A manufacturer of semiconductor devices takes a random sample of size n of chips

Question

A manufacturer of semiconductor devices takes a random sample of size n of chips and tests them, classifying each chip as defective or nondefective. Let X0 if the chip is X1X2+... + Xn nondefective and x,-1 if the chip is defective. The sample fraction defective is pi- After collecting a sample, we are interested in computing the error in estimating the true value p. For each of the sample sizes and estimates of p, compute the error at the 95% confidence level. Round your error answers to 3 decimal places. 0.10 (a)n = 60 and p = (b)n = 70 and = 0.10 (c)n = 120 and = 0.10 (d) Compare your results from parts (a)-(c) and comment on the effect of sample size on the error in estimating the true value of p and the 95% confidence level (e) Repeat parts (a)-(c), this time using a 99% confidence level (f) Examine your results when the 95% confidence level and then the 99% confidence level are used to compute the error and explain what happens to the magnitude of the error as the percentage confidence increases.

Explanation / Answer

SolutionA:

n=60,p^=0.1

margin of error=z crit(sqrt(p^(1-p^)/n)

z crit for 95%=1.96

margin of error=1.96 sqrt(0.1)(1-0.1)/60

=0.076

SolutionB:

n=70 p^=.1

margin of error=1.96 sqrt(0.1)(1-0.1)/70

=0.070

SolutionC:

n=120 p^=0.1

margin of error=1.96 sqrt(0.1)(1-0.1)/120

=0.054

as sample size increases marigin of error decreases.

Solutione:

for z crit=99%=2.576

n=60 p^=0.1

moe=2.576 sqrt(0.1*0.9/60=0.099

(ii)

moe=2.576 sqrt(0.1*0.9/70)

=0.092

(iii)

moe=2.576 sqrt(0.1*0.9/120)

=0.071

AS CONFIDENCE LEVEL INCREAES MARGIN OF ERROR INCREASES

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