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Unformatted text preview: 7.6 7.12 _ I E __ _ﬁ. 1.0. .__H _. H _The test statistic isz = I; (155. 256) 19 2 2 01 a 9 16
__= +_: _ _=./_25=.5
“Iix2 —1 :22 100 + 100 The sampling distribution of 351 — E2 is approximately normal
by the Central Limit Theorem since n1 2 30 and n2 2 30. nil—32 = #1  #2 =10 fl — I: =15.5 ~ 26.6 = —11.1 w11.1 contradicts the null hypothesis H9: til — p2 = 10. Yes, it appears that $1 — I; The rejection region requires all = .025 = .052 in each tail of the zdistribution. From
Table IV, Appendix B, 2:025 = 1.96. The rejection region is z < 1.96 or z > 1.96. Ha::“1_r”2#10 =' —42..2'.f' 
.5 _ _ .  meanness is :2 '<'" '—'i.9é'olr'%">' 1:96; (Refer have) Since the observed value of the test statistic falls in the rejection region (2 __= 5422
< — 1.96), H9 is rejected. There is sufﬁcient evidence to indibate the differénce in the
9013mm normalto 10 3:0: = '  '  ' " ' ' The form of conﬁdence interval is: .. .Forconﬁdence..c0efﬁcient .953: = 1 — .95 =".os andeta? .0512 = .025. From Table IV, Appendix E; z "= 1:96. Theeonﬁdence interval is:: . —26..6' . ..._ _  .11.' 2+. .—12.0 ,— .12
(15.5 )i_l9'6100'+_100 1.... 28=f( “Sr 10 ). We are 95% conﬁdent that the difference in the two means is between “12.08 and —10.12. The conﬁdence interval gives more information. The ﬁrst pepulation .is the set of responses. for all business 5114de who have access to lecture
notes andthe secondpopulation is the setof'responses forall business students not having
access to lecture notes; _ ' To determine if there is a difference in the mean response of the two groups, we test: _ as. r.» ..
Ha; _p._1_ .0' 7.18 The test statistic is_z._= = (8'48.  3780) O = J9 . ' “The rejection region and": .le2 .005 of theiz'edistribntion. From
“ Table IV, Appendix 5 27.58; ' The rejection region is z .<. , —2.58_ 0rsz > 2.58. Since the observed value of the test statistic does not fall in the rejection region (2: = 2.19 a
2.58);“HD is not rejected.' There is insufﬁcient evidence to indicate a difference in mean sf the. felts: .=3'. '01:. Brzms =. 2.535. ’ITseImﬁdence'interv‘alis: " ‘ ooriooaﬁoom .99,_,o e .01 ageon '='.0u2*= .005. From Table Iv, Appendix +' =5 (3.43... 130) i 253. 2.99 n: 86 35
 '".='>'.68'i'.801''==v'('.12i,2'l.481) _ ' We are conﬁdent that the difference in the mean response between the trio groups is
between —.121 and 11.481. A 95 % conﬁdence interval would be smaller than the 99% conﬁdence intervat. The 2: value
used in the 95% conﬁdence interval is {025 = 1.96 compared with the z value used in the  99 % conﬁdence intei'val: ofzms _= 2.58. To determine if the mean annual percentage turnover for U.S. plants exceeds that for Japanese"
plants, we test: H0: pl—ng=0
H3: n1—n2>0 The test statistic is r = 4.46 (from printout). The rejection region requires or = .05 in the upper tail of the r~disrribution with df 2 321 + it: h 2 = 5 + 5 — 2 = 8. From Table VI, Appendix B, {05 = 1.860. The rejection
region is t > 1.860. Since the observed value of the test statistic falls in the rejection region (I = 4.46 > 1.860), H0 is rejected. There is sufﬁcient evidence to indicate the mean annual percentage turnover
for U.S. plants exceeds that for Japanese plants at o: = .05. The observed signiﬁcance is .0031l2 = .00155. Since the pvalue is so small, there is evidence to reject H0 for o: > .005. The necessary assumptions are: 1. Both sampled populations are approximately normal.
