Mind Expanding Exercises
8-95 a.) αχλχ αα −=<<−
1)2( 2
2,2
2
2,2
1 rr rTP
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛<<=
−
rr TTP
rr
22
2
2,2
2
2,2
1 αα χλ
χ
Then a confidence interval for λ
μ 1= is ⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛
−
2
2,2
1
2
2,2
2,2
rr
rr TTαα χχ
b) n = 20 , r = 10 , and the observed value of Tr is 199 + 10(29) = 489.
A 95% confidence interval for λ1
is )98.101,62.28(59.9
)489(2,17.34
)489(2=⎟
⎠⎞
⎜⎝⎛
8-96 dxez
dxez
xx2
212
2
1
211
21
1−−
∫∞−
−=∫∞
=ππ
αα
α
Therefore, )(111 αα zΦ=− .
To minimize L we need to minimize subject to )1()1( 211 αα −Φ+−Φ − ααα =+ 21 .
Therefore, we need to minimize . )1()1( 111 ααα +−Φ+−Φ −
2
21
2
21
2)1(
2)1(
11
1
11
1
αα
α
παα∂α∂
πα∂α∂
−
=+−Φ
−=−Φ
−
−
z
z
e
e
Upon setting the sum of the two derivatives equal to zero, we obtain 2
21
2
21 ααα zz
ee =−
. This is solved by . Consequently, and z zα α α1 = − 1 ααααα =−= 111 2, 221
ααα == .
8.97 a) n=1/2+(1.9/.1)(9.4877/4)
n=46
b) (10-.5)/(9.4877/4)=(1+p)/(1-p) p=0.6004 between 10.19 and 10.41.
8-98 a)
ni
ni
i
allXP
allXP
XP
)2/1()~(
)2/1()~(
2/1)~(
=≥
=≤
=≤
μ
μ
μ
8-39
1
21
212
21
21
)()()()(−
⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
∩−+=∪nnnn
BAPBPAPBAP
n
ii XXPBAP ⎟⎠⎞
⎜⎝⎛−=<<=∪−
211))max(~)(min()(1 μ
b) αμ −=<< 1))max(~)(min( ii XXP The confidence interval is min(Xi), max(Xi)
8-99 We would expect that 950 of the confidence intervals would include the value of μ. This is due to
9the definition of a confidence interval. Let X bet the number of intervals that contain the true mean (μ). We can use the large sample approximation to determine the probability that P(930 < X < 970).
Let 950.01000950
==p 930.01000930
1 ==p and 970.01000970
2 ==p
The variance is estimated by 1000
)050.0(950.0)1(=
−n
pp
9963.0)90.2()90.2(006892.0
02.0006892.0
02.01000
)050.0(950.0)950.0930.0(
1000)050.0(950.0)950.0970.0()970.0930.0(
=−<−<=⎟⎠⎞
⎜⎝⎛ −
<−⎟⎠⎞
⎜⎝⎛ <=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛−
<−⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛−
<=<<
ZPZPZPZP
ZPZPpP
8-40