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Unformatted text preview: Business Statistics
Fifth Edition Ken Black Chapter 8:
Statistical Inference:
Estimation for Single Populations These notes are not to be reproduced without the written permission of F. B. A/t Learning objectives for this chapter 0 Obtaining a point estimate of a population parameter 0 Obtaining a conﬁdence interval for a mean when population standard
deviation is known 0 Obtaining a conﬁdence interval for a mean when population standard
deviation is unknown 0 Obtaining a conﬁdence interval for a proportion 0 Determining the sample size required to estimate a mean 0 Determining the sample size required to estimate a proportion o Specifying the underlying assumptions for conﬁdence interval estimation These notes are not to be reproduced without the written permission of F. B. Alt 2 8.1 Estimating the Population Mean Using the Z Statistic (0' Known) 0 Point Estimation Concept [Black, page 255] 0 Examples of population parameters: 0 Population parameters are usually unknown. 0 Population parameters can be estimated by a statistic. 0 Rule of thumb for estimating population parameters: Use the sample counterpart S eciﬁc cases: Population
Parameter An estimate is the speciﬁc value obtained from the data. Example: Consider a bottling plant which speciﬁes that the average content
of its containers should be 16 02. A random sample of size 25 yielded the following output.
16.3425, 15.4072, 16.0421, 16.0696, 15.9998, , 15.2664
Descriptive Statistics: OUNCES Variable N Mean SE Mean StDev Median
OUNCES 25 15.961 0.0496 0.248 15.936 These notes are not to be reproduced without the written permission of F. B. Alt 3 o Interval Estimation of a Mean, Known Standard Deviation o A conﬁdence interval is a range of probable values for a parameter. 0 A conﬁdence interval has a conﬁdence level. Typical conﬁdence
levels: .95 or .99 or .90. I In general, the conﬁdence level is 1  a 0 Procedure for constructing a CI. for y (0' known) 1) Start with a random sample from a normal distribution. 2) Estimator for ,u is 3) _is normally distributed with mean [1 y : 1” and standard error 0'; = 0/\/; 4) Standardize the sample mean: X  rt aA/Z 5) Specify the conﬁdence level, say 95% 6) From Table A5, .95 = P[—1.96 < Z < +1.96] 7) Convert this statement about Z into a statement about the sample mean: z: X a/x/n— 8) Rearrange the quantity in brackets so that y is isolated .95 = P[X —1.96(a/\/Z) < ,u < 2? +1.96(a/JZ)] .95=P[1.96<Z: <+196] 0 Conﬁdence interval end points: Lower end point: X '1.96(0'/\/;) ; Upper end point: X +1.96(0'/\/;2) These notes are not to be reproduced without the written permission of F. B. Alt 4 o Shorthand expression for a 95 % conﬁdence interval for ,U : Yi1.96(a/\/Z) o Percentiles of the Zdistribution I a/ 2 denotes the area in the righthand tail. 95% :> Za/2 =Z_025 = 1.96 99% :> Za/2 =Z_()05 =2.575 0 General expression for a 100(1  a)% conﬁdence interval for ,U : f i za/2(0'/\/;z—) [Black, page 256] These notes are not to be reproduced without the written permission of F. B. A/t Examgle: The data below specify how much a sample of 20 executives paid in federal
income taxes, as a percentage of gross income: 16.0 18.1 18.6 20.2 21.7 22.4 22.4 23.1 23.2 23.5
24.1 24.3 24.7 25.2 25.9 26.3 27.9 28.0 30.4 33.7 [x = 23 .985]
Assume that the standard deviation for the underlying population is 4.0. a) Calculate a 95% conﬁdence interval for the population mean.
