Data on oxide thickness of semiconductors are as follows: 425 431 418 419 421 43
ID: 3375639 • Letter: D
Question
Data on oxide thickness of semiconductors are as follows: 425 431 418 419 421 437 418 410 431 433 424 426 409 437 437 427 410 426 407 438 422 431 413 416 Consider this data as a sample of the population (a) Calculate a point estimate of the mean oxide thickness for all wafers in the population. (Round your answer to 3 decimal places.) (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 decimal places.) (a) The point estimate of the mean oxide thickness is 423 (b) The point estimate of the standard deviation of oxide thickness issurr. Angstroms. (c) The point estimate of the standard error of the mean isAngstroms (d) The point estimate of the median oxide thickness is Angstroms . Angstroms (e) The estimate of the proportion requested isExplanation / Answer
Solution:- Given that values 425,431,418,419,421,437,418,410,431,433,424,426,409,437,437,427,410,426,407,438,422,431,413,416
(a) The point estimate of the mean oxide thickness is 423.583 Angstroms.
=> mean = sum of terms /number of terms = 10166/24 = 423.583
(b) The point estimate of the standard deviation of oxide thickness is 9.66
(c) The point estimate of the standard error of the mean is 1.96 Angstroms.
=> Standard error = s/sqrt(n) = 9.66/sqrt(24) = 1.97
(d) The point estimate of the median oxide thickness is 424.5
(e) The estimate of the proportion requested is 0.2917
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