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Unformatted text preview: W CHAPTER q_ Hygomezsrs TESTS (ONE goﬁuMTzoN) W E ALII‘Dmoﬁ'uc gradud" I‘exeargé $12.91.” Wanbé f0 fave Mm” Me
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APP ROACH STEPS OF HYPOTHESIS TESTING Step 1. Develop the null and alternative hypotheses. ‘
Step 2. Specify the level of Signiﬁcance
Step 3. Collect the sample data and compute the value of the test statistic. i pValue Approach Step 4. Use the value of the test statistic to compute the pivalue.
Step 5. Reject H0 if the pvalue E a. Critical Value Approach Step 4. Use the level of significance to determine the critical value and the re jection rule
Step 5. Use the value of the test statistic and the rejection rule to determine whether to reject H” l
l HYPOTHESIS‘ T557: {330m Popuznwow HEAN{,u) t DISTRIBUTION H VP 0TH 6555 TEST
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Do 10.3 lnferences About the Difference Between Two
Population Means: Matched Samples Suppose employees at a manufacturing company can use two different methods to perform
a production task. To maximize production output, the company wants to identify the
method with the smaller population mean completion time. Let it, denote the population
mean completion time for production method 1 and lit: denote the population mean com?
pletion time for production method 2. With no preliminary indication of the preferred prtt
duction method. we begin by tentatively assuming that the two production methods have
the same population mean completion time. Thus. the null hypothesis is HG: ﬂl i it: 3 or
If this hypothesis is rejected, we can conclude that the population mean completion timES
differ. in this case, the method prm'iding the smaller mean completion time would be fCC'
ommended. The null and alternative hypotheses are written as follows. Hugo, — ‘11:, i 0
Ham”! 7 in: #5 O In choosing the sampling procedure that will he used to collect production time data and
test the hypotheses, we consider two alternative designs One is based on independent 53"“
ples and the other is based on matched samples. 1. Independent sample design: A simple random sample of workers is selected and
each werker in the sample uses method 1. A second independent simple random
sample of workers is selected and each worker in this sample uses method 2‘ The
test of the difference between population means is based on the procedures in
Section 10.2, 2. Matched sample design: One simple random sample of workers is selected. Each
worker first uses one method and then uses the other method. The order of the two
methods is assigned randomly to the workers, with sotne workers performing method I first and others performing method 2 first. Each worker provides a pair of data values, one value for method 1 and another value for method 2, In the matched sample design the two production methods are tested under similar con
ditions (i.e., with the same workers); hence this design often leads to a smaller sampling
error than the independent sample design. The primary reason is that in a matched sample
design, variation between workers is eliminated because the same workers are used for both
production methods. Let us demonstrate the analysis of a matched sample design by assuming it is the
method used to test the difference between pOpulation means for the two production methods.
A random sample of six workers is used. The data on completion times for the six workers
are given in Table 10.2. Note that each worker provides a pair of data values, one for each
production method. Also note that the last column contains the difference in completion
times d; for each worker in the sample. The key to the analysis of the matched sample design is to realize that we consider only
the column of differences. Therefore. we have six data values (.6. .2, .5. .3, .0. and .6)
that will be used to analyze the difference between population means of the two production methods.
Let lad = the mean of the dtﬂerence in values for the population of workers. With this notation, the null and alternative hypotheses are rewritten as follmvs. ; = _ .H new 2“
Him 0 Us)“ 0 new“,ng If H0 is rejected, we can conclude that the population mean completion times differ. Other than the use of the d notation. the formulas for
the sampie mean and
sample standard deviation
are the same ones used
previously in the text. The d notation is a reminder that the matched sample provides difference data. The E]
sample mean and sample standard deviation for the six difference values in Table 10.2 follow. _ Ed. 1.8
d=—‘:—:.
