2(.4207) = .8414 d.  x 5000 / 100  500 x   1000 z    2.00  x 500P(40,979  x  42,979) = P(-2  z  2) = 2(.4772) = .9544N  n  47. a.  x  N 1 nN = 20002000  50 144  x   20.11 2000 1 50N = 50005000  50 144  x   20.26 5000 1 50 N = 10,0007 - 15 Chapter 710,000  50 144  x   20.31 10,000 1 50Note: With n / N  .05 for all three cases, common statistical practice would be to ignore 144 the finite population correction factor and use  x   20.36 for each case. 50 b. N = 200025 z   1.24 Area = .3925 20.112(.3925) = .7850N = 500025 z   1.23 Area = .3907 20.262(.3907) = .7814N = 10,00025 z   1.23 Area = .3907 20.312(.3907) = .7814All probabilities are approximately .78 500 48. a.  x    20 n n n = 500/20 = 25 and n = (25)2 = 625 b. For  25,25 z   1.25 Area = .3944 202(.3944) = .788849. Sampling distribution of x7 - 16 Sampling and Sampling Distributions   x   n 300.05 0.05 x  1.9 2.11.9 + 2.1  = = 2 2The area between  = 2 and 2.1 must be .45. An area of .45 in the standard normal table shows z = 1.645.Thus,2.1 2.0    1.645  / 30Solve for (.1) 30    .33 1.64550. p = .305 a. Normal distribution with E( p ) = p = .305 and p(1 p ) .305(1  .305)     .0326 p n 200 p p .04 b. z    1.23  p .0326P(.265  p  .345) = P(-1.23  z  1.23) = 2(.3907) = .7814 p p .02 c. z    .61  p .0326P(.285  p  .325) = P(-.61  z  .61) = 2(.2291) = .45827 - 17 Chapter 7 p(1 p ) (.40)(.60) 51.     .0245 p n 400P ( p  .375) = ?.375 .40 z   1.02 Area = .3461 .0245P ( p  .375) = .3461 + .5000 = .8461 p(1 p ) (.71)(1  .71) 52. a.     .0243 p n 350 p p .05 z    2.06 Area = .4803  p .02432(.4803) = .9606 p p .75  .71 b. z    1.65 Area = .4505  p .0243P ( p  .75) = .5000 - .4505 = .049553. a. Normal distribution with E ( p ) = .15 and p(1 p ) (.15)(.85)     .0292 p n 150 b. P (.12  p  .18) = ?.18 .15 z   1.03 Area = .3485 .02922(.3485) = .6970 p(1 p) .25(.75) 54. a.    .0625 p n nSolve for n.25(.75) n   48 (.0625) 2 b. Normal distribution with E( p ) = .25 and  x = .0625 c. P ( p  .30) = ?7 - 18 Sampling and Sampling Distributions.30 .25 z   .80 Area = .2881 .0625P ( p  .30) = .5000 - .2881 = .21197 - 19

Sampling and Sampling Distributions

Details

Download

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

Support