Chapter 20: Additional Topics in Sampling 511
Chapter 20:
Additional Topics in Sampling
20.1 Answers should refer to each of the steps outlined in Figure 20.1
20.2 Answers should refer to each of the steps outlined in Figure 20.1
20.3 Answers should refer to each of the steps outlined in Figure 20.1
20.4 Answers should refer to each of the steps outlined in Figure 20.1
20.5 Answers should refer to each of the steps outlined in Figure 20.1
20.6 Answers should deal with issues such as (a) the identification of the correct population, (b) selection (nonresponse) bias, (c) response bias
20.7 Answers should deal with issues such as (a) the identification of the correct
population, (b) selection (nonresponse) bias, (c) response bias
20.8 Answers should deal with issues such as (a) the identification of the correct population, (b) selection (nonresponse) bias, (c) response bias
20.9 While the recall method will lower the number of nonresponses, a bias may be built in since this technique will reduce the participation of individuals who are absent on Thursday evenings (e.g., night students, second shift workers, shopping mall employees, etc.)
20.10 Within Minitab, go to Calc è Make Patterned Data… in order to generate a simple set of numbers of size ‘n’ or ‘N’. Enter first value as 1, last value as the total number of stocks traded on the New York Stock Exchange. Go to Calc è Random Data è Sample from Columns… in order to generate a simple random sample of size ‘n’ from the number of all stocks. “Sample ____ rows from column(s):” For Exercise 20.10 enter 20 as the number of rows to sample from. The results will be the observation numbers in the list to include in the sample.
20.11 Within Minitab, go to Calc è Make Patterned Data… in order to generate a simple set of numbers of size ‘n’ or ‘N’. Enter first value as 1, last value as the total number of houses advertised for sale in your city. Go to Calc è Random Data è Sample from Columns… in order to generate a simple random sample of size ‘n’ from the number of all houses advertised for sale in your city. “Sample ____ rows from column(s):” For Exercise 20.11 enter 15 as the number of rows to sample from. The results will be the observation numbers in the list to include in the sample.
20.12 Within Minitab, go to Calc è Make Patterned Data… in order to generate a simple set of numbers of size ‘n’or ‘N’. Enter first value as 1, last value as 12,723. Go to Calc è Random Data è Sample from Columns… in order to generate a simple random sample of size ‘n’. “Sample ____ rows from column(s):” Enter 100 as the number of rows to sample from. The results will be the observation numbers in the list to include in the sample.
20.13 Within Minitab, go to Calc è Make Patterned Data… in order to generate a simple set of numbers of size ‘n’ or ‘N’. Enter first value as 1, last value as 984 which is the total number of pages in the text. Go to Calc è Random Data è Sample from Columns… in order to generate a simple random sample of size ‘n’. “Sample ____ rows from column(s):” Enter 50 as the number of rows to sample from. The results will be the observation numbers in the list to include in the sample.
20.14 , = .7519
9.7 ± 1.96 (.7519) (8.2262, 11.1738)
20.15 a.
b. = 28.9216
c. 127.43 ± 1.645 () (118.5834, 136.2766)
d. [137.43 – 117.43]/2 = 10 = , solving for z: 1.86
tabled value of .9686 yields a confidence level of 93.72% or an of .0628
20.16 = .6936
7.28 ± 2.58 (.6936) (5.4904, 9.0696)
20.17 a. false: as n increases, the confidence interval becomes narrower for a given N and s2
b. true
c. true: the finite population correction factor is larger to account for the fact that a smaller proportion of the population is represented as N increases relative to n.
d. true
20.18
20.19 99% confidence interval:
where,
= 142.1167
1833.30
1466.6390 < < 2199.9610
20.20 95% confidence interval: using
, where
= 4,409.8619
104,492.6
95,849.2706 < < 113,135.9294
20.21 90% confidence interval:
, where
= 86.7054
910
767.3696 < < 1,052.6304
20.22 = 143/35 = 4.0857
90% confidence interval:
, where
= 52.9210
490.2857
403.2307 < < 577.3407
20.23 = x/n = 39/400 = .0975
= .0125
95% confidence interval: .0975 ± 1.96(.0125): .073 up to .1220
20.24 = 56/100 = .56
= .0435
90% confidence interval: .56 ± 1.645(.0435): .4884 up to .6316
20.25 = 37/120 = .3083
= .0309
95% confidence interval: .3083 ± 1.96(.0309): .2477 up to .3689
20.26 = 31/80 = .3875
= .0493
90% confidence interval: .3875 ± 1.645(.0493): .3064 up to .4686
128.688 < Np < 196.812 or between 129 and 197 students intend to take the final.
20.27 a. = = 342.089
b.
,
c. 95% confidence interval: 342.089 ± 1.96
335.609 up to 348.569
20.28 a. ,
90% confidence interval: 43.3 ± 1.645 : 40.806 up to 45.794
b. = = 37.3306
c. ,
90% confidence interval: 37.3306 ± 1.645 : 36.0313 up to 38.6299
95% confidence interval: 37.3306 ± 1.96 : 35.7825 up to 38.8787
20.29 a.