t1 6 30
— i—ﬂd‘iaz— if 335
5d \ rt—l _ 5 —' TABLE 10.2 TASK COMPLETION TIMES FOR A MATCHED SAMPLE DESIGN Completion Time Completion Time Difference in for Method 1 for Method 2 Completion
Worker (minutes) (minutes) Times (di)
1 6.0 5.4 .6
2 5.0 5.2 —.2
3 7.0 6.5 .5
4 6.2 5.9 .3
5 6.0 6.0 .0
6 6.4 5.8 .6 it is not necessan to make
the assumption that the
population has a normal
distribution ifthg sompir’
size is large. Sample size
guidelinesfor using the t distribution were
presented in Chapters 8
and 9. OnCe the dﬁerence dam
are computed, the t distribution procedure for
matched sampies is the
same as the onepopulation
estimation and hypothesis
testing procedures
described in Chapters 8
and 9. With the small sample of n = 6 workers. we need to make the assumption that the popu,
lation of differences has a normal distribution. This assumption is necessary so that we
may use the tdistribution for hypothesis testing and interval estimation procedures. Based
on this assumption, the following test statistic has a t distribution with n 7 1 degrees of freedom. TEST STATISTIC FOR HYPOTHESIS TESTS INVOLVING MATCHED SAMPLES [Mo = Sgt/f” {10.9) Let us use equation ([09) to test the hypotheses H0: ,ud I 0 and H3: ,ud as 0, using a = _05_
Substituting the sample results d : .30, sa. : .335, and it : 6 into equation (10.9), we com pute the value of the test statistic. d_ _ 4a
t:—(&i)0= 30 O=2.20 sd/vﬁ .335 N8 Now let us compute the prvalue for this twotailed test. Because t = 2.20 > 0, the test
statistic is in the upper tail of the t distribution. With t = 2.20, the area in the upper tail to
the right of the test statistic can be found by using the t distribution table with degrees of
freedom = n i l = 6 i l = 5. Information from the 5 degrees offreedorn row ofthe tdis tribution table is as follows: Area in Upper TaiI‘—; .20 .10 .05 .025 .01 .005
0.920 L476 2015 2.571 3.365 4.032 \ t : 2.20 t Value (5 df) l Thus, we see that the area in the upper tail is between ‘05 and 025‘ Because this testis «1
twortailed test, we double these values to conclude that the pvalue is between .i0 and .05
This plvalue is greater than a : .05. Thus? the null hypothesis H0: lad = O is not rejected,
Using Excel or Minitab and the data in Table 10.2, we find the exam In addition we can obtain an interval estimate ofthe difference between the two popli' lation means by using the single population methodology of Chapter 8. At 95% conﬁdenCC
the calculation follows. (151/. Cr _—. (l i [m ‘75]. CI _— ,3 : 25747373:?) ‘iE—j. (:1 z .3 : .35 = f o. 05, o. 65) Thus, the margin of error is .35 and the 95% conﬁdence interval for the difference betWeen
the pOpulation means of the two production methods is ~05 minutes to .65 minutes. Sith 95+ CI : (—005, 065) Con+a"“5 Waist? DD NOT Rﬂjf’c‘f' H0 2. The manager of an automobile dealership is considering a new bonus plan desi :
increase sales volume. Currently, the mean sales volume is 14 automobiles per mun '
manager wants to conduct a research study to see whether the new bonus plan inc '
sales volume. To collect data on the plan, a sample of sales personnel will be allow
sell under the new bonus plan for a onemonth period.
a. Develop the null and altemativeh'ypotheses most appropriate forthis research situti: c:
b. Comment on the conclusion when Ill}J cannot be rejected.
c. Comment on the conclusion when Ho can be rejected. 6. The CHHPTER 3 w 50 cameo PEDBLEMS
Mf— ‘1 a, .. label on a 3~quart container of orange juice claims that the orange juice contains an average of 1 gram of fat or less. Answar the following questions for a hypothesisfcst that
could be used to test the claim on the label. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error?
e. What is theType 11 error in this situation? What are the consequences of making this error? ‘ 10. Consider the following hypothesis test: Hazy > 25 .
l A sample of 40 provided a sample mean of 264. The population standard deviation is 6. a.
b. c.
d. Compute the value of the test statistic.
