Statistics for Business and Economics chapter 07 Sampling and Sampling Distributions

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 dev

Sampling and Sampling Distributions

Learning Objectives

1 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:

sample statistic finite population correction factor

sampling without replacement central limit theorem

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

7 108, 290, 201, 292, 322, 9, 244, 249, 226, 125, (continuing at the top of column 9) 147, and 113

8 Random numbers used: 13, 8, 27, 23, 25, 18

The second occurrence of the random number 13 is ignored

Companies selected: ExxonMobil, Chevron, Travelers, Microsoft, Pfizer, and Intel

9 102, 115, 122, 290, 447, 351, 157, 498, 55, 165, 528, 25

10 a Finite population A frame could be constructed obtaining a list of licensed drivers from the New

York State driver’s license bureau

b Infinite population Sampling from a process The process is the production line producing boxes

18 .4540

p  

b Six of the 40 funds in the sample are high risk funds Our point estimate is

6.1540

p  

c The below average fund ratings are low and very low Twelve of the funds have a rating of low and 6 have a rating of very low Our point estimate is

18.4540

18 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 with

The normal distribution forxis 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

51,800

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 With a sample of size 60 4000 516.40

z   P(z < -.95) = 1711

probability = 8289 - 1711 =.6578

The probability of being within 10 of the mean on the Mathematics portion of the test is exactly the same as the probability of being within 10 on the Critical Reading portion of the SAT This is because the standard error is the same in both cases The fact that the means differ does not affect the probability calculation

c x / n100 / 100 10.0 The standard error is smaller here because the sample size is larger

504 494

1.0010.0

z   P(z ≤ 1.00) = 8413

484 494

1.0010.0

z   P(z < -1.00) = 1587

probability = 8413 - 1587 =.6826

The probability is larger here than it is in parts (a) and (b) because the larger sample size has madethe standard error smaller

Within 25 means x- 939 must be between -25 and +25.

The z value for x- 939 = -25 is just the negative of the z value for x- 939 = 25 So we just show

the computation of z for x- 939 = 25

b A larger sample increases the probability that the sample mean will be within a specified

distance of the population mean In the automobile insurance example, the probability of

being within 25 of  ranges from 4246 for a sample of size 30 to 9586 for a sample of

c In part (b) we have a higher probability of obtaining a sample mean within $10,000 of the

population mean because the standard error is smaller

1.442.09

P(103  x 109) = P(-1.44 ≤ z ≤ 1.44) = 9251 - 0749 = 8502

The probability the sample means will be within 3 strokes of the population mean of 106 is 8502

d The probability of being within 3 strokes for female golfers is higher because the sample size is larger

29  = 2.34  = 20

a n = 30

.03

.82/ 20 / 30

.03 1.06/ 20 / 50

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

.031.96

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

P(z < -1.54) = 0618

Probability = 9382 - 0618 = 8764

31 a E( p ) = p = 40

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

(1 ) 40(.60)

.0346200

p

(.55)(.45) .0352200

p

(.55)(.45)

.0222500

p

(.55)(.45) .01571000

p

The standard error of the proportion,p,decreases as n increases

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  p  40) = ?

.40 30

2.18.0458

E( p ) = 56

(1 ) (.56)(.44)

.0248400

e The probability of the sample proportion being within 04 of the population mean was reduced from 9500 to 8098 So there is a gain in precision by increasing the sample size from 200 to 450

If the extra cost of using the larger sample size is not too great, we should probably do so

40 a E ( p ) = 76

(1 ) 76(1 76)

.0214400

z   P(z ≤ 2.13) = 9834

Author’s note: The universities identified are: Clarkson U (122), U of Arizona (99), UCLA (25),

U of Maryland (55), U of New Hampshire (115), Florida State U (102), Clemson U (61)

43 a Normal distribution because n = 50

E( x) = 6883

2000 282.8450

P(z < -1.05) = 1469

P(26,175  x 28,175) = P(-1.05  z  1.05) = 8531 - 1469 = 7062

d x 7400 / 100 740

10001.35740

Note: With n / N  05 for all three cases, common statistical practice would be to ignore

the finite population correction factor and use x 144 

50 20 36 for each case

b N = 2000

25 1.2420.11

z   P(z ≤ 1.24) = 8925

P(z < -1.24) = 1075

Probability = P(-1.24  z  1.24) = 8925 - 1075 = 7850

N = 5000

251.2320.26

.30 25 .80

.0625

z   P(z ≤ 80) = 7881

P ( p  30) = 1 - 7881 = 2119