SBE_SM07

Chapter 7

Chapter 7

Sampling and Sampling Distributions

Chapter 7

Sampling and Sampling Distributions

Learning 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 () and the sampling distribution of the sample proportion ().

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

consistency

Solutions: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, 209

Numbers 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, 348

4.a.6, 8, 5, 4, 1

Nasdaq 100, Oracle, Microsoft, Lucent, Applied Materials

b.

5.

283, 610, 39, 254, 568, 353, 602, 421, 638, 164

6.

2782, 493, 825, 1807, 289

7.

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, 5

The second occurrences of random numbers 13 and 25 are ignored.

Maryland, Iowa, Florida State, Virginia, Pittsburgh, Oklahoma

9.

102, 115, 122, 290, 447, 351, 157, 498, 55, 165, 528, 25

10.

finite, infinite, infinite, infinite, finite

11.a.

b.

= (-4)2 + (-1)2 + 12 (-2)2 + 12 + 52 = 48

s =

12.a.

= 75/150 = .50

b.

= 55/150 = .3667

13.a.

b.

94+11

100+749

85-864

94+11

92-1 1

Totals4650116

14.a.149/784 = .19

b.251/784 = .32

c.Total receiving cash = 149 + 219 + 251 = 619

619/784 = .79

15.a.

years

b.

years

16.

= 1117/1400 = .80

17.a.595/1008 = .59

b.332/1008 = .33

c.81/1008 = .08

18.a.

b.

c.Normal with E () = 200 and = 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 is within a specified from 19.a.The sampling distribution is normal with

E () = = 200

For 5, (- ) = 5

Area = .3413

2(.3413) = .6826

b.For 10, (- ) = 10

Area = .4772

2(.4772) = .9544

20.

The standard error of the mean decreases as the sample size increases.

21.a.

b.n / N = 50 / 50,000 = .001

Use

c.n / N = 50 / 5000 = .01

Use

d.n / N = 50 / 500 = .10

Note: Only case (d) where n /N = .10 requires the use of the finite population correction factor.

22.a.

The normal distribution is based on the Central Limit Theorem.

b.For n = 120, E () remains $51,800 and the sampling distribution of can still be approximated by a normal distribution. However, is reduced to = 365.15.

c.As the sample size is increased, the standard error of the mean, , 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, should decrease as the sample size is increased.

23.a.

Area = .3340

2(.3340) = .6680

b.

Area = .4147

2(.4147) = .8294

24.a.Normal distribution,

b.Within $250

P(4010 (( 4510)

Area = .4750

2(.4750) = .95

c.Within $100

P(4160 ( ( 4360)

Area = .2852

2(.2852) = .5704

25.a.E() = 1020

b.

Area =.3078

Area =.3078

probability = .3078 + .3078 =.6156

c.

Area = .4582

Area = .4582

probability = .4582 + .4582 = .9164

26.a.

Error = - 34,000 = 250

n = 30

2(.2517) = .5034

n = 50

2(.3106) = .6212

n = 100

2(.3944) = .7888

n = 200

2(.4616) = .9232

n = 400

2(.4938) = .9876

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.

= 10994

Area = .4987

2(.4987) = .9974

b.

Area = .2734

2(.2734) = .5468

c. The sample with n = 40 has a very high probability (.9974) of providing a sample mean within ( NT$1,000. However, the sample with n = 40 only has a .5468 probability of providing a sample mean within ( NT$250. A larger sample size is desirable if the ( N$250 is needed.

28.a.

= 3909

b.Within NT$570

P(3339 ( ( 4476)

Area = .4982

2(.4982) = .9964

c.Within NT$142

P(3980 ( ( 3838)

Area = .2673

2(.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

P(1.43 ( ( 1.49) = P(-1.10 ( z ( 1.10) = 2(.3643) = .7286

b.n = 50

P(1.43 ( ( 1.49) = P(-1.41 ( z ( 1.41) = 2(.4207) = .8414

c.n = 100

P(1.43 ( ( 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 factor

Without the finite population correction factor

Including the finite population correction factor provides only a slightly different value for than when the correction factor is not used.

c.

Area = .4382

2(.4382) = .8764

31.a.E () = p = .40

b.

c.Normal distribution with E () = .40 and = .0490

d.It shows the probability distribution for the sample proportion .