What is the p—value‘? '
At a = .01, what is your conclusion? What is the rejection rule using the critical value? What is your conclusion? 15. Individuals ﬁling federal income tax returns prior to March 3! rCCeived an average refund the
a. of $ l056. Consider the population of “lastminute“ ﬁlers who mail their tax return during last ﬁve days of the income tax period (typically April 10 to April [5). r
A researcher suggests that a reason individuals wait until the last ﬁve days is that 0“ 3
average these individuals receive lower refunds than do early ﬁlers. Develop appro.
priate hypotheses such that rejection of H,, will Support the researchers contention i
For a sample 01’4th individuals who filed a tax return between April [0 and 15, the
sample mean refund was $9”). Based on prior experience a population standard dc. J
viatiorl oftr = $1600 may be asstuned. What is the pvalue‘.’ I"
At a = .05. what is your conclusion? ‘ Repeat the preceding hypothesis test using the critical value approach. I
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E
i
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l
1
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:
Hu:ﬂ525 i
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i 24. Consider the following hypothesis test: " H041 =13 ,
Ham # 18 A sample of 48 provided a sample mean :2 a l? and a sample standard deviation .9 = 4.5. a.
b.
C.
_ d. 26. Consider the following hypothesis test: A sample of 65 is used. identify the p—value and state your conclusion for each of 5*» V
fol lowing sample results. Use a = .05. 8. b. i = 96.5 ands = 11.0 C. Compute the value of the test statistic. USe the r distribution table (Table 2 in Appendix B) to compute a range for the pvalue. '
At a = .05, what is your conclusion? What is the rejection rule using the critical value? What is your conclusion? How = 100
Hazy a lot) i =103ands= 11.5 i = 102 ands = 10.5 34. 36. Joan's Nursery specializes in customdesigned landscaping for residential areas. The call»
matedlabor cost associated with a particular landscaping proposal is based on the number
of plantings of trees, shrubs, and so on to be used for the project. For costestimating
purposes. managers use two hours of labor time for the planting of a mediumsized tree. ?
Actual times from a sample of to plantings during the past month follow (times in hours). 9 i
1.7 1.5 ' 2.6 2.2 2.4 2.3 2.6 3.0 1.4 2.3;I With a .05 level of signiﬁcance, test to see whether the mean treeplanting time differs it
from two hours. i a. State the null and alternative hypotheses. 5]
b. Compute the sample mean. i
c. Compute the sample standard deviation.
d. What is the p—value? e. What is your conclusion? a
3 Consider the following hypothesis test: 7 ' 30:13 a .75
Ha:p < 75 A sample of 300 items was selected. Compute the pvalue and state your conclusion tag .
each of the following sample results. Use a = .05. ' a. p=.68 c. gar—.70 1). 5:32 d. p=.77 ' 38. A study by Consumer Reports showed that 64% of supermarket shoppers believe super . u on hie Drowsy market brands to be as good as national name brands. To investigate whether this result ap ] plies to its own product, the manufacturer of a national name—brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national ' brand ketchup. 7 7 a. Formulate the hypotheses that could be used to determine whether the percentage of ‘
supermarket shoppers who believe that the supermarket ketchup was as good as the
national brand ketchup differed from 64%. b. If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good  as the national brand, what is the p—value? c. At a = .05, what is your conclusion? :1. Should the national brand ketchup manufacturer be pleased with this conclusion?
Explain. E l
; 44. In a cover story, BusinessWeek published information about sleep habits of Ameri : (BusinessWaek, January 26, 2004). The article noted that sleep deprivation causes a nu ber of problems. including highway. deaths. Fiftyone percent of adult drivers admit to I
ving while drowsy. A reSearcher hypothesized that this issue was an even bigger proble for night shift workers.
a. Formulate the hypotheses that can be used to heIp determine whether more than 51 of the population of night shift Workers admit to driving while drowsy. '
b. A sample of 500 night shift workers identiﬁed those 'who admitted to driving w drowsy. What is the sample proportion? What is the p—value?
c. At a = .01, what is your conclusion? Pas: CHRPTER ‘1 BNSMJERS T0 SUMESTEZD Pﬂpgﬁgﬁg [W]
W . 2. a. Hows l4 Ha:y>14 ‘ 36_a_z=ﬁ£p0—:Mzuzgo
b. No evidence that the new plan increases [pea _ p0) _75(1 2 '75)
sales _ r n 300
c; The research hypothesis [1 > 14 15 SUPPOITCd; the new pvalue = .0026
plan increases sates ' pvalue S .05; reject HfJ
.72 — .75
b. “ m_ = r 1.20 6. a. H020 s 1 W. Z _ u
. Ha: 'u > 1 5 .75(1 .75)
300 b. Claimingy > 1 when it is not true I j “51
c. Claiming}; E 1 when it is not true pva ue _ ' .
