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CHAPTER 9
Section 9.
a. E X Y E X E Y 4. 1 4. 5 . 4 , irrespective of sample sizes.
b. (^)
2 2 2
2
2
1
m n
V X Y V X V Y
, and the s.d. of
X Y . 0724 . 2691.
c. A normal curve with mean and s.d. as given in a and b (because m = n = 100, the CLT
implies that both X and Y have approximately normal distributions, so X Y does
also). The shape is not necessarily that of a normal curve when m = n = 10, because the
CLT cannot be invoked. So if the two lifetime population distributions are not normal,
the distribution of
X Y
will typically be quite complicated.
2. The test statistic value is
n
s
m
s
x y
z
2
2
2
1
, and H 0 will be rejected if either z 1. 96 or
z 1. 96. We compute
2 2
z
. Since 4.85 >
1.96, reject H 0 and conclude that the two brands differ with respect to true average tread lives.
3. The test statistic value is
n
s
m
s
x y
z
2
2
2
1
, and H 0 will be rejected at level .01 if
z 2. 33. We compute
2 2
z
, which is
not > 2.33, so we don’t reject H 0 and conclude that the true average life for radials does not
exceed that for economy brand by significantly more than 500.
a. From Exercise 2, the C.I. is
1. 96 2100 1. 96 433. 33 2100 849. 33
2
2
2
1
n
s
m
s
x y
1250. 67 , 2949. 33
. In the context of this problem situation, the interval is
moderately wide (a consequence of the standard deviations being large), so the
information about 1
and 2
is not as precise as might be desirable.
b. From Exercise 3, the upper bound is
5700 1. 645 396. 93 5700 652. 95 6352. 95 .
a. Ha says that the average calorie output for sufferers is more than 1 cal/cm
2 /min below that
for non-sufferers.
2 2 2
2
2
1
m n
, so
z . At level .01, H
0 is rejected if^
z 2. 33 ;
since –2.90 < -2.33, reject H 0.
b. P ^ ^2.^90 ^ .^0019
c.^1 ^.^92 ^.^8212
d.
2
2
m n , so use 66.
a.
1 Parameter of interest: ^
1 2
the true difference of mean tensile strength of the
1064 grade and the 1078 grade wire rod. Let
1
1064 grade average and
2
1078 grade average.
2 H 0 : 10
1 2
3 Ha: 10
1 2
n
s
m
s
x y
n
s
m
s
x y
z
o
2
2
2
1
2
2
2
1
5 RR: p ^ value ^
2 2
z
7 For a lower-tailed test, the p-value = ^ ^28.^57 ^ ^0 , which is less than any^ ^ ,
so reject H 0. There is very compelling evidence that the mean tensile strength of the
1078 grade exceeds that of the 1064 grade by more than 10.
b. The requested information can be provided by a 95% confidence interval for 1 2
1. 96 16 1. 96 . 210 16. 412 , 15. 588
2
2
2
1
n
s
m
s
x y.
a. point estimate x^ ^ y ^19.^9 ^13.^7 ^6.^2. It appears that there could be a
difference.
b.
H 0 : 0
1 2
,Ha: 0
1 2
2 2
z
, and the
p-value = 2[P(z > 1.14)] = 2( .1271) = .2542. The p value is larger than any reasonable
, so we do not reject H 0. There is no significant difference.
c. No. With a normal distribution, we would expect most of the data to be within 2 standard
deviations of the mean, and the distribution should be symmetric. 2 sd’s above the mean
is 98.1, but the distribution stops at zero on the left. The distribution is positively
skewed.
d. We will calculate a 95% confidence interval for μ, the true average length of stays for
patients given the treatment. 19.^99.^9 ^10.^0 ,^21.^8
a. The hypotheses are H 0 : 5
1 2
and H
a:^
1 2
. At level .001, H
0 should be
rejected if z 3. 08. Since
z , H 0 cannot be
rejected in favor of Ha at this level, so the use of the high purity steel cannot be justified.
b.^1
1 2
o
, so . 53 . 2891
11. (^)
n
s
m
s
X Y z
2
2
2
1
/ 2
. Standard error =
n
s
. Substitution yields
2
2
2
/ 2 1
x y z SE SE
. Using ^ .^05 , z ^ / 2 ^1.^96 , so
2 2 . We are 95% confident that the
true average blood lead level for male workers is between 0.99 and 2.41 higher than the
corresponding average for female workers.
