Sampling and Sampling Distributions s1

Chapter 7465 9. 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 4$45,500 11. a. x= S x/ n = = $4,550 i 10S(x - x )2 9,068,620 b. s =i = = $1003.80 n -1 10 - 115. a. The sampling distribution is normal withE (x ) =  = 200 sx = s /n = 50 / 100 = 5For 5, 195＃x 205Using Standard Normal Probability Table: x - m 5 At x = 205, z = = = 1 P( z 1) = .8413 s x 5 x -m -5 At x = 195, z = = = -1 P( z < - 1) = .1587 s x 5P(195＃ x 205) = .8413 - .1587 = .6826Using Excel: =NORMDIST(205,200,5,TRUE)-NORMDIST(195,200,5,TRUE) = .6827 b. For  10, 190＃x 210Using Standard Normal Probability Table: x   10 At x = 210, z    2 P( z 2) = .9772  x 57 - 1 Chapter 7 x -m -10 At x = 190, z = = = -2 P( z < - 2) = .0228 s x 5P(190＃ x 210) = .9772 - .0228 = .9544Using Excel: =NORMDIST(210,200,5,TRUE)-NORMDIST(190,200,5,TRUE) = .954517. 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. 25.  = 2.34  = .20 a. n = 30 x - m .03 z = = = .82 s /n .20 / 30P(2.31  x  2.37) = P(-.82  z  .82) = .7939 - .2061 = .5878Using Excel: =NORMDIST(2.37,2.34,.20/SQRT(30),TRUE)- NORMDIST(2.31,2.34,.20/SQRT(30),TRUE)=.5887 b. n = 50 x - m .03 z = = = 1.06 s /n .20 / 50P(2.31  x  2.37) = P(-1.06  z  1.06) = .8554 - .1446 = .7108Using Excel: =NORMDIST(2.37,2.34,.20/SQRT(50),TRUE)- NORMDIST(2.31,2.34,.20/SQRT(50),TRUE)=.7112 c. n = 100 x - m .03 z = = = 1.50 s /n .20 / 1007 - 2 Sampling and Sampling DistributionsP(2.31  x  2.37) = P(-1.50  z  1.50) = .9332 - .0668 = .8664Using Excel: =NORMDIST(2.37,2.34,.20/SQRT(100),TRUE)- NORMDIST(2.31,2.34,.20/SQRT(100),TRUE)=.8664 d. None of the sample sizes in parts (a), (b), and (c) are large enough. At z = 1.96 we find P(-1.96  z  1.96) = .95. So, we must find the sample size corresponding to z = 1.96. Solve .03 = 1.96 .20 / n骣.20 n =1.96琪 = 13.0667 桫.03 n = 170.73Rounding up, we see that a sample size of 171 will be needed to ensure a probability of .95 that the sample mean will be within $.03 of the population mean.31. a. 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 P(z ≤ 2.18) = .9854 .0458P(z < -2.18) = .0146P(.20 ≤ p ≤ .40) = .9854 - .0146 = .9708Using Excel:=NORMDIST(.40,.30,.0458,TRUE)-NORMDIST(.20,.30,.0458,TRUE) = .9710 c. P (.25  p  .35) = ?7 - 3 Chapter 7.35 .30 z   1.09 P(z ≤ 1.09) = .8621 .0458P(z < -1.09) = .1379P(.25 ≤ p ≤ .35) = .8621 - .1379 = .7242Using Excel:=NORMDIST(.35,.30,.0458,TRUE)-NORMDIST(.25,.30,.0458,TRUE) = .725033. a. Normal distributionE ( p ) = .50 p(1 p ) (.50)(1  .50)     .0206 p n 589 p p .04 b. z    1.94 P(z ≤ 1.94) = .9738  p .0206P(z < -1.94) = .0262P(.46 ≤ p ≤ .54) = .9738 - .0262 = .9476Using Excel:=NORMDIST(.54,.50,.0206,TRUE)-NORMDIST(.46,.50,.0206,TRUE) = .9478 p p .03 c. z    1.46 P(z ≤ 1.46) = .9279  p .0206P(z < -1.46) = .0721P(.47 ≤ p ≤ .53) = .9279 - .0721 = .8558Using Excel:=NORMDIST(.53,.50,.0206,TRUE)-NORMDIST(.47,.50,.0206,TRUE) = .8547 p p .02 d. z    .97 P(z ≤ .97) = .8340  p .0206P(z < -.97) = .1660P(.48 ≤ p ≤ .52) = .8340 - .1660 = .6680Using Excel:=NORMDIST(.52,.50,.0206,TRUE)-NORMDIST(.48,.50,.0206,TRUE) = .66847 - 4 Sampling and Sampling Distributions35. a. Normal distribution with E( p ) = p = .25 and p(1 p ) .25(1  .25)     .0137 p n 1000 p p .03 b. z    2.19 P(z ≤ 2.19) = .9857  p .0137P(z < -2.19) = .0143P(.22  p  .28) = P(-2.19  z  2.19) = .9857 - .0143 = .9714Using Excel:=NORMDIST(.28,.25,.0137,TRUE)-NORMDIST(.22,.25,.0137,TRUE) = .9715 p p .03 z    1.55 c. .25(1 .25) .0194 P(z ≤ 1.55) = .9394 500P(z < -1.55) = .0606P(.22  p  .28) = P(-1.55  z  1.55) = .9394 - .0606 = .8788Using Excel:=NORMDIST(.28,.25,.0194,TRUE)-NORMDIST(.22,.25,.0194,TRUE) = .8780Chapter 8x 80 13. a. x i   10 n 8 b. 2 xi (xi  x ) (xi  x ) 10 0 0 8 -2 4 12 2 4 15 5 25 13 3 9 11 1 1 6 -4 16 5 -5 25 84(x  x )2 84 s i   3.464 n 1 7 c. t.025 ( s / n ) 2.365(3.464 / 8)  2.97 - 5 Chapter 7 d. x t.025 ( s / n )10 ± 2.9 or 7.1 to 12.915. x t / 2 ( s / n )90% confidence df = 64 t.05 = 1.66919.5 ± 1.669 (5.2 / 65)19.5 ± 1.08 or 18.42 to 20.5895% confidence df = 64 t.025 = 1.99819.5 ± 1.998 (5.2 / 65)19.5 ± 1.29 or 18.21 to 20.7917. For the Miami data set, the output obtained using Excel’s Descriptive Statistics tool follows:RatingMean 6.34 Standard Error 0.3059 Median 6.5 Mode 8 Standard Deviation 2.1629 Sample Variance 4.6780 Kurtosis -1.1806 Skewness -0.1445 Range 8 Minimum 2 Maximum 10 Sum 317 Count 50 Confidence Level(95.0%) 0.6147The 95% confidence interval is x margin of error6.34 0.6147 or 5.73 to 6.95Sx 2600 21. x =i = = 130 liters of alcoholic beverages n 20S(x - x )2 81244 s =i = = 65.39 n -1 20 - 17 - 6 Sampling and Sampling Distributions t.025 = 2.093 df = 1995% confidence interval: x  t.025 (s / n)130  2.093 (65.39 / 20) 130  30.60 or 99.40 to 160.60 liters per year35. a. p = 281/611 = .4599 (46%) p(1 p ) .4599(1  .4599) b. z 1.645  .0332 .05 n 611 c. p ± .0332.4599  .0332 or .4267 to .4931 p(1 p ) (.53)(.47) 43. a. Margin of Error = z 1.96  .0253  / 2 n 150095% Confidence Interval: .53  .0253 or .5047 to .5553(.31)(.69) b. Margin of Error = 1.96 = .0234 150095% Confidence Interval: .31  .0234 or .2866 to .3334(.05)(.95) c. Margin of Error = 1.96 = .0110 150095% Confidence Interval: .05  .0110 or .039 to .061 d. The margin of error decreases as p gets smaller. If the margin of error for all of the interval estimates must be less than a given value (say .03), an estimate of the largest proportion should be used as a planning value. Using p* .50 as a planning value guarantees that the margin of error for all the interval estimates will be small enough.Chapter 91. a. H0:   600 Manager’s claim.Ha:  > 600 b. We are not able to conclude that the manager’s claim is wrong. c. The manager’s claim can be rejected. We can conclude that  > 600.3. a. H0:  = 32 Specified filling weightHa:   32 Overfilling or underfilling exists7 - 7 Chapter 7 b. There is no evidence that the production line is not operating properly. Allow the production process to