n = 20, 0' = 4%, 2025 = 1.96 Xiza/2(0'/x/n) 23.985 i 1.96(4.0/726) = [2223,2574] This is a 95% conﬁdence interval for the average income tax paid by all
executives. Minitab output for part (a):
OneSample Z: Tax(%)
The assumed stande deviation = 4
Variable N Mean StDev SE Mean 95% CI
Tax(%) 20 23.9850 4.1783 0.8944 (22.2320, 25.7380) b) Calculate a 99% conﬁdence interval for the population mean. 99% ‘9 251/2 =Z‘00522.575
23985 i 2.575(4.0A/26) = [21.68.2629] This is a 99% conﬁdence interval for the average income tax paid by all
executives. Minitab output for part (b):
OneSample Z: Tax(%)
The assumed standard deviation = 4
Variable N Mean StDev SE Mean 99% CI
Tax(%) 20 23.9850 4.1783 0.8944 (21.6811, 26.2889) These notes are not to be reproduced without the written permission of F. B. Alt 6 Example:
0) Does the conﬁdence interval [22.23, 25.74] contain ,u ? d) Give a careful verbal interpretation of the conﬁdence interval in part (a). 25.74 Sample No. 1
Samnle No. 2 Sample No. 3 Real number line Trueu=? o Assumptions necessary to use the Z  interval: I If the data came from a population Where X is normally distributed, then 2? is normally distributed for sample size. I If the data came from a population where the distribution of X is not normal, the sampling distribution of A7 is nearly normal as long as the sample size is large enough because of the I The most important assumption is that These notes are not to be reproduced without the written permission of F. 3. Alt 7 e) Is it reasonable to conclude that the data came from a normal distribution? Probability Plotof TaxP/a)
Normal til 95
9B
39
3'6
66
Si}
46
38
29 H
m 0 W Since the NPP (is, is not) linear, it (is, is not ) reasonable to conclude that the data came from a normal distribution. Suppose the data did not come from a normal distribution. Unless the distribution of the population is markedly skewed, a
sample of size 20 should be large enough for the CLT to apply. 0 Procedure to obtain Z—interval using Minitab
Click on Stat > Basic Statistics > lsample;
In “Samples in Column” box, enter column where data is stored;
In “Standard deviation” box, enter 4.0;
Click on “Options” and enter “95.0” in “Conﬁdence Level” box; Click on “OK” These notes are not to be reproduced without the written permission of F. B. Alt 8 8.2 Estimating the Population Mean Using the t Statistic (0 Unknown)
0 Recall 2:11 0'/\/; has a standard normal distribution. 0 Gosset (pseudonym: Student) determined the distribution of Z when “s” is used as an estimate of 0' : X—# t:— s/x/ii 0 The sample standard deviation S = \/S72 has (nl) degrees of freedom.
0 The probability distribution of t is shown below. t distribution with a Normal
Distribution superimposed _____ Normal These notes are not to be reproduced without the written permission of F. 8. Alt 9 0 Characteristics of Student’s t Distribution [Black, page 263] o The t distribution is symmetric about 0. o The t distribution has heavier tails. I The t random variable has 2 sources of variation: 0 There are many different t distributions. I We specify a particular one by its “degrees of freedom,” d.f. and 0 As 11 increases, the distribution oft approaches the distribution of a standard normal. 0 Reading the t Distribution Table (Table A.6) df a = .1 a = .05 a = .025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
8 1.397 1.860 2.306
9 1.383 1.833 2.262
10 1.372 1.812 2.228 For example, with n=10, the d.f. = 9, and P(t9 > )= .025 It is customary to say ____ = £0259 a = .01
31.821
6.965 2.896
2.821
2.764 a = .005
63.657
9.925 O 3.355
3.250
3.169 These notes are not to be reproduced without the written permission of F. B. A/t a = .001
318.309
22.327 4.501
4.297
4.144 10 0 Conﬁdence Intervals to Estimate the Population Mean Using the t Statistic [Black, page 264] 0 When 0' is known, the CI. for ,U is given by [l7 i 201/2 (0 / J?)
