stach8

8/6/2019 stach8

1/32

Chapter 8: Statistical Inference: Estimation for Single Populations 1

Chapter 8Statistical Inference: Estimation for Single

Populations

LEARNING OBJECTIVES

The overall learning objective of Chapter 8 is to help youunderstand estimating

parameters of single populations, thereby enabling you to:

1. Know the difference between point and interval estimation.

2. Estimate a population mean from a sample mean when is known.

3. Estimate a population mean from a sample mean when is unknown.

4. Estimate a population proportion from a sample proportion.

5. Estimate the population variance from a sample variance.

6. Estimate the minimum sample size necessary to achievegiven statistical goals.

8/6/2019 stach8

2/32


CHAPTER TEACHING STRATEGY

Chapter 8 is the student's introduction to intervalestimation and estimation of sample size. In this chapter, theconcept of point estimate is discussed along with the notion that

as each sample changes in all likelihood so will the pointestimate. From this, the student can see that an intervalestimate may be more usable as a one-time proposition than thepoint estimate. The confidence interval formulas for large sample meansand proportions can be presented as mere algebraic manipulations of formulasdeveloped in chapter 7 from the Central Limit Theorem.

It is very important that students begin to understand the differencebetween mean and proportions. Means can be generated by averaging some sortof measurable item such as age, sales, volume, test score, etc. Proportions arecomputed by counting the number of items containing a characteristic of interestout of the total number of items. Examples might be proportion of people

carrying a VISA card, proportion of items that are defective, proportion of marketpurchasing brand A. In addition, students can begin to see that sometimes singlesamples are taken and analyzed; but that other times, two samples are taken inorder to compare two brands, two techniques, two conditions, male/female, etc.

In an effort to understand the impact of variables on confidence intervals,it may be useful to ask the students what would happen to a confidence interval ifthe sample size is varied or the confidence is increased or decreased. Suchconsideration helps the student see in a different light the items that make up aconfidence interval. The student can see that increasing the sample size, reducesthe width of the confidence interval all other things being constant or that itincreases confidence if other things are held constant. Business students probably

understand that increasing sample size costs more and thus there are trade-offs inthe research set-up.In addition, it is probably worthwhile to have some discussion with

students regarding the meaning of confidence, say 95%. The idea is presented inthe chapter that if 100 samples are randomly taken from a population and 95%confidence intervals are computed on each sample, that 95%(100) or 95 intervalsshould contain the parameter of estimation and approximately 5 will not. In mostcases, only one confidence interval is computed, not 100, so the 95% confidenceputs the odds in the researcher's favor. It should be pointed out, however, that theconfidence interval computed may not contain the parameter of interest.

This chapter introduces the student to the tdistribution to estimate

population means from small samples when is unknown. Emphasize that thisapplies only when the population is normally distributed. The student willobserve that the tformula is essentially the same as thezformula and that it is the

table that is different. When the population is normally distributed and isknown, the z formula can be used even for small samples. In addition, note that

some business researchers always prefer to use the tdistribution when isunknown.

8/6/2019 stach8

3/32


A formula is given in chapter 8 for estimating the population variance.Here the student is introduced to the chi-square distribution. An assumptionunderlying the use of this technique is that the population is normally distributed.

The use of the chi-square statistic to estimate the population variance is extremelysensitive to violations of this assumption. For this reason, exercise extremecaution is using this technique. Some statisticians omit this technique fromconsideration.

Lastly, this chapter contains a section on the estimation of sample size.One of the more common questions asked of statisticians is: "How large of asample size should I take?" In this section, it should be emphasized that samplesize estimation gives the researcher a "ball park" figure as to how many to sample.The error of estimation is a measure of the sampling error. It is also equal tothe + error of the interval shown earlier in the chapter.

CHAPTER OUTLINE

8.1 Estimating the Population Mean Using the zStatistic.

Finite Correction Factor

Confidence Interval to Estimate When isUnknown

Confidence Interval to Estimate When thePopulation Standard

Deviation is Unknown and n is Large.