.1 _ p—value > .05; do not reject H.J 10.a.z=x"“°=M=1.48 j c.z=—M=—2.oo
UN; 5/% 1.750 — .75)
b. Using normal table with z = 1.48zpva1ue = 300
1.0000 — .9306 = .0694 pvalue = .0228
c. pnvalue > .01, do not reject H0 5' puvalue E .05; reject H0
d. Reject H0 ifz 3 2.33 .77 — .75 d.z= ‘280 175(1 — .75) _
300 p—value = .7881
p—value > .05; do not reject H0 2.. i—u—————=—1.83 ' 38.3.H:p=.64
ONE WOO/11400 _ H; p 2 .54 pvalue = .0336 b  = 52,100 I 2
c. pfvglue :I .05, reject H0; the mean refund of j ' p c _ '5 52 _ 64
“fastminute" ﬁlers is less than $1056 z = —pee£°—  a = —2.50 1pm — p.) _ (.640 — .64)
1.54 n 100 lnvalue : 2(.0062) = .0124 1.48 < 2.33. do not reject 11O i b. Degre I n — l = 47 c. pvalue E .05; reject HO 1 i
1 i 1 E E 15. a. Hazy 21056 M
Ha: M < 1056 b _i~,u0_910—1056 ‘ _17—18 ' between .05 and .10 Proportion differs from the reported .64
is between .10 and .20 (1. Yes, because 13 = .52 indicates that fewer believe the
supermarket brand is as good as the name brand
c. reject H0
d. _ 1‘ 44 3 H03? 5 '51
eject H0 if: 5 —2.0120 2 2.012 I Ha: p > 51
t = —1.54; do not reject H0 'b. 1'7 = 58:P‘Valuc = ‘0026 c. Reject H0 2.6. a. Bet . 9'7;rejectH0 1
13 Between .01 and .0 —value = .0125; reject Ho 5
c. Between .10 an = .1285; do not "' 1g  —1 Ll‘
reject HD I (a) z 5”?! gig/ﬁg
34. a. Hniﬂ=2 l .. H :ﬂ :15 2 4 _, .4. ' '
13.2.2 (19) P” V411“ " 2x P
“‘52 :ZxO0618 (1. Between .20 and .40 Exact pvalue = .2535 5 O ‘ 0 1
e. Do not reject Hu ...———""' J1
CC) P—Vahu = 0.0335 5 ad 5 0.05“ awe»  1 (a) 222.; eP»Vabu=ZEP’Zéz"ﬂ Do nor 120‘” H” % =a£1~onszﬂ Cd) 2 I :Z‘ _..__j,9g  = . scanas ""L' ’7 i
RHCC’LH" “(Mr E Since 2:454 > Ev: J96 ‘
 (b) Z=_2.§7.=> [34mm ._ 2x0.00$! :o.otaz :po no+ Rged’ Ho . :00102. éct=0'”"
gawk!" 501(2) ::3 
C9) 2:169! => Pvaluz = aka— 0.?382) : 0.1235 i
" 1 at: 0.05‘ $0 "01' WadiLowe '2 04135, > i
' was «1.59 2. co Monastic co . em ;,:. CHaPTER to  Suaaesrw PRO BLEMS‘ Consider the following hypothesis test. Hair“: ‘1‘250
HaZﬂt —#2>0 The following results are for two independent samples taken from the two populating Sample Satanic 1 “1 = 40 112 3 it = 25.2 :2 = 223
‘ or! = 5.2 oz 2 6.0 a. What is the value of the test statistic?
What is the p—value?
With a — ‘05. what is your hypothesis testing conclusion? 5"?" 6. The nation’s 40,000 mortgage brokerages are some of the most proﬁtable small businesses
in the United States. These lowuproﬁle companies ﬁnd loans for customers in exchange for
conunissions. Mortgage Bankers Association of America provides data on the average size 
of loans handled by mortgage brokerages (The Wall Street Journal, February 24, 2003). The CD ﬁle named Mortgage contains data from a sample of 250 loans made in 2001 and
a sample of 270 loans made in 2002 that are consistent with these data. Based on hiStori i
cal loan data the population standard deviations for the loan amounts can be assumed i
known at $55,000 in 2002 and $50,000 in 2001. Do the sample data indicate an increase ‘
in the mean loan amount between 2001 and 2002? Use a = .05. 12. The US. Department of Transportation provides the number of miles that residents of the
litan areas travel per day in a car. Suppose that for a simple random 75 largest metropo
sample of 50 Buffalo residents the mean is 22.5 miles a day and the standard deviation is 8.4 miles a day, and for an independent simple random sample of 40 Boston residen ' mean is 18.6 miles a day and the standard deviation is 7.4 miles a day. a. What is the point estimate of the difference between the mean number of mil
Buffalo residents travel per day and the mean number of miles that Boston res travel per day?