2. The population variances are equal.
3. The samples are randomly and independently sampled. There is no indication that the populations are not normal. Both sample variances are similar, so there is no evidence the population variances are unequal. There is no indication the
assumptions are not valid. 7.22 a. 13 #D=#1p2 c. For conﬁdence coefﬁcient .95, o: = .05 and 0:10 = .025. From Table VI, Appendix B, with
df = nD — 1 = 6  1 = 5, 11025 = 2.571. The conﬁdence interval is: S
— o
It) i tar: = 2 i 2.571é =9 2 i 1.484 = (.516, 3.484) a: rt (1. H0: #1) = 0
Ha: [in 3‘5 0 . . . ID 2 _
The test statistic ts t z = — 3.46 EDIE «EH/6. The rejection region requires orfz = .05t’2 = .025 in each tail of the zdistribution with df = 3113 — 1 = 6  1 = 5. From Table VI, Appendix B, {025 = 2.571. The rejection region is
I < —2.5710rr > 2.571. Since the observed value of the test statistic falls in the rejection region (3.46 > 2.571), H0 is rejected. There is sufﬁcient evidence to indicate that the mean difference is different from 0 at
o: = .05. 7.28 a. To determine whether male students’ attitudes toward their fathers differ from their attitudes
toward their mothers, on average, we test: in. From the printout. the value of the test statistic is I = ﬁ1.08. The pvalue is p = .3033.
Since the pvalue is not less than tr = .05, H0 is not rejected. There is insufﬁcient evidence to indicate male students’ attitudes toward their fathers differ from their attitudes toward their
mothers. on average at o: = .05. c. In order for the above test to be valid, we must assume that I. The population of differences is normal
2. The differences are randomly selected A sternandleaf display of the differences is: Character StemandLeaf Display Stemandleaf of Diff N
Leaf Unit = 0.10 13 1 2 0
(6) 1 000000
6 0 00 4 1 0000 We know that the data are not normal, because there are only 4 different values for the
differences. Although the stemandleaf display does not look particularly normal, it is rather
moundshaped. The rtest procedure works pretty well even if the data are not exactly normal. 7.36 First, find the sample sizes needed for width 5, or bound 2.5.
For conﬁdence coefﬁcient .9, o: : 1 — .9 = .1 and (#2 = .1f2 = .05. From Table IV, Appendix
B, 2.05 : 2'2 2
Z ll 0 + 0' 2 7 2
n1: r12 = 1 2) 2 (L645) (10 +10) = 86.59 a: 87
32 2.52
Thus, the necessary sample size from each pepulation is 87 . Therefore, sufﬁcient funds have not
been allocated to meet the speciﬁcations since :11 = n2 = 100 are large enough samples.
7.46 a. To determine if the variance for population 2 is greater than that for population 1, we test:
Ho: 0i = 0%
Iii: 0% ‘z 0%
52 92
The test statistic is F = .3“. = 2‘972 = 4.29
512 1.43592
The rejection region requires a = .05 in the upper tail of the Fdistribution with v1 = n2  l
= 5 — I = 4and v2 = r11 — 1 = 6 — 1 = 5. From Table IX, Appendix B. Fm = 5.19.
The rejection region is F > 5.19. .
Since the observed value of the test statistic does not fall in the rejection region (F = 4.29 r
5.19), H0 is not rejected. There is insufﬁcient evidence to indicate the variance for population
2 is greater than that for p0pulation l at or = .05.
b. The pvalue is P(F a 4.29). From Tables VIIIand IX, with v1 = 4 and v2 = 5,
.05 < P(F 2 4.29) < .10
There is no evidence to reject H0 for or < .05 but there is evidence to reject H0 for o: = .10.
E51; 3 Let a? = variance of the solid I waste generation rates for industrialized countries and 0% =
variance of the solidwaste generation rates for middle~income countries. . . To determine if the
variances differ, we test: 7.76 02
1 H0: 1—1
Er2
2 H U1¢1 a‘_2'
02 The test statistic is F = 7.03 (from printout). The pvalue is p = 0.0800 (from printout). Since the pvalue is not less than or (p = .0800
a .05). H0 is not rejected. There is insufﬁcient evidence to indicate the variances for the two
groups differ at a = .05. From the test, there is insufﬁcient evidence to indicate the variances are different at a z .05.