12. The C.I. is (^)
2
2
2
1
n
s
m
s
x y
11. 23 , 6. 31
. With 99% confidence we may say that the true difference between
the average 7-day and 28-day strengths is between -11.23 and -6.31 N/mm
2 .
1 2
, d = .04, ^ .^01 ,^ .^05 , and the test is one-tailed, so
2
n , so use n = 50.
14. The appropriate hypotheses are H 0 : 0 vs. Ha: 0 , where
1 2
2 . ( 0 is
equivalent to 1 2
2 , so normal is more than twice schizophrenic) The estimator of is
2 X Y
, with^
m n
Var VarX Var Y
2
2
2
1
is the square root of
Var , and^
ˆ is obtained by replacing each
2
i
with
2
i
S. The test statistic is then
ˆ
(since
o
), and H 0 is rejected if z 2. 33. With
and
2 2
z ; Because –1.05 > -2.33, H
0 is not rejected.
d.
2 2
2 2
18. With H 0 : 0
1 2
vs. Ha: 0
1 2
, we will reject H 0 if p^ ^ value ^ .
2 2
2 2
, and the test statistic
5
. 240
6
. 164
2 2
t
leads to a p-value of 2[ P(t > 6.17)] < 2(.0005)
=.001, which is less than most reasonable ' s , so we reject H 0 and conclude that there is a
difference in the densities of the two brick types.
19. For the given hypotheses, the test statistic
6
- 38
6
- 03
2 2
t
and the d.f. is
2 2
2
, so use d.f. = 9. We will reject H 0 if
. 01 , 9
t t since –1.20 > -2.764, we don’t reject H 0.
20. We want a 95% confidence interval for 1 2
. 2.^262
. 025 , 9
t , so the interval is
13. 6 2. 262 3. 007 20. 40 , 6. 80 . Because the interval is so wide, it does
not appear that precise information is available.
21. Let
1
the true average gap detection threshold for normal subjects, and
2
the
corresponding value for CTS subjects. The relevant hypotheses are H 0 : 0
1 2
vs. Ha:
1 2
, and the test statistic 2.^46
t
. Using
d.f.
2 2
2
, or 15, the rejection region is
. 01 , 15
t t
. Since –2.46 is not 2. 602 , we fail to reject H 0. We have
insufficient evidence to claim that the true average gap detection threshold for CTS subjects
exceeds that for normal subjects.
b.
- 5 1. 5 2. 5
Comparative Box Plot for High Quality and Poor Quality Fabric
Quality
Poor
Quality
High
extensibility (%)
The comparative boxplot does not suggest a difference between average extensibility for
the two types of fabrics.
c. We test :^0
0 1 2
H vs. : 0
1 2
a
H . With degrees of freedom
2
, which we round down to 10, and using significance
level .05 (not specified in the problem), we reject H 0 if 2.^228
. 025 , 10
t t . The test
statistic is
t , which is not
in absolute value, so we
cannot reject H 0. There is insufficient evidence to claim that the true average
extensibility differs for the two types of fabrics.
a. 95% upper confidence bound:
x
- t .05,65-1 SE = 13.4 + 1.671(2.05) = 16.83 seconds
b. Let μ 1 and μ 2 represent the true average time spent by blackbirds at the experimental and
natural locations, respectively. We wish to test H 0 : μ 1 – μ 2 = 0 v. Ha: μ 1 – μ 2 > 0. The
relevant test statistic is 2 2
- 05 1. 76
t = 1.37, with estimated df =
4 4
2 2 2
≈ 112.9. Rounding to t = 1.4 and df = 120, the tabulated P -value is
very roughly .082. Hence, at the 5% significance level, we fail to reject the null
hypothesis. The true average time spent by blackbirds at the experimental location is not
statistically significantly higher than at the natural location.
c. 95% CI for silvereyes’ average time – blackbirds’ average time at the natural location:
2 2
- 76 5. 06 = (17.96 sec, 39.44 sec). The^ t -value 2.00 is
based on estimated df = 55.
25. We calculate the degrees of freedom
2 2
2 2
, or about 54
(normally we would round down to 53, but this number is very close to 54 – of course for this
large number of df, using either 53 or 54 won’t make much difference in the critical t value)
so the desired confidence interval is
2 2
3. 2 2. 931 . 269 , 6. 131
. Because 0 does not lie inside this interval, we can be
reasonably certain that the true difference 1 2
is not 0 and, therefore, that the two
population means are not equal. For a 95% interval, the t value increases to about 2.01 or so,
which results in the interval 3. 2 3. 506. Since this interval does contain 0, we can no
longer conclude that the means are different if we use a 95% confidence interval.