90% confidence interval: 2.5± 1.645 : 2.3102 up to 2.6897
b. = = 3.2387
c. ,
,
90% confidence interval: 3.2387 ± 1.645 : 3.1177 up to 3.3596
95% confidence interval: 3.2387 ± 1.96 : 3.0947 up to 3.3827
20.30 a. ; 3.12 ± 1.96 :
2.8435 up to 3.3965
b. ; 3.37 ± 1.96 : 3.1431 up to 3.5969
c. ; = 3.2339
3.2339 ± 1.96 : 3.0513 up to 3.4166
20.31 a. 90% confidence interval:
, where
= 315.3409
9006.4
8487.6642 < < 9525.1358
b. from Exercise 20-28: 37.3306,
(487)(37.3306) = 18,180.0022
90% confidence interval: 18,180.0022 ± 1.645(384.6620):
17,547.2332 < N < 18,812.7716
20.32 a. = 237(120) + 198(150) + 131(180) = 81,720
b. , ,
95% confidence interval: 181.6(450) ± 1.96(450) : 77,542.3153 < N < 85,897.6847
20.33 a.
b.
95% confidence interval: .1638 ± 1.96 : .0931 up to .2345
20.34 a.
b.
,
90% confidence interval: .3467 ± 1.645 : .2550 up to .4383
95% confidence interval: .3467 ± 1.96 : .2375 up to .4559
20.35 a.
b.
,
90% confidence interval: .7214 ± 1.645 : .6649 up to .7779 95% confidence interval: .7214 ± 1.96 : .6541 up to .7887
20.36 a. = 56 observations
b. = 68 observations
20.37 a. = 37 observations
b. = 32 observations
20.38 a. = 55 observations
b. = 60 observations
20.39 a. observations
b. = 52 observations
20.40 a. = 74 observations
b. = 88 observations
20.41 , = 262 observations
20.42 How large n?
= 57.988, take 58 observations
20.43
= 217 observations
20.44 , = 211 observations
20.45 Proportional allocation:
= 225 observations
Optimal allocation:
= 196 observations
20.46 Proportional allocation:
= 498 observations
Optimal allocation:
= 471 observations
20.47 a.
b.
90% confidence interval: 24.979 ± 1.645 : 22.8627 up to 27.0953
20.48 a.
b. = 66.409
99% confidence interval: 91.6761 ± 2.58
70.6920 up to 112.6602
20.49 a.
b.
90% confidence interval: .4636 ± 1.645 : .4069 up to .5203
20.50 a.
b.
95% confidence interval: .4507 ± 1.96: .38 up to .5214
20.51
90% confidence interval: .231 ± 1.645 : .2146 up to .2489
20.52 , = 127 observations.
Additional sample observations needed are 127-20 = 107
20.53
= 196
observations. Additional sample observations needed: 196-60 = 136
20.54
= 160
observations. Additional sample observations needed: 160-30 = 130
20.55 - 20.57 Discussion questions – various answers
20.58 a. , s = 11.44,
90% confidence interval: 74.7 ± 1.645 : 69.089 up to 80.311
b. The interval would be wider; the z-score would increase to 1.96
20.59 a.
99% confidence interval: 492.36 ± 2.58 :
442.9139 up to 541.7501
b. 95% confidence interval: 492.36 ± 1.96 :
454.8175 up to 529.9025
c. The 90% interval is narrower; the z-score would decline to 1.645
20.60 a. ,
90% confidence interval: .623 ± 1.645: .559 up to .687
b. If the sample information is not randomly selected, the resulting conclusions may be biased
20.61 ,
95% confidence interval: .6 ± 1.96: .504 up to .696
20.62 a.
, ,
99% confidence interval for managers in subdivision 1:
9.2 ± 2.575 : 6.997 up to 11.403
b. 99% confidence interval for all managers:
11.5845 ± 2.575: 9.8247 up to 13.3444
20.63 a.
,
95% confidence interval for mean number of errors per page in the book: 1.084 ± 1.96 : .8747 up to 1.2933
b.
where,
, 325.2
99% confidence interval for the total number of errors in the book:
242.4279 up to 407.9721 or from 243 total errors up to 408 total errors.
20.64
90% confidence interval is:
or 0.5147 up to 0.7453
20.65 a. = 30 observations
b. = 23 observations
20.66 a. observations
b. = 22 observations
20.67 Refer to section 20.4 – Stratified Random Sampling
20.68 , = 76 observations
20.69 , = 178 observations
20.70 Various answers. Answers should include a discussion of the potential for stratification of the population. Because different counties utilize different ballots and ballot techniques, a stratification by county may prove reasonable. The method used by the county could also be utilized in the stratification – e.g., butterfly ballots and hand-punched vs. electronic ballots.