This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 6.12 6.20 6.26 a. A Type I error is rejecting the null hypothesis when it is true. In this problem, we would be
concluding that the individual is a liar when, in fact, the individual is telling the truth. A Type II error is accepting the null hypothesis when it is false. In this problem, we would be
concluding that the individual is telling the truth when, in fact, the individual is a liar. The probability of a Type I error would be the probability of concluding the individual is a liar
when hefshe is telling the truth. From the problem, it stated that the polygraph would indicate
that of 500 individuals who were telling the truth, 185 would be liars. Thus, an estimate of
the probability of a Type I error would be 185500 = .37. The probability of a Type II error would be the probability of concluding the individual is
telling the truth when hefshe is a liar. From the problem, it stated that the polygraph would
indicate that of 500 individuals who were liars, 120 would be telling the truth. Thus. an
estimate of the probability of a Type II error would be 120500 = .24. Let a = average fullservice fee (in thousands of dollars) of U.S. funeral homes in 1999. To
determine if the average fullservice fee exceeds $5,020, we test: H0; a = 5.02
H: a > 5.02 3 Using MINTAB, the output is: Descriptive Statistics Variable It Mean Median TrHean StDev SE Mean
Fee 36 5.519 5.300 5.35? 1.265 0.211
Variable Minimun Maxims 01 03 Fee 3.900 10.300 1.325 6.100 Z‘Test Test of rru = 5.020 vs mu > 5.020
The asstmed sigma = 1.26 Variable N Mean stDev SE Mean 2 P
Fee 36 5.519 1.265 0.211 2.37 0.0090
HO: ILL = 5.02
Ha: n > 5.02 The test statistic is z = 2.37. The pvalue is .0090. Since the p—value is less than or = .05, reject H0. There is sufﬁcient evidence to indicate that the average tall—service fee of U.S. funeral homes in 1999 exceeds
$5,020 at or = .05. No. Since the sample size {n = 36) is greater than 30, the Central Limit Theorem applies.
The distribution of f is approximately normal regardless of the population distribution. To determine if the mean lifetime of the new cartridges exceeds that of the old, we test: H0: u = 1,5025
H: a > 1,502.5 8. The test Statistic is z : I _ p0 = M = 3,74 “3 35.7;«225 The rejection region requires or = .005 in the upper tail of the zdistribution. From Table IV,
Appendix B, 2005 = 2.58. The rejection region is z > 2.58. Since the observed value of the test statistic falls in the rejection region (2 = 3.74 > 2.513),
H0 is rejected. There is sufﬁcient evidence to indicate the mean lifetime of the new cartridges
exceeds that of the old at o: = .005. c. No. The mean lifetime of the old cartridges was 1,511.4 pages and the mean of the new
cartridges is 1,502.5 pages. In practical terms, there is not much difference between these mo numbers. d. Yes. There is much less variability among the new cartridges. Thus, they are more similar to
each other and the quality is higher. 6.32 First, ﬁnd the value of the test statistic: x'llo 107—10: 2: = 0. x sis/5F p‘value = P(z s —l.60 orz 2 1.60) = 2P(z a 1.60) = 2(.5 — .4452) = 2(.0548) = .1096.
(using Table IV, Appendlx Bi 1.60 There is no evidence to reject H0 for or 5 .10. 6.44 a. H0: ,u = 6
Hat ,0. < 6
E  _
The test statistic is r = ”’0 = iii : 42.064 . .'—‘ '—
stttrt 1.3w’5
The necessary assumption is that the p0pulation is normal. The rejection region requires or = .05 in the lower tail of the Idistribution with df 2 it — l
= 5 — 1 = 4. From Table VI, Appendix B, ['05 = 2.132. The rejection region is r < —2.132. Since the observed value of the test statistic does not fall in the rejection region {ft = —2.064
t —2.132). H0 is not rejected. There is insufﬁcient evidence to indicate the mean is less than
6 at 0’ = .05. b. H0:,u=6
Ha:,tt#6 The ten statistic is r = —2.064 (from a).
The assumption is the same as in a.
The rejecrion region requires £3.92 2 .0512 = .025 in, each tail of the tdistribution with df = n — 1 = 5 w 1 = 4. From Table VI, Appendix B. {025 = 2.3776. The rejection regime
is t < —2.??6 or: > 2.??6. Since the observed value of the test statistic does not fall in the rejecrion region (r = 2.064 1: —2.7?6), H0 is not rejected. There is insufﬁcient evidence to indicate the mean is different
from 6 at or = .05. c. For part a, the pvalue = Pa 5 —2.064).
From Table V1, with df = 4, .05 < PH 3 —2.064) < .10 or .05 < pvalue < .10.