26. Let
1
the true average potential drop for alloy connections and let
2
the true
average potential drop for EC connections. Since we are interested in whether the potential
drop is higher for alloy connections, an upper tailed test is appropriate. We test
0 1 2
H vs. : 0
1 2
a
H . Using the SAS output provided, the test statistic,
when assuming unequal variances, is t = 3.6362, the corresponding df is 37.5, and the p-value
for our upper tailed test would be ½ (two-tailed p-value) = . 0008 . 0004
2
1
. Our p-
value of .0004 is less than the significance level of .01, so we reject H 0. We have sufficient
evidence to claim that the true average potential drop for alloy connections is higher than that
for EC connections.
27. The approximate degrees of freedom for this estimate are
2 2
2 2
, which we round down to 8, so
. 025 , 8 t and the desired interval is
2 2
18. 9 12. 607 6. 3 , 31. 5
. Because 0 is not contained in this interval, there is
strong evidence that 1 2
is not 0; i.e., we can conclude that the population means are not
33. 4 42. 8 2. 719 9. 4 2. 576 11. 98 , 6. 83 26
2 2
We are 99% confident that the true average load for carbon beams exceeds that for
fiberglass beams by between 6.83 and 11.98 kN.
b. The upper limit of the interval in part a does not give a 99% upper confidence bound.
The 99% upper bound would be ^9.^4 ^2.^434 ^.^9473 ^ ^7.^09 , meaning that the
true average load for carbon beams exceeds that for fiberglass beams by at least 7.09 kN.
a.
The most notable feature of these boxplots is the larger amount of variation present in the
mid-range data compared to the high-range data. Otherwise, both look reasonably
symmetric with no outliers present.
b. Using df = 23, a 95% confidence interval for ^ mid range ^ high range is
438. 3 437. 45 2. 069. 85 8. 69 7. 84 , 9. 54 11
2 2
Since plausible values for ^ mid range ^ high range are both positive and negative (i.e.,
the interval spans zero) we would conclude that there is not sufficient evidence to suggest
that the average value for mid-range and the average value for high-range differ.
32. Let
1
the true average proportional stress limit for red oak and let
2
the true
average proportional stress limit for Douglas fir. We test H^ 0 :^ 1 ^ 2 ^1 vs.
1 2
a
H . The test statistic is
- 818
. 2084
- 48 6. 65 1 1. 83
10
- 28
14
. 79
2 2
t
. With
degrees of freedom
- 85 13
13 9
. 2084
2
10
- 28
2
14
. 79
2
2 2
, the p-value = P(t > 1.8)
= .048. We would reject H 0 at significance levels greater than .046 (e.g., the standard 5%
significance level). At α = .05, there is sufficient evidence to claim that true average
proportional stress limit for red oak exceeds that of Douglas fir by more than 1 MPa.
mid range high range
470
460
450
440
430
420
m
id
r
a
n
g
e
Comparative Box Plot for High Range and Mid Range
35. There are two changes that must be made to the procedure we currently use. First, the
equation used to compute the value of the t test statistic is:
m n
s
x y
t
p
where sp is
defined as in Exercise 34 above. Second, the degrees of freedom = m + n – 2. Assuming
equal variances in the situation from Exercise 33, we calculate sp as follows:
2. 5 2. 544
^
p
s . The value of the test statistic is, then,
t
. The degrees of freedom = 16, and the p-
value is P ( t < -2.2) = .021. Since .021 > .01, we fail to reject H 0.
Section 9.
36. d 7. 25 , 11. 8628
D
s
1 Parameter of Interest:
D
true average difference of breaking load for fabric in
unabraded or abraded condition.
2 :^0
0
D
H
3 :^ ^0
a D
H
s n
d
s n
d
t
D D
D
5 RR:
. 01 , 7
t t
t
7 Fail to reject H 0. The data does not indicate a significant mean difference in
breaking load for the two fabric load conditions.
a. This exercise calls for paired analysis. First, compute the difference between indoor and
outdoor concentrations of hexavalent chromium for each of the 33 houses. These 33
differences are summarized as follows: n = 33, d . 4239 , .^3868
d
s , where d =
(indoor value – outdoor value). Then t^. 025 , 32 ^2.^037 , and a 95% confidence interval
for the population mean difference between indoor and outdoor concentration is
. 4239. 13715 . 5611 ,. 2868
. We can
be highly confident, at the 95% confidence level, that the true average concentration of
hexavalent chromium outdoors exceeds the true average concentration indoors by
between .2868 and .5611 nanograms/m
3 .
b. A 95% prediction interval for the difference in concentration for the 34
th house is
1 . 4239 2. 037 . 3868 1 1. 224 ,. 3758 33
1 1
. 025 , 32
d n
d t s.