8.2 Estimating the Population Mean Using the tStatistic.

The tDistribution

Robustness

Characteristics of the tDistribution.

Reading the tDistribution Table

Confidence Intervals to Estimate When isUnknown and

8/6/2019 stach8

4/32


Sample Size is Small

8.3 Estimating the Population Proportion

8.4 Estimating the Population Variance

8.5 Estimating Sample Size

Sample Size When Estimating

Determining Sample Size When Estimating p

KEY WORDS

Bounds Point EstimateChi-square Distribution RobustDegrees of Freedom(df) Sample-Size EstimationError of Estimation tDistributionInterval Estimate tValue

8/6/2019 stach8

5/32


SOLUTIONS TO PROBLEMS IN CHAPTER 8

8.1 a) x = 25 = 3.5 n = 60

95% Confidence z.025 = 1.96

x + zn

= 25 + 1.9660

5.3= 25 + 0.89 = 24.11 < < 25.89

b) x = 119.6 s = 23.89 n = 75

98% Confidence z.01 = 2.33

x + zn

s= 119.6 + 2.33

75

89.2= 119.6 6.43 = 113.17

8/6/2019 stach8

6/32


8.2 n = 36 x = 211 = 2395% C.I. z.025 = 1.96

x zn

= 211 1.96362 = 211 7.51 = 203.49 < < 218.51

8.3 n = 81 x = 47 s = 5.8990% C.I. z.05=1.645

x zn

s= 47 1.645

81

89.5= 47 1.08 = 45.92 < < 48.08

8.4 n = 70 2 = 49 x = 90.4

x = 90.4 Point Estimate

94% C.I. z.03 = 1.88

x + zn

= 90.4 1.8870

49= 90.4 1.57 = 88.83 < < 91.97

8.5 n = 39 N= 200 x = 66 s = 11

96% C.I. z.02 = 2.05

x z1

N

nN

n

s= 66 2.05

1200

39200

9

11

=

66 3.25 = 62.75 < < 69.25

x = 66 Point Estimate

8.6 n = 120 x = 18.72 s = 0.873599% C.I. z.005 = 2.575

8/6/2019 stach8

7/32



x + zn

s= 18.72 2.575

120

8735.0= 8.72 .21 = 18.51 < < 18.93

8.7 N= 1500 n = 187 x = 5.3 years s = 1.28 years95% C.I. z.025 = 1.96

x = 5.3 years Point Estimate

x z1

N

nN

n

s= 5.3 1.96

11500

1871500

187

28.1

=

5.3 .17 = 5.13 < < 5.47

8.8 n = 32 x = 5.656 s = 3.22990% C.I. z.05 = 1.645

x zn

s= 5.656 1.645

32

229.3= 5.656 .939 = 4.717 < < 6.595

8.9 n = 36 x = 3.306 s = 1.16798% C.I. z.01 = 2.33

x zn

s= 3.306 2.33

36

167.1= 3.306 .453 = 2.853 < < 3.759

8/6/2019 stach8

8/32


8.10 n = 36 x = 2.139 s = .113


90% C.I. z.05 = 1.645

x zn

s= 2.139 1.645

36

)113(.= 2.139 .03 = 2.109 < < 2.169

8.11 = 27.4 95% confidence interval n = 45

x = 24.533 s = 5.1239

z= + 1.96

Confidence interval: x + zn

s= 24.533 + 1.96

45

1239.5=

24.533 + 1.497 = 23.036 < < 26.030

8.12 The point estimate is 0.5765. n = 41

The assumed standard deviation is 0.1394

99% level of confidence: z= + 1.96

Confidence interval: 0.5336

8/6/2019 stach8

9/32


8.13 n = 13 x = 45.62 s = 5.694 df = 13 1 = 12

95% Confidence Interval

/2=.025

