“An Aggie does not lie, cheat, or steal or tolerate those who do.”
Answer Key Homework 8 (Due date: April 8, 2005, Friday in class)

1.
If the  is not given numerically in any question use 0.05.
Consider a normal population distribution with the value of  known.
_
(a) What is the confidence level for the interval x  2.31

? =P(z<2.31)-P(z<-2.31)=0.9792
n
_
(b) What value of
z / 2 in the confidence interval formula, x  z / 2

results in a confidence level
n
of 99.42%?
P( z / 2  z  z / 2 )  1  2P( z   z / 2 ) =0.9942 then
P( z   z / 2 ) = (1-0.9942)/2=0.0029 where the table gives you z / 2  2.76
2.
Exercise 7.5 (a)(c)
(a) 95% C.I. for . is
_
( x  z / 2

n
_
, x  z / 2

n
)  (4.85  1.96
0.75
20
,4.85  1.96
0.75
20
)  (4.52,5.18)
 2  z / 2   
 2(1.96)(0.75) 
(c) n  
 
  54.02 (round it up) then n=55.
w
0.40




2
3.
2
Answer complete Exercise 7.7 only by changing 25 to 64.
 2  z / 2   
The sample size can be computed using n  
 .
w


2
 2  z / 2   
 2  z / 2   
If you reduce the size of the width by 1/2, n'  
  4
 =4n. 4 times the
w
 w/ 2 


2
2
sample size should be increased.
 2  z / 2   
 2  z / 2   
If the sample size is increased by a factor of 64, 64n  64
 
 .
w


 w/8 
2
Width is reduced by factor 1/8.
4.
Exercise 7.29 (c) (e) (f)
(c) 1-=0.99 then t / 2:df  t 0.005; 20 =2.845
(e) =0.01 then t :df  t 0.01; 25 =2.485
(f) =0.025 then t :df  t 0.025;5 =2.571
5.
Exercise 7.30 (c) (d)
(c) 1-=0.99 then t / 2:df  t 0.005;15 =2.947
(d) 1-=0.99 and df=5-1=4 then t / 2:df  t 0.005; 4 =4.604
6.
Exercise 8.1 (c)(d) (e)
(c) No. The sample standard deviation should not be used to hypothesize
(d) Yes
(e) No. The sample mean should not be used to hypothesize
2
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
7.
Exercise 8.12(a)(b)(c)
(a)  : the true average breaking distance at 40 mph under specified conditions for the car.
H 0 :   120 versus H a :   120
(b) X the breaking distance ~ N(  ,2=100)
n=36
It is a lower tailed test, R2 should be the possible rejection region.
_
 115.20

(c) = P x



  120   P z 
when

115.20  120 
 =P(z  -2.88)=0.002
10 / 36 
_
If =0.001 then reject H0 when
x  120
10 / 36
_
 3.08 . Then reject H0 when x  114.87
8.
Exercise 8.15
(a) =P(Z1.88)=0.0310
(b) =P(Z-2.75)=0.003
(c) =P(Z-2.88)+ P(Z2.88)=0.002+0.002=0.004
9.
Exercise 8.16
(a) =P(t3.733)=0.001 using df=15 and table A5. Table A8 gives you P(t3.7)=0.001
(b) =P(t-2.5)= P(t 2.5) by symmetry. =0.01 using df=24-1=23 and table A5. Table A8 gives you
P(t2.5)=0.01
(c) =P(t-1.697)+ P(t1.697)=2 P(t1.697) by symmetry. =2(0.05)=0.10 using df=31-1=30 and
table A5. Table A8 gives you 2P(t1.7)=2(0.05)=0.1
10. Exercise 8.20 (Notice that the test statistics and the P-value is given in the minitab output)
H 0 :   750 versus H 0 :   750
_
z
x 
738.44  750
 2.14
038.20 / 50
Reject H0 if Z <  Z   Z 0.05  1.645 . Z=-2.14 < -1.645 then H0 is rejected. Conclude that the
/ n

true average lifetime of a light bulb is smaller than 750.
Reject H0 if Z <  Z   Z 0.01  2.33 . Z=-2.14 > -2.33 then H0 is not rejected. Conclude that the
true average lifetime of a light bulb is not smaller than 750.
Small  is preferred.
11. Exercise 8.21 (a)(b)
Two tailed test
(a) |t=1.6|=1.6 < t 0.025;12  2.179 then fail to reject H0. The true average diameter of ball bearings
of a certain type is 0.5 in.
(b) |t=-1.6|=1.6 < t 0.025;12  2.179 then fail to reject H0. The true average diameter of ball bearings
of a certain type is 0.5 in.
12. Exercise 8.22
(a) The data look more positively skewed by looking at the rectangle and more negatively skewed by
looking at the whiskers. It suggests skewed data with mean higher than 200.
_
(b)
H 0 :   200 versus H a :   200 then Test statistics, t 
x 
s/ n
(i) Since |t|=5.81 > t / 2;n 1  t 0.025; 29  2.045 , Reject H0.
(ii) Since P-value=2P(t>5.81)=2(0.0005)=0.001 < =0.05, Reject H0.

206.73  200
6.35 / 30
=5.81
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
Yes, it indicates that the true average weight differs from 200.
13. Exercise 8.25 (a)
(a) H 0 :   5.5 versus
H a :   5.5
_
Test statistics,
z
x 
/ n

5.25  5.5
=-3.3333
0.3 / 16
(i) Since |z|=|-3.3333|=3.3333 > z / 2  z 0.005  2.58 , Reject H0.
(ii) Since P-value=2P(z>3.3333)=2(0.0004)=0.0008 < =0.01, Reject H0.
Yes, it indicates that the true average percentage differs from 5.5.
14. Exercise 8.29 (a)
_
(a) n=8, x  3.72 , s=1.25
H 0 :   3.5 versus H a :   3.5
_
t
x 
3.72  3.5
 0.4978
1.25 / 8
(i) Reject H0 if t > t  t 0.05;7  1.895 . Since 0.4978 < 1.895, fail to reject H0.
s/ n

(ii) P-value =P(t>0.4978) > 0.10 by looking at the t-table with df=7. Since the P-value > =0.05, fail
to reject H0.
Conclude that the average mileage of shaft wear is not larger than 3.5.
15. Exercise 8.45 (a)(c)
(a) P-value=0.084 > = 0.05 then fail to reject H0.
(c) P-value=0.498 > = 0.05 then fail to reject H0.
16. Exercise 8.46 (a)(e)
(a) P-value = P(Z> 1.42)=0.0778
(e) P-value = P(Z > -0.11)=1-0.4562=0.5438
17. Exercise 8.47 (a)(c)
(a) P-value = 2P(Z>2.10)=2(0.0179)=0.0358
(c) P-value = 2P(Z>|-0.55|)=2(0.2912)=0.5824