A fabrication plant has just completed a contract to supply memory chips to a co

Question

A fabrication plant has just completed a contract to supply memory chips to a computer manufacturer that requires the defective rate of these chips to be no more than 5%. The plant has just installed new equipment to produce these chips. An initial production run of 400 chips will be obtained and one of three actions will be taken depending on the results of this run. If it is shown that more than 5% of the chips are defective, then the equipment will be recalibrated to reduce the defective rate; if it is shown that fewer than 5% of chips are defective, then cheaper raw material will be used to reduce costs; otherwise, the equipment will be unchanged and production will begin. Suppose that among the initial production of 400 chips it was found that 16 chips were defective.

What action should plant management take at the 5% level of significance?

What is the probability that the null hypothesis would be rejected if the population defective rate is actually 8%?

Estimate the current defective rate with a 95% confidence interval.

In the future plant management would like to estimate the defective rate to within 2% using 95% confidence intervals. What sample size would be required to accomplish this if it is assumed that the defective rate would be no more than 8%?

Explanation / Answer

a)

16/400 = 0.04 i.e. 4%

As the defect rate is less than 5%, management can choose to use cheaper raw material to reduce the cost

b)

z = (0.08 - 0.05)/sqrt(0.05*0.95/400) = 2.753

P(z > 2.753) = 0.00295

Required probability is 0.003

c)

n = 400

p = 16/400 = 0.04

z-value of 95% CI = 1.9600

SE = sqrt(p*(1-p)/n) = 0.00980

ME = z*SE = 0.01920

Lower Limit = p - ME = 0.02080

Upper Limit = p + ME = 0.05920

95% CI (0.0208 , 0.0592 )

d)

Given CI Level = 95%

p = 0.08

ME = 0.02

z-value of 95% CI = 1.9600

n = (z/ME)^2*p*(1-p) = 707

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