A certain manufacturer produces computer chips; each chip is independently acceptable with some unknown probability p. To obtain an approximate 99 percent confidence interval for p, whose length is approximately .05, an initial sample of 30 chips has been taken. If 26 of these chips are of acceptable quality, then the preliminary estimate of p is 26/30. Using this value, a 99 percent confidence interval of length

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approximately .05 would require an approximate sample of size n = 4(2.005) 26 (.05)2 30 (1-30) = 4(2.58)2 26 4 (.05)2 30 30 = 1,231 Hence, we should now sample an additional 1,201 chips and if, for instance, 1,040 of them are acceptable, then the final 99 percent confidence interval for p is 1,066 1,066 1 1,231 (1- 1,066 2.005 1,066 + 1,066 1,231 1,231' 1,231 (1- - 1,066 2.005 1,231 1,231 10 or pe (.84091,.89101)