0 When 0' is unknown, it seems reasonable to replace 0' by s. Yiza/z (s/x/n) 0 Also need to replace ZM by to“2 I? i am (W?) I The expression for a 100(1(1) % conﬁdence interval for u (6 unknown). . If 0' is unknown, always use this expression regardless of the sample size. 0 Requirements: Random Sample from a Normal Distribution. These notes are not to be reproduced without the written permission of F. 8. Alt 11 Examgle: A random sample of 20 tastetesters rates the quality of a proposed new
product on a 0100 scale. The ordered scores are: 16 23 31 35 40 43 48 52 53 55 57 59 60 60 61 65 67 72 81 92 Descriptive Statistics: Scores
Variable N N* Mean SE Mean StDev Minimum Median Maximum
Scores 20 0 53.50 4.16 18.63 16.00 56.00 92.00 a) Find the 95% conﬁdence interval for the population mean score. n=20, A7 = , s= , ta/2,n—l
A100(1~ a)% CI. for ,u is:
— s
X ita/2,n—l 7—— =
n
[ , ] is a 95% CI. for the mean score of all tasters
The Minitab output follows:
OneSample T: Scores
Variable N Mean StDev SE Mean 95% CI
Scores 20 53.50 18.63 4.16 (44.78, 62.22) b) In the past, the company has found that any product whose average taste score was not at least 70 was not successﬁJl. Should the company put the product on
the market ? These notes are not to be reproduced without the written permission of F. B. Alt 12 c) Were the assumptions met to construct a t—interval? Do we have a random sample? Did the data come from a normally distributed population? Probability Plot of Scores $ New 53. 5
stew 18.53
N 23
AB 5. 22?
[Waive 05285 95
5'6
8‘3
X?
60
59
4Q
30
20 0 Using the Computer to Construct t Conﬁdence Intervals for the Mean Click on Stat ) Basic Statistics ) 1—Sample t;
Enter variable in “Samples in Column” box;
Click on Options; Enter .95 for conﬁdence level; Click on OK. These notes are not to be reproduced without the written permission of F. B. Alt 13 8.3 Estimating the Population Proportion [Black, page 269] is approximately standard normal. [Page 710] 0 By pivoting on the above expression and simplifying, a 100(1 — a )% CI.
for p is obtained: P i Za/2 o This expression is based on the premise that a sample proportion can be approximated by a normal random variable. 0 In order to say sample proportions are approximately normal, we require: i) np 2 9, and
ii) nq 2 9 i) np 2 5, and
ii) nq 2 5 o This is okay when 0.4 < p < 0.6 0 To use the conﬁdence interval expression for p, the above two conditions must be satisﬁed. These notes are not to be reproduced Without the written permission of F. 8. Alt 14 Examgle: As part of a market research study, in a sample of 125, 84 individuals are
aware of a certain product. a) Calculate a 90% conﬁdence interval for the proportion of individuals in
the population who are aware of the product. p = Proportion of individuals in the population who are aware of product. /\ n = 125, XB =84, p = = 0.672 igi Za/2 V120 ‘27)”?
(.672) i (1 .645)‘/(.672)(.328)/125 =[ a l This is a 90% conﬁdence interval for the population proportion who are
aware of the product. 0 When would such a product awareness study be undertaken?
One possibility would be prior to the start of an advertising campaign.
Such a study would also be undertaken after the advertising campaign to determine the effectiveness of the advertising campaign 0 Minitab Output
Test and CI for One Proportion Sample X N Samplep 90% CI
1 84 125 0.672000 (0.602929, 0.741071) These notes are not to be reproduced without the written permission of F. B. Alt 15 b) Was the normal approximation underlying the conﬁdence interval justiﬁed? Conditions for using the normal approximation: np 2 9 and nq 2 9 Is (125)(84/125) 2 9 ? Is (125)(41/125) 2 9 ? The conditions required to use the expression for a conﬁdence interval for p based on the normal approximation (are, are not) satisﬁed. 0 Using the Computer to Construct Conﬁdence Intervals for a Proportion Using Minitab Click on Stat 9 Basic Statistics '9 1 Proportion;
Select "Summarized Data" and enter 125 and 84 for "Number of Trials" and "Number of Successes."