This prediction interval means that the indoor concentration may exceed the outdoor
concentration by as much as .3758 nanograms/m
3 and that the outdoor concentration may
exceed the indoor concentration by a much as 1.224 nanograms/m
3 , for the 34
th house.
Clearly, this is a wide prediction interval, largely because of the amount of variation in
the differences.
a. The median of the “Normal” data is 46.80 and the upper and lower quartiles are 45.
and 49.55, which yields an IQR of 49.55 – 45.55 = 4.00. The median of the “High” data
is 90.1 and the upper and lower quartiles are 88.55 and 90.95, which yields an IQR of
90.95 – 88.55 = 2.40. The most significant feature of these boxplots is the fact that their
locations (medians) are far apart.
High: Normal:
90
80
70
60
50
40
Comparative Boxplots
for Normal and High Strength Concrete Mix
1 Parameter of interest: d
denotes the true average difference of spatial ability in
brothers exposed to DES and brothers not exposed to DES. Let
d exp osed un exp osed.
2 H 0 :^ D ^0
3 Ha :^ D ^0
s n
d
s n
d
t
D D
D
5 RR: P-value < .05, df = 9
t , with corresponding p-value .028 (from Table
A.8)
7 Reject H 0. The data supports the idea that exposure to DES reduces spatial ability.
a. Although there is a “jump” in the middle of the Normal Probability plot, the data follow a
reasonably straight path, so there is no strong reason for doubting the normality of the
population of differences.
b. A 95% lower confidence bound for the population mean difference is:
. 05 , 14
n
s
d t
d .
We are 95% confident that the true mean difference between age at onset of Cushing’s
disease symptoms and age at diagnosis is greater than -49.14.
c. A 95% upper confidence bound for the corresponding population mean difference is
44. We need to check the differences to see if the assumption of normality is plausible. A normal
probability plot validates our use of the t distribution. A 95% upper confidence bound for μD
is ^ ^2635.^63222.^91
. 05 , 15
n
s
d t
d = 2858.54.
We are 95% confident that the true mean difference between modulus of elasticity after 1
minute and after 4 weeks is at most 2858.54.
45. From the data, n = 12, (^) d = –0.73, sd = 2.81.
a. Let μd = the true mean difference in strength between curing under moist conditions and
laboratory drying conditions. A 95% CI for μd is (^) d ± t .025,11 sd/ (^) n = –0.73 ±
2.201(2.81)/ 10 = (–2.52 MPa, 1.05 MPa). In particular, this interval estimate includes
the value zero, suggesting that true mean strength is not significantly different under
these two conditions.
b. Since n = 12, we must check that the differences are plausibly from a normal population.
The normal probability plot below strongly substantiates that condition.
Differences
P
e
r
c
e
n
t
-7.5 -5.0 -2.5 0.0 2.5 5.
99
95
90
80
70
60 50 40 30
20
10
5
1
Normal Probability Plot of Differences
Normal
46. With , 6 , 5
1 1
x y , , 15 , 14
2 2
x y , ^ ,^ ^ ^1 ,^0 3 3
x y , and , 21 , 20
4 4
x y ,
d 1 and ^0
d
s (the d
I’s are 1, 1, 1, and 1), while s 1 = s 2 = 8.96, so sp = 8.96 and t = .16.
Section 9.
47. H 0 will be rejected if z^ ^ z. 01 ^2.^33. With ˆ.^150
1
p , and ˆ. 300
2
p ,
p , and q ˆ^ . 737. The calculated test statistic is
600
1
200
1
z
. Because ^4.^18 ^2.^33 , H 0 is
rejected; the proportion of those who repeat after inducement appears lower than those who
repeat after no inducement.
a. H 0 will be rejected if z^ ^1.^96. With. 2100
1
p , and
2
p ,. 2875
p ,
180
1
300
1
z
. Since 4. 84 1. 96
, H 0 is rejected.
b. p . 275 and (^) . 0432 , so power =
^
