Question

A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation σ = 15. (a) How large must n be if the length of the 95% CI is to be not greater than 30? (b) How large must n be if the length of the 99% CI is to be not greater than 30?

Answer 1

Answer:

(a) n≥4 (b) n≥7

Step-by-step explanation:

The confidence interval is defined as

$\bar{X}\pm z*\frac{s}{\sqrt{n} }$

So the difference between upper limit (UL) and lower limit (LL) must be

$UL-LL=(\bar{X}+ z*\frac{s}{\sqrt{n} })-(\bar{X}- z*\frac{s}{\sqrt{n} })\\\\UL-LL=2*z*\frac{s}{\sqrt{n} }$

We can clear the number of observations rearranging the last equation

$UL-LL=2*z*\frac{s}{\sqrt{n} }\\\\\sqrt{n}=\frac{2*z*s}{(UL-LL)}\\ \\n=(\frac{2*z*s}{(UL-LL)})^{2}$

(a) UL-LL must be less than 30. For a 95% CI, z=1.96.

$n=(\frac{2*z*s}{(UL-LL)})^{2}= (\frac{2*1.96*15}{30} )^{2}=(\frac{58.8}{30})^{2} =1.96^{2}=3.8416$

The sample needs to be at least 4 observations.

(b) UL-LL must be less than 30. For a 99% CI, z=2.576.

$n=(\frac{2*z*s}{(UL-LL)})^{2}= (\frac{2*2.576*15}{30} )^{2}=(\frac{77.28}{30})^{2} =2.576^{2}=6.6358$

The sample needs to be at least 7 observations.