continue. c. Conclude   32 and that overfilling or underfilling exists. Shut down and adjust the production line.5. a. The Type I error is rejecting H0 when it is true. This error occurs if the researcher concludes that young men in Germany spend more than 56.2 minutes per day watching prime-time TV when the national average for Germans is not greater than 56.2 minutes. b. The Type II error is accepting H0 when it is false. This error occurs if the researcher concludes that the national average for German young men is  56.2 minutes when in fact it is greater than 56.2 minutes. x   14 12 23. a. t 0   2.31 s/ n 4.32 / 25 b. Degrees of freedom = n – 1 = 24Upper tail p-value is the area to the right of the test statisticUsing t table: p-value is between .01 and .025Using Excel: p-value = TDIST(2.31,24,1) = .0149 c. p-value  .05, reject H0. c. With df = 24, t.05 = 1.711Reject H0 if t  1.7112.31 > 1.711, reject H0. x   44 45 25. a. t 0   1.15 s/ n 5.2 / 36Degrees of freedom = n – 1 = 35Lower tail p-value is the area to the left of the test statisticUsing t table: p-value is between .10 and .20Using Excel: p-value = TDIST(1.15,35,1) = .1290 p-value > .01, do not reject H0 x   43 45 b. t 0   2.61 s/ n 4.6 / 36Lower tail p-value is the area to the left of the test statistic7 - 8 Sampling and Sampling DistributionsUsing t table: p-value is between .005 and .01Using Excel: p-value = TDIST(2.61,35,1) = .0066 p-value  .01, reject H0 x   46 45 c. t 0   1.20 s/ n 5/ 36Lower tail p-value is the area to the left of the test statisticUsing t table: p-value is between .80 and .90Using Excel: p-value = TDIST(1.20,35,1) = .8809 p-value > .01, do not reject H031. H0:   47.50Ha:  > 47.50 x   51 47.50 t 0   2.33 s/ n 12 / 64Degrees of freedom = n - 1 = 63Upper tail p-value is the area to the right of the test statisticUsing t table: p-value is between .01 and .025Using Excel: p-value = TDIST(2.33,91,1) = .0110Reject H0; Atlanta customers are paying a higher mean water bill. 37. a. H0: p  .125Ha: p > .12552 b. p = = .13 400 p- p .13- .125 z =0 = = .30 p0(1- p 0 ) .125(1 - .125) n 400Upper tail p-value is the area to the right of the test statisticUsing normal table with z = .30: p-value = 1.0000 - .6179 = .3821Using Excel: p-value = 1-NORMSDIST(.30) = .38217 - 9 Chapter 7 c. p-value > .05; do not reject H0. We cannot conclude that there has been an increase in union membership.43. a. H0: p ≤ .10Ha: p > .10 b. There are 13 “Yes” responses in the Eagle data set.13 p = = .13 100 p- p .13- .10 z =0 = = 1.00 c. p0(1- p 0 ) .10(1 - .10) n 100Upper tail p-value is the area to the right of the test statisticUsing normal table with z = 1.00: p-value = 1 - .8413 = .1587Using Excel: p-value = 1-NORMSDIST(1) = .1587 p-value > .05; do not reject H0.The statistical results do not allow us to conclude that p > .10. But, given that p = .13, management may want to authorize a larger study before deciding not to go national.Chapter 1054 42 11. a. x   9 x   