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rusak2 [61]
4 years ago
10

A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally dist

ributed 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?
Mathematics
1 answer:
Alenkinab [10]4 years ago
3 0

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.

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