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