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Unformatted text preview: 7.6 7.12 2 2
_ C’1 “2 _ 9 16 =./25 = 5
”Erin" J's—1 a m *no' ' The sampling distribution of 351 — E2 is approximately normal by the Central Limit Theorem since n1 2 30 and n2 2 30.
$1.32 = :11 — #2 = 10 x1x211.1 11322 10 ’71"2 r. — :2 =15.5 ~ 26.6 = —11.1 Yes. it appears that E1 — x2 w11.1 contradicts the null hypothesis H9: #1 — p2 = 10. The rejection region requires (IQ = .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:i“1_r”2¢10 _ . .5 31.1.0. _ . .
_The test statistic isz = £‘_x2_)__.;£1_55__2.5‘W—_ = —42..2.=
E "+_' .___2__. smashiregios is :3 '<'" 41950.}; 1.96. (Refer "minds Since the observed value of the test statistic falls in the rejection region (_2= _42_. 2 < —l. 96) H9 is rejected There' is sufﬁcient evidence to indicate the difference in the
populanonmeansisnoteqnaltomata— ..05_ "  The form of the conﬁdence interval is: . For confidence. .coefﬁCient 95 n: = 1 — .95 = .05 and (Liz. = .052’2 = .025. From Table IV AppendixB',z_1023— 1:96. Theconﬁdenceintervalis: ___+ 9 16
— 26.6 .6  11+_. —,1.20 — ._12 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 students who have access to lecture notes and the second population is the set .of responses for all business students not having
access to lecture notes: To determine if there is a difference in the menu response of the two groups, we test: Hem =...0.
n2¢0 7.18 The test statistic is_z._= (i—l—i)—_ = (8'48.  3780) O = J9 . ' "The rejection region requires and": .01f2.= .005 ineaeh tail of theiz'edistribntion. From
:' Table IV, Appendix 3353005 5 27.58; ' The rejection region is z .<. . —2.58_ or__'z" > 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 the mean response sf that“ was its: .=3'. '01:. Brzms =. 2.535. the".conndence.iaterva1is:. " ‘ resonance coefﬁcwm .994: ep01aeciai2'ezoiir2'e .005. From Table Iv, Appendix —— '=" .4 ' 7.‘ i 2. 8' ' .
+ n: (8 8 80) 5 86 + 35 ='>'.68' a 1801219”('—.I2t,:'l.481) _ ' We are 99% 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 interval: ofzms _= 2.58. To determine if the mean annual percentage turnover for U.S. plants exceeds that for Japanese"
plants, we test: H0: #1‘HQ=0
H3: ul—n2>0 The test statistic is r = 4.46 (from printout). The rejection region requires or = .05 in the upper tail of the r~distribution with df 2 321 + it; _ 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 .0031f2 = .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 9'5 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 = ”D — 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 7.46 351’; 3 Let a? = variance of the solid A stemandleaf 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 Heat procedure works pretty well even if the data are not exactly normal. First, find the sample sizes needed for width 5, or bound 2.5. For conﬁdence coefﬁcient .9, o: : 1 — .9 = .1 and M2 = .1f2 = .05. From Table IV, Appendix
B, 2.05 : 1.645. , 2 (1.6459003 + 102)
32 2.52 . 2 2'
— W = 86.59 .._.. 87 Thus, the necessary sample size from each p0pulation is 87 . Therefore, sufﬁcient funds have not
been allocated to meet the speciﬁcations since :11 = n2 = 100 are large enough samples. a. To determine if the variance for population 2 is greater than that for population 1, we test: Ho: 0i = 0%
Iii: 0% <1 0%
52 92
The test statistic is F = .31. = 2‘972 = 4.29
512 1.43592
The rejection region requires a = .05 in the upper tail of the Fdistribution with r1 = n2  l
= 5 — I = 4and r2 = r11 — 1 = 6 — 1 = 5. From Table IX, Appendix B. F.05 = 5.19. The rejection region is F > 5.19. Since the observed value of the test statistic does not fail in the rejection region (F = 4.29 :l 5.19), H0 is not rejected. There is insufﬁcient evidence to indicate the variance for population
2 is greater than that for population 1 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 o: < .05 but there is evidence to reject H0 for o: = .10. 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: Hg: —E=1
”a
s H ”I 3. _;—e1
2
”a The test statistic is F = $.03 (freni printeut). The pvalue is p = 0.0300 {frern printeut). Since the pvalue is net less than n: (p = .0300
r. .05}. Hg is net rejected. There is insufﬁcient evidence te indicate the variances fer the twe greups differ at ct = .05. Frent the test, there is insufﬁcient evidence te indicate the variances are different at er = .05.
Hewever, the pvalue ef .0800 is quite clese te e: = .05. If we cenclude that the variances
are equal, we have a chance ef cernrnitting a Type II errer. Witheut specifying a fitted
alternate value fer the ratie ef the variances, we cannet centpute the prehahilityr ef cennnirting
a Type II errer, 5. if we weuld inﬂate the value ef er. (say te the value cf .25 er higher).
then we knew that the value ef B will decrease. If we still de net reject He with the inﬂated
value ef es. then we can have cenﬁdence that the variances are equal. In this prehletn.
hewever, the p~value is quite small (p I .0800). Thus. the assutnptien ef equal variances is questienahle and the twesantple ttest might net be apprepriate. ...
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 Spring '07
 Safarzadeh

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