32.a.E () = .40

Area = .3078

2(.3078) = .6156

b.

Area = .4251

2(.4251) = .8502

33.

decreases as n increases

34.a.

Area = 2(.3078) = .6156

b.

Area = 2(.3907) = .7814

c.

Area = 2(.4744) = 0.9488

d.

Area = 2(.4971) = .9942

e.With a larger sample, there is a higher probability will be within .04 of the population proportion p.

35.a. The normal distribution is appropriate because np = 100(.30) = 30 and n(1 - p) = 100(.70) = 70 are both greater than 5.

b.P (.20 .40) = ?

Area = .4854

2(.4854) = .9708

c.P (.25 .35) = ?

Area = .3621

2(.3621) = .7242

36.a.Normal distribution,= .49

b.Within .03

Area = .3485

2(.3485) = .6970

c.For n = 600,

Area = .4292

2(.4292) = .8584

For n = 1000,

Area = .4713

2(.4713) = .942637.a.Normal distribution

E () = .50

b.

2(.4738) = .9476

c.

2(.4279) = .8558

d.

2(.3340) = .6680

38.a.Normal distribution

E() = .25

b.

Area = .3365

2(.3365) = .6730

c.

Area = .4484

2(.4484) = .8968

39.a.Normal distribution with E() = p = .25 and

b.

P(.22 ( ( .28) = P(-2.19 ( z ( 2.19) = 2(.4857) = .9714

c.

P(.22 ( ( .28) = P(-1.55 ( z ( 1.55) = 2(.4394) = .8788

40.a.E () = .76

Normal distribution because np = 400(.76) = 304 and n(1 - p) = 400(.24) = 96

b.

Area = .4192

Area = .4192

probability = .4192 + .4192 = .8384

c.

Area = .4726

Area = .4726

probability = .4726 + .4726 = .9452

41.a.E() = .25

b.

Area = .4049

Area = .4049

probability = .4049 + .4049 = .8098

c.

Area = .4678

Area = .4678

probability = .4678 + .4678 = .935642.

112, 145, 73, 324, 293, 875, 318, 618


E() = 3

b.

Area = .4292

2(.4292) = .8584


E() = 31.5

b.

Area = .2224

2(.2224) = .4448

c.

Area = .4616

2(.4616) = .9232

45.a.E() = 34944

Area = .3729

2(.3729) = .7458

b.

Area = .4887

2(.4887) = .9774

46.

( = 56953 ( = 16104

a.

b.

P(>56953) = P(z > 0) = .50

c.

P(53753 (( 60153) = P(-1.41 ( z ( 1.41) = 2(.4207) = .8414

d.

P(53,753((60,153) = P(-1.99 ( z ( 1.99) = 2(.4767) = .9534

47.a.

N = 2000

N = 5000

N = 10,000

Note: With n / N .05 for all three cases, common statistical practice would be to ignore the finite population correction factor and use for each case.

b.N = 2000

Area = .3925

2(.3925) = .7850

N = 5000

Area = .3907

2(.3907) = .7814

N = 10,000

Area = .3907

2(.3907) = .7814

All probabilities are approximately .78

48.a.

= 500/20 = 25 and n = (25)2 = 625

b.For 25,

Area = .3944

2(.3944) = .7888

49.

Sampling distribution of

The area between = 2 and 2.1 must be .45. An area of .45 in the standard normal table shows

z = 1.645.

Thus,

Solve for

50.

p = .305

a.Normal distribution with E() = p = .305 and

b.

P(.265 ( ( .345) = P(-1.23 ( z ( 1.23) = 2(.3907) = .7814

c.

P(.285 ( ( .325) = P(-.61 ( z ( .61) = 2(.2291) = .4582

51.

P ( .375) = ?

Area = .3461

P ( .375) = .3461 + .5000 = .8461

52.a.

Area = .4726

2(.4726) = .9452

b.

Area = .4370

P ( .65) = .5000 - .4370 = .0063

53.a.Normal distribution with E () = .15 and

b.P (.12 .18) = ?

Area = .3485

2(.3485) = .6970

54.a.

Solve for n

b.Normal distribution with E() = .25 and = .0625

c.P ( .30) = ?

Area = .2881

P ( .30) = .5000 - .2881 = .2119

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