7 1 6 2 62 (xi  x1 ) b. s1   2.28 n1 12 (xi  x2 ) s2   1.79 n2 1 c. x1 x 2 = 9 - 7 = 22 2 s2 s 2  2.282 1.79 2  1 2    n1 n 2  6 6  d. df 2 2  2 2  9.5 1s2  1  s 2 1  2.28 2  1  1.79 2  1   2      n11 n 1  n 2  1  n 2  5 6  5  6 Use df = 9, t.05 = 1.8332.282 1.79 2 x x 1.833  1 2 6 67 - 10 Sampling and Sampling Distributions2 2.17 (-.17 to 4.17)xi 111.6 13. a. x1    9.3 n1 122 (xi  x1 ) 71.12 s1    2.54 n1 1 12  1xi 42 x2    4.2 n2 102 (xi  x2 ) 18.4 s2    1.43 n2 10.1 b. x1 x 2 = 9.3 - 4.2 = 5.1 tonsMemphis is the higher volume airport and handled an average of 5.1 tons per day more than Louisville. Memphis handles more than twice the volume of Louisville.2 2 s2 s 2  2.542 1.43 2  1 2    n1 n 2  12 10  c. df 2 2  2 2  17.8 1s2  1  s 2 1  2.54 2  1  1.43 2  1   2      n11 n 1  n 2  1  n 2  11 12  9  10 Use df = 17, t.025 = 2.1102 2 s1 s 2 (x1 x 2 )  t .025  n1 n 22.542 1.43 2 5.1 2.110  12 105.1  1.82 (3.28 to 6.92)15. 1 for 2001 season2 for 1992 seasonH0: 1  2  0Ha: 1  2  0 b. x1 x 2 = 60 - 51 = 9 days9/51(100) = 17.6% increase in number of days.7 - 11 Chapter 7 x1 x 2   0 (60 51)  0 t    2.48 c. s2 s 218 2 15 2 1 2  n1 n 2 45 382 2 s2 s 2  182 15 2  1 2    n1 n 2  45 38  df 2 2  2 2  81 1s2  1  s 2 1  18 2  1  15 2  1   2      n11 n 1  n 2  1  n 2  44 45  37  38 Using t table, p-value is between .005 and .01.Exact p-value corresponding to t = 2.48 is .0076 p-value  .01, reject H0. There is a greater mean number of days on the disabled list in 2001. d. Management should be concerned. Players on the disabled list have increased 32% and time on the list has increased by 17.6%. Both the increase in inquiries to players and the cost of lost playing time need to be addressed.31. a. p1 = 150/250 = .46 Republicans p2 = 98/350 = .28 Democrats b. p1 p 2 = .46 - .28 = .18Republicans have a .18, 18%, higher participation rate than Democrats. p1(1 p 2 ) p 2 (1  p 2 ) c. z.025  n1 n 2.46(1 .46) .28(1  .28) 1.96  .0777 250 350 d. Yes, .18  .0777 (.1023 to .2577)Republicans have a 10% to 26% higher participation rate in online surveys than Democrats. Biased survey results of online political surveys are very likely.33. a. p1 = 256/320 = .80 b. p2 = 165/250 = .66 c. p1 p 2 = .80 - .66 = .14 p1(1 p 1 ) p 2 (1  p 2 ) .14 z.025  n1 n 27 - 12 Sampling and Sampling Distributions.80(1 .80) .66(1  .66) .14 1.96  320 250.14  .0733 (.0667 to .2133)35. a. H0: p1 - p2 = 0Ha: p1 - p2  0 p1 = 63/150 = .42 p2 = 60/200 = .30 n p n p 63 60 p 1 1 2 2   .3514 n1 n 2 150  200 p p .42 .30 z 1 2   2.33 1 1  1 1  p1 p  .3514 1 .3514     150 200  n1 n 2    p-value = 2(1.0000 - .9901) = .0198 p-value  .05, reject H0. There is a difference between the recall rates for the two commercials. p1(1 p 1 ) p 2 (1  p 2 ) b. p1 p 2  z .025  n1 n 2.42(1 .42) .30(1  .30) .42 .30  1.96  150 200.12  .1014 (.0186 to .2214)Commercial A has the better recall rate.7 - 13

Sampling and Sampling Distributions s1

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