t.025,12 = 2.179

n

stx = 45.62 2.179

13

694.5= 45.62 3.44 = 42.18

8/6/2019 stach8

10/32


8.16 n = 15 x = 2.364 s2 = 0.81 df = 15 1 = 14

90% Confidence interval

/2=.05

t.05,14 = 1.761

n

stx = 2.364 1.761

15

81.0= 2.364 .409 = 1.955

8/6/2019 stach8

11/32


8.19 n = 20 df = 19 95% CI t.025,19 = 2.093

x = 2.36116 s = 0.19721

2.36116 + 2.093201972.0 = 2.36116 + 0.0923 = 2.26886 < < 2.45346

Point Estimate = 2.36116

Error = 0.0923

8.20 n = 28 x = 5.335 s = 2.016 df = 28 1 = 27

90% Confidence Interval /2=.05

t.05,27 = 1.703

n

stx = 5.335 1.703

28

016.2= 5.335 + .649 = 4.686

8/6/2019 stach8

12/32


8.22 n = 14, 98% confidence, /2 = .01, df = 13

t.01,13 = 2.650

from data: x = 152.16 s = 14.42

confidence interval:n

stx = 152.16 + 2.65

14

42.14=

152.16 + 10.21 = 141.95 < < 162.37

The point estimate is 152.16

8.23 a) n = 44 p =.51 99% C.I. z.005 = 2.575

n

qpzp

= .51 2.575

44

)49)(.51(.= .51 .194 = .316

8/6/2019 stach8

13/32


8.24 a) n = 116 x = 57 99% C.I. z.005 = 2.575

p =116

57=

n

x= .49

n

qpzp

= .49 2.575

116

)51)(.49(.= .49 .12 = .37 < p < .61

b) n = 800 x = 479 97% C.I. z.015 = 2.17

p =800

479=

n

x= .60

n

qpzp

= .60 2.17800

)40)(.60(. = .60 .038 = .562 < p < .638

c) n = 240 x = 106 85% C.I. z.075 = 1.44

p =240

106=

n

x= .44

n

qpzp

= .44 1.44

240

)56)(.44(.= .44 .046 = .394 < p < .486

d) n = 60 x = 21 90% C.I. z.05 = 1.645

p =60

21=

n

x= .35

n

qpzp

= .35 1.645

60

)65)(.35(.= .35 .10 = .25 < p < .45

8/6/2019 stach8

14/32


8.25 n = 85 x = 40 90% C.I. z.05 = 1.645

p =

85

40=

n

x = .47

n

qpzp

= .47 1.645

85

)53)(.47(.= .47 .09 = .38 < p < .56

95% C.I. z.025 = 1.96

n

qpzp

= .47 1.96

85

)53)(.47(.= .47 .106 = .364 < p < .576

99% C.I. z.005 = 2.575

n

qpzp

= .47 2.575

85

)53)(.47(.= .47 .14 = .33 < p < .61

All things being constant, as the confidence increased, the width of the intervalincreased.

8.26 n = 1003 p = .245 99% CI z.005 = 2.575

n

qpzp

= .245 + 2.575

1003

)755)(.245(.= .245 + .035 = .21

8/6/2019 stach8

15/32


8.27 n = 560 p = .47 95% CI z.025 = 1.96

n

qpzp

= .47 + 1.96

560

)53)(.47(.= .47 + .0413 = .4287

8/6/2019 stach8

16/32


p =89

48=

n

x= .54

n

qpzp

= .54 1.44

89

)46)(.54(.= .54 .076 = .464 < p < .616

8.31 p = .63 n = 672 95% Confidence z = + 1.96

n

qpzp

= .63 + 1.96

672

)37)(.63(.= .63 + .0365 = .5935

8/6/2019 stach8

17/32


645.45 < 2 < 1923.10

d) n = 17 s2 = 18.56 80% C.I. df = 17 1 = 16

2.90,16 = 9.31223

2.10,16 = 23.5418

5418.23

)56.18)(117( < 2

stach8

Documents