However, the pvalue of .0800 is quite close to or = .05. If we conclude that the variances
are equal, we have a chance of committing a Type II error. Without specifying a ﬁxed
alternate value for the ratio of the variances, we cannot compute the probability of committing
a Type II error, .6. If we would inflate the value of or, (say to the value of .25 or higher),
then we know that the value of .8 will decrease. If we still do not reject H0 with the inﬂated
value of a, then we can have conﬁdence that the variances are equai. In this problem.
however, the pvalue is quite small (p I .0800). Thus, the assumption of equal variances is
questionable and the twosample ttest might not be appropriate. df for Error is 41  6 = 35 SSE = SS(Total) — SST = 46.5 — 17.5 = 29.0 _ SST 17.5 _ SSE 290
MST _ = _ — 2.9167 MSE = = _.__ =
P_1 6 n_p 35 .8236
MST 2.9167
F = _ = = 3.
MSE .8286 52 The ANOVA table is: Source df SS MS F
Ireatment 6 “.5 2.9167 3.5?
Error 35 29.0 .8286 ioﬁi 31 36.5 The number of treatments is p. We knowP — 1 = 6 =>p = 7. To determine if there is a difference among the population means, we test: Ho: #1=#2=”‘=F/7
Ha: At least one of the population means differs from the rest The test statistic is F = 3.52. The rejection region requires a = .10 in the upper tail of the Fdistribution with numerator df = p — 1 = 6 and denominator df = n — p = 35. From Table VIII, Appendix B, F10 =
1.98. The rejecti0n region is F > 1.98. ’ Since the observed vaiue of the test statistic falls in the rejection region (F = 3.52 > 1.98}, H0 is rejected. There is sufficient evidence to indicate a difference among the population
means at [X = .10. 7.78 b. The observed signiﬁcance level is P(F a 3.52). With numerator df = 6 and denominator
df = 35, and Table XI, P(F a 3.52) < .01. H03 #1 = #2
Ha: ,ul 95 pig
3 —E _
The test statistic isr = 1—2 = 3'7 4'1 = 316
1 1
MSE i + i 3286[g + 3]
I11 "2 The rejection region requires (:12 = .1072 = .05 in each tail of the Idistribution with df =
n — p = 35. From Table VI, Appendix B, {05 3 1.697. The rejection region is r <
—1.697 or r > 1.697. Since the observed value of the test statistic does not fall in the rejection region (I = u .76 t *1.697), H0 is not rejected. There is insufﬁcient evidence to indicate that it] and p2 differ at
a = .10. For conﬁdence coefﬁcient .90, o: = .10 and 0272 = .05. From Table VI, Appendix B, with
df = 35, {05 2 1.697. The couﬁdence interval is: = (3.7 — 4.1) i 1.692[.8286[l + l]
6 6
= —.4 i .892 = (1.292, .492) The conﬁdence interval is:
ii i rosthSEm n 3.7 i 1.697V‘.8286!6 n 3.7 i .631 = (3.069, 4.331) a. For diagram 3, _Ex1~7+8+9+9+10+11 54
Il— —....._.—_. :F___=9
n 6 6
J—Ezzg’e:12+13+14+14+15+nszsftﬂ4
n 6 6 For diagram b, 51:20=5+5+7+11+13+13=ﬁ=9
rt 6 6
J—C2=}:’~’2=10+10+12+16+13+18aria“4
n 6 6 For diagram 3, 2
SST = Zﬂi(EI — if = 6(9 _ }1_5)2 + 6(14 _115)2 = 75 i=1 (E = L" = _.._54 + 34 = 11.5)
n 12 For diagram 1), 2
SST = Ema — if = 6(9 —11.s::2 + 604 —11.5>2 = 75 i=1 I For diagram a, '2
x 2
2102121) 49634.
