Data on oxide thickness of semiconductors are as follows 425 430 417 420 420 434
ID: 3306559 • Letter: D
Question
Data on oxide thickness of semiconductors are as follows 425 430 417 420 420 434 416 412 430 432 422 426 410 435 436 427 412 426 407 437 422 431 411 416 (a) Calculate (b) Calculate a point estimate of the standard deviation of oxide thickness for all wafers in the population. (Round your answer to 2 decimal places.) (c) Calculate the standard error of the point estimate from part (a). (Round your answer to 2 decimal places.) (d) Calculate a point estimate of the median oxide thickness for all wafers in the population. (Express your answer to 1 decimal places.) (e) Calculate a point estimate of the proportion of wafers in the population that have oxide thickness greater than 430 angstrom. (Round your answer to 4 declmal places.) apoint estimate of the mean oxide thickness for all wafers in the population. (Round your answer to 3 decimal places.) (a) The point estimate of the mean oxide thickness is Angstroms (b) The point estimate of the standard deviation of oxide thickness is :-=-Lmmi.) Angstroms the tolerance is +/-2% (c) The point estimate of the standard error of the mean is (d) The point estimate of the median oxide thickness is (e) The estimate of the proportion requested is Angstroms Angstroms.Explanation / Answer
A)Point estimate of mean = 18553.783
B) point estimate of standard deviation = 86949.77
C) standard error of point estimate = sqrt(summation of(X - mean)^2) / n - 2) = sqrt(166.32/ 21) = 0.6141
D) meadian = 422
E) estimate of proportion = (432 + 434 + 435 + 436) / 426737 = 0.0041
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