For part b. the pvalue = Pt! 3 —2.064) + P0 2 2.064). From Table VI. with df = 4. 2(.05) < pvalue < 2(.10) or .10 < pvalue < .20. 6.46 6.48 Some preliminary calculations are: 3: = XI = 33—6 = 105.14
H 78696 — (736):
2 ' n ‘7
——}?—_'1— ——7 ‘ 1—— 2 218.470: Ln
II
II = 121891162 2143809 a. To determine if the mean consumption rate of salad dressings in the Southeastern US. is different than the mean national consumption rate. we test: H02 ,(L =100
Ha: p. 95100 b. Since the sample size is so small, we must assume that the population being sampled is normal. In addition, we must assume that the sample is random. Eso 105.14 A 100 2 92 c. The test statistic is t = = 3M7? 14.730975 The rejection region requires 0112 = .051‘2 = .025 in each tail of the tdistribution. From
Table VI, Appendix B, with df = n — l = 7 A 1 = 6. {025 = 2.44”}. The rejection region
ist > 2.447 ort < —2.44'i'. Since the value of the test statistic does not fail in the rejection region (r = .92 a 2.447), H0
is not rejected. There is insufﬁcient evidence to indicate the mean consumption rate of salad dressings in the Southeastern U.S. is different than the mean national consumption rate at
05 = .05. d. The observed signiﬁcance level is pvalue = P(r a .92) + P(t s —.92). Since we did not
reject H0 in part c. we know that the pvalue must be greater than .05. Using Table VI,
Appendix B, with (if = n — 1 x '7’  l = 6,pvalue = P0 2 .92) + P(t s —.92) > .1 +
.1 = .2 Thus, with this table, we only know that the pvalne is greater than .2. Some preliminary calculations: f = 25 — 25587 — 2,558.7
n 10
2 2
2x2 — (2") 65,474,113 — 25,1557
52 = n = ___._ : 517.3444
7: ﬂ 1 10 — i S = 115113444 = 22.7452 To determine whether the true mean pouring temperature differs from the target setting, we test: H0: p. = 2,550
H3: is FE 2,550 3 — _
The test statistic is I = #0 = M = 1.21 311/; 22.74am? 6.54 6.58 The rejection region requires on? = .0112 = .005 in each tail of the tdistribution with df = n — 1
10 — l = 9. From Table VI. Appendix B, {005 = 3.250. The rejection region is: < —3.250 or
t > 3.250. Since the observed value of the test statistic does not fall in the rejection region (r = 1.21 r 3.250),
H0 is not rejected. There is insufﬁcient evidence to indicate the true mean pouring temperature
differs from the target setting at a = .01. 13. First, check to see if n is large enough. Po i sagepo :3tp0q°=.70 i 3.3%? a .70 i .14==(.56, .84)
n Since the interval iies within the intervai if]. % .~. the normal approximation will be adequate.
ago: 19 = .70
Ha: p < .70
Pr) ﬁp‘ an." .
The test statistic is z = 0 = .— 9 = .1_J__i_ = 1.33
‘3; ' 1 ”0(30}
Pogo 1 100
l n “ The rejection region requires at = .05 in the lower tail of the zdistribution. From Table IV,
Appendix B, {05 = 1.645. The rejection region is z < #1645. Since the observed value of the test statistic does not fall in the rejection region (v1.53 1::
—l.645), H0 is not rejected. There is insufﬁcient evidence to indicate that the proportion is less than .70 at o: = .05.
c. pvalue = Ptz E _1_53) = .5 — .4370 = .0630 Since p is not less than a = .05, H0 is not rejected. a. To determine if the normal approximation is appropriate. we check: 1—_..___ Po i 301,3 =~ pg 1 3J§iq9 = .5 i 310165} = .5 i .237 = {.‘263, .737)
J . I? Since the interval falls completely in the interval (0, l), the normal distribution will be
adequate. b. j: = 24sec = .6 To determine if the proportion of shoplifters turned over to police is greater than .5. we test: HO: p = .5
Ha: p > .5
The test statistic is z = P  p0 = .3; = 1.26
.5(.5)
[P040 [T
n The rejecrion region requires or = .05 in the upper tail of the zdistribution. From Table IV,
Appendix B, 3.05 = 1.643. The rejection region is z > 1.645. Since the observed value of the test statistic does not fall in the rejection region (5 I 1.316 f
1.645). H0 is not rejected. There is insufﬁcient evidence to indicate the proportion of
shoplifters turned over to police is greater than .5 at a = .05. The observed signiﬁcance level of the test is pvalue = PL»: 2 1.26) = .5 — .3962
= .1038. Any value of or that is greater than the pvalue would lead one to reject H0. Thus, for this
problem, we would reject H0 for any value of a > .1038. ...
View
Full Document
 Spring '07
 Safarzadeh

Click to edit the document details