1). What is the 95% conﬁdence interval for the difference between the two .4}: .‘ tion means? 18. Educational testing companies provide tutoring, classroom learning, and practice tests in an: ' fort to help students perform better on tests such as the Scholastic Aptitude Test (SAT). The:
preparation companies claim that their courses will improve SAT score performances by
erage of 120 points (The Wall Street Journal. January 23, 2003). Aresearcher is uncertain of _
claim and believes that 120 points may be an overstatement in an effort to encourageituden '
take the test preparation course. in an evaluation study of one test preparation service, th A
searcher collects SAT score data for 35 students who took the test preparation course and 48 s
dents who did not take the course. The CD ﬁle named SAT contains the scares for this size '
a. Formulate the hypotheses that can be used to test the researcher's belief that the '
provement in SAT scores may be less than the stated average of 120 points.
b. Using a: = .05, what is your conclusion?
c. What is the point estimate of the improvement in the average SAT scores pro '
by the test preparation course? Provide a 95% conﬁdence interval estimate 0 ’ improvement. D
d. What advice would you have_for the researcher after seeing the conﬁdence in  i
i


i
i “ 22.
c» Eamings2005  Earnings 26. ' Ptovide a 95% conﬁdence interval estimate of the difference between the population I, Per'shm earnings data ComPaIiﬂg 1316 cunent quarter’s earnings with the previous  ter are in the CD ﬁle entitled Earnings 2005 (The Wall Street Journal, January 27, for the current quarter versus the previous quarter. Have earnings increased? Streetlnsidemom reported 2002 earnings per share data for a sample of major companies
(Februaty 12, 2003). Prior to 2002. ﬁnancial analysts predicted the 2002 earnings per share
for these same companies (Barron ’3, September 10, 2001). Use the following data to com
ment on differences between actual and estimated earnings per share. Company Actual Predicted
AT&T 1.29 ‘ _ 0.33
Americanﬁxpress‘ 7 2.01 7 ' ' 2.31
Citig'mpp  ' ‘ 2.591 3143
r CocaiCola 1.60  1.78
DuPotit_ _ ' . 1.84  2.18
ExxonMpbil 2.72 2.197
= Generat Electric 7 1.51 trit
Johnson & Johnson 2.28 a  2.13 '
McDonald’s 0.77 1.55
,watMan 7 ‘ _ 1.31 . 1.74 a. Use a = .05 and test for any difference between the population mean actual and popu—
., . lation r'neanestimated earnings per share. What isthepwalue? What is your conclusion;
b. What is the point estimate of the difference between the two means? Did the analysts tend to underestimate or overestimate the earnings? _
c. At 95% conﬁdence, what is the margin of error for the estimate in part (b)? What would you recornmend based on this information?  E
J
l
! —....._.___——..Mﬁ__———~M"_______,_H_____n.mm. ‘ CHAPTER 10 , ANSWERS To SUGGESTED PROBLEHS' (ii ‘ fz)  Du (25.2  7.2.8)  O 2. a. r. g: + 0—5 16.2): + £6):
\ 40 50 n3 n,_
b. p—value : 1.0000 — .9788 = .0212
c. pvalue S .05; reject H(] = 0.015 18. a. Haunt#1 2 £20 Hazyl — #2 < 120 —2.10 ______——"‘_""——____— !
ﬂap—value=£+95‘OOqu c. 32113118 d. Larger sample size 22. $0.10 +0 $0.32 26. a. £= —.60
Using t table, p»va‘lue is greater than .40 Exactpvalue = .5633
Do not reject HG h. —.103 c. .39; larger sample size b. ...
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This note was uploaded on 03/03/2012 for the course MGMT 305 taught by Professor Priya during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Priya

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