Question

( C L Value UL- 2 SIVUS 0/0 LUuence. 3. A confidence interval estimate is desired for the mean gain in a circuit on a semiconductor device. Assume that gain is normally distributed with a standard deviation of a - 20.

Answer 1

The ME is given by :

ME = Zcritical *

Squaring both sides we get: (ME)² = (Z critical)² * $\sigma$ ²/n

_________________________________________________

(f) Given ME = 20 (Since length is 40), $\sigma$ = 20, $\alpha$ = 0.05

The Zcritical at $\alpha$ = 0.05 is 1.96

Therefore n = (Zcritical * $\sigma$ /ME)² = (1.96 * 20/40)² = 3.84

Therefore n = 4 (Taking it to the next whole number)

_________________________________________________

(f) Given ME = 20, $\sigma$ = 20, $\alpha$ = 0.01

The Zcritical at $\alpha$ = 0.05 is 2.576

Therefore n = (Zcritical * $\sigma$ /ME)² = (2.576 * 20 / 20)² = 6.64

Therefore n = 7 (Taking it to the next whole number)

_________________________________________________