Konfidensintervall
Konfidensintervall
Konfidensintervall
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Armin Haalilovic:<br />
EXTRRA<br />
ÖVNINGA AR<br />
x =1.5225<br />
= λ<br />
( λα / 2 =<br />
=(0.1399,<br />
2.91)<br />
Svar a) (0.139, 2.991)<br />
b) Intervvallets<br />
läng<br />
Hälften = 1.39<br />
2 =<br />
gd=d=2.77<br />
d<br />
.<br />
= λ = 2.<br />
3263 )<br />
( λα / 2 =<br />
Från forrmeln<br />
för koonfidensinte<br />
ervall får vii<br />
σ<br />
2<br />
2λα / 2 = 1.<br />
399<br />
⇒ 2⋅<br />
2.<br />
32 263 ⋅ = 1. 39 ⇒<br />
n<br />
n<br />
2<br />
⎛ 2 2⋅<br />
2.<br />
3263⋅<br />
2 ⎞<br />
n = ⎜<br />
⎟ ⇒<br />
⎝ 1.<br />
39 ⎠<br />
n ≈ 455<br />
Svar b) : Det behövvs<br />
45 mätnin ngar ( 8 +377).<br />
c)<br />
( = λλ<br />
λ α<br />
Svar c)<br />
0 . 025<br />
= 1 . 96)<br />
Konfideensintervall:<br />
σ<br />
(x − λαα / 2 , x +<br />
n<br />
0 . 01<br />
0 . 10<br />
= 1.<br />
28816<br />
Konfideensintervall:<br />
σ<br />
( −∞,<br />
x + λα ) =<br />
n<br />
(−∞,<br />
2.<br />
433<br />
)<br />
λα / 2<br />
(−∞,<br />
σ<br />
) = ( 1.525 − −1.<br />
96<br />
n<br />
1.525 + 1.<br />
22816<br />
6 av 6<br />
2<br />
, 1.525 + 1.<br />
96 9<br />
8<br />
2<br />
)<br />
8<br />
2⋅ 2.<br />
3263⋅<br />
2<br />
n =<br />
⇒<br />
1.<br />
39<br />
= (−∞,<br />
2.<br />
43 4 )<br />
2<br />
)<br />
8<br />
Konfiden nsintervall