Click on "Options" and enter 90.0 for the “Conﬁdence level. Choose "Tests and interval based on normal distribution." Click on “OK” These notes are not to be reproduced without the written permission of F. B. Alt 16 8.5 Estimating Sample Size [Black, page 277] 0 General format of a conﬁdence interval:
Point Estimate i Margin of Error
0 Choosing an appropriate sample size controls the magnitude of the margin of error. 0 Sample Size When Estimating ,U Example:
An insurance company wishes to investigate the mean value of the personal property held by urban apartment renters. A previous study suggested that
the population standard deviation should be roughly $5,000. You want to be
95% conﬁdent that your estimate is within $2,000 of the actual ﬁgure. How large a sample must be taken ? 0 Find the sample size (n) so that the bound on the error of estimation (B)
will hold with a high probability (1  a ).
o FrompagegizaQUT/Vn) o E = za/2(0'/\/;) _ 2 2 2
o n—za/20' /E 0 Need to specify E, l  a and 0' to ﬁnd 11. These notes are not to be reproduced without the written permission of F. 8. Alt 17 Examgle:
An insurance company wishes to investigate the mean value of the personal property held by urban apartment renters. A previous study suggested that
the population standard deviation should be roughly $5,000. You want to be
95% conﬁdent that your estimate is within $2,000 of the actual ﬁgure. How large a sample must be taken ? E4; ,1—(1: ,6: _ 2 2 2
”—20:00 /E 0 Finding 3 Rough Estimate of 0' (Sometimes!) Examgle: A researcher wants to determine the sample size necessary to conduct a
study to estimate the population mean to within 5 units of the actual mean.
The range of population values is 80 and the researcher plans to use a 95%
level of conﬁdence. What should t mple size be ? I To estimate 6, use Range 4 !! Why? Exalee: Estimate of o = E] _ 2 2 2
”—26:00 /E = These notes are not to be reproduced without the written permission of F. B. Alt 18 0 Sample Size When Estimating p Examgle:
An electrical utility offers reduced rates to homeowners who have installed “peak hours” meters. These meters effectively shut off highconsumption
electrical appliances (primarily dishwashers and clothes dryers) during the
peak electrical usage hours between 9 am. and 3 pm. daily. The utility
wants to inspect a sample of these meters to determine the proportion that are not working, either because they were bypassed or because of equipment
failure. There are 45,300 meters in use and the utility isn’t about to inspect them all. How large a sample s the utility take if it wants to estimate the true
proportion to within i .02 with 95% conﬁdence. a) How many meters must be sampled, if one can make no particular assumption
about the true proportion? b) How many meters must be sampled if the utility believes that the true
proportion is 0.15? o From page 14, 0 Good news! We have an expression to ﬁnd 11. 0 Bad news! The expression depends on p, which we are trying to ﬁnd. These notes are not to be reproduced without the written permission of F. B. Alt 19 > Approach 1 (Worse case scenario): Set p = 1/2. (Z a/2_____)_____2 Pq_ n: _. E2 Example:
a) How many meters must be sampled, if one can make no particular assumption
about the true proportion? (Z 0) 1’1: E2 (4): > Approach 2: Use a prior estimate of p, denoted p0, if available. n_ (Za a/i:Poq0
Example: b) How many meters must be sampled if the utility believes that the true
proportion is 0.15? 11— (Z a/2)2 Pogo : E2 These notes are not to be reproduced without the written permission of F. 8. Alt 20 Summary of Chapter 8
0 Inductive inference — estimating a popul ' parameter
0 How to obtain a point estimate of a population parameter How to obtain and use a conﬁdence interval estimate for the mean when a is speciﬁed How to obtain and use a conﬁdence interval estimate for the mean when a is unknown How to obtain and use a conﬁdence interval estimate for a proportion 0 How to check the underlying assumptions for conﬁdence interval
estimation 0 How to determine the sample size required to estimate a mean 0 How to determine the sample size required to estimate a proportion o A ﬂow chart to assist in using the correct conﬁdence interval follows. These notes are not to be reproduced without the written permission of EB. Alt 21 ...
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