92: nl = =2
1 n—l 6—1
1
' '2
x 2
£122 ‘ (E 2') lamEL
2_ n2 _ 6 =2
Sz—"ﬁE—TTT
2 SSE = (In — 1):? + (:22 1)32={6—1)2 + (6 w 02 = 20 For diagram 13, . x 2
Ex? — 1} 558  5i
s§=_wn_l‘ 2—6_l =14.4
1
_ I x 2 2
Ex; " (EH 2) 1248  E1
2_ 2 _ 6 _
52 — —_ — .—..—.—.——_ — HZ 1 6 1 SSE 2021* U521 +012 — 0:2 = {6 — U144 + {6 a 014.4 =1441 For diagram 3, SS(T0tal) = SST + SSE = 75 + 20 = 95 ssr is “ESL x 100% = g x 100% = 73.95% ofSS(T0tal) SS(Total) For diagram 1), SS(TOtal) = SST + SSE = 75 + 144 = 219 SST is i x 100% = x 100% : 34.25% of SS(Total)
SS(Total} 219
For diagram 3. MST = ﬂ = i = 75
p — 1 2 v 1
MSE= SSE = 20 =2 'F=__MST=3§=375
n — p 12 — 2 SE 2
For diagram b. MST = ﬂ = i = 75
p — 1 2 — 1
MSE= SSE = 144 =14.4 F=ﬂsl=i=521
n ‘ p 12 — 2 MSE 14.4 The rajection region for both diagrams requires a: = .05 in the upper tail of the Fdistributinn
with 012;) +1: 2 —1=1and v2 = n —p =12  2 =10. From Table 1X. Appendix
8, F05 = 4.96. The rcjccdon region is F > 4.96. 7.86 180 a. For diagram 3, the observed value of the test statistic falls in the rejection region (F = 3?.5
b 4.96). Thus, H0 is rejected. There is sufﬁcient evidence to indicate the samples were
drawn from populations with different means at o: = .05. For diagram 1), the observed value of the test statistic falls in the rejection region (F = 5.21 > 4.96}. Thus. H0 is rejected. There is sufﬁcient evidence to indicate the samples were
drawn from populations with different means at o = .05. g. We must assume both populations are normally distributed with women variances.
Refer to Exercise 7.78, the ANOVA table is: For diagram 3: Source df SS MS F
Treatment 1 75 75 7 . 5
Error 1 0 20 2 Total 1 l 95. For diagram b: Source df SS MS F
Treatment 1 75 75 5 . 21
Error 10 144 14 .4 Total 1 1 21 9 To determine if differences exist among the mean sorption rates of the three organic solvent
types, we test: Ho: #1 = M = #3
Ha: At least two treatment means differ where ,ui represents the mean sorption rate for solvent type t'. The test statistic is F = 24.51. The rejection region requires or = .10 in the upper tail of the F—distribution with v1 = p — 1 = 3 —1=2andv2 = n —p = 32 — 3 = 29. UsingTableWIl,AppendixB.
F110 = 2.50. The rejection region is F > 2.50. Since the observed value of the test statistic falls inthe rejection region (F = 24.51 > 2.50),
H0 is rejected. There is sufﬁcient evidence to indicate that differences exist among the mean
sorption rates of the three organic solvent types at a = .10. The assumptions required for the analysis to be valid are: l. The samples are selected randome and independently. 2. The probability distributions of sorption rates for each type are normal. 3. The variances of the sorption rate probability distributions for each type are equai. To check the assumptions, we will look at the stem—andleaf displays and the dot plots. The
stemandleaf displays are: Stemand—leaf of Sorption Type
Leaf Unit = 0.10 II
—>
2 Ii
0 2 0 67
(3) 0 889
4 1 0011 Stem~andleaf of Sorption Type
Leaf Unit = 0.10 II
N
2 II
0: NmbbNN
D
m
0 StemandLeaf of Sorption Type = 3 N = 15
Leaf Unit = 0.10 6 0 000111
(2) 0 23
7 0 4&555
2 0 66 These stemandleaf displays do not look particularly moundshaped. Thus the assumption that
each of the samples is selected from a normal distribution may not be valid. The dot plots for the three types are: Dolpr for Sorption Sorplion The spread of the sorptiou rates for type 2 appears to be greater than the spreads for the other
two types Thus‘ the assumption of equal variances may not be met. ...
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This note was uploaded on 09/11/2008 for the course ECON 317 taught by Professor Safarzadeh during the Spring '07 term at USC.
 Spring '07
 Safarzadeh

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