Question

Data on oxide thickness of semiconductors are as follows:

424

430

416

417

421

434

417

409

432

431

423

426

412

437

436

426

413

426

408

439

422

426

412

417

Consider this dataset as the population, not a sample.

(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.)

Answer 1

a.

b.

Create the following table.

data	data-mean	(data - mean)2
424	0.91669999999999	0.84033888999998
430	6.9167	47.84073889
416	-7.0833	50.17313889
417	-6.0833	37.00653889
421	-2.0833	4.34013889
434	10.9167	119.17433889
417	-6.0833	37.00653889
409	-14.0833	198.33933889
432	8.9167	79.50753889
431	7.9167	62.67413889
423	-0.0833	0.00693889
426	2.9167	8.50713889
412	-11.0833	122.83953889
437	13.9167	193.67453889
436	12.9167	166.84113889
426	2.9167	8.50713889
413	-10.0833	101.67293889
426	2.9167	8.50713889
408	-15.0833	227.50593889
439	15.9167	253.34133889
422	-1.0833	1.17353889
426	2.9167	8.50713889
412	-11.0833	122.83953889
417	-6.0833	37.00653889

Find the sum of numbers in the last column to get.

So standard deviation is

c. Standard error is

d. The median is the middle number in a sorted list of numbers. So, to find the median, we need to place the numbers in value order and find the middle number.

Ordering the data from least to greatest, we get:

408   409   412   412   413   416   417   417   417   421   422   423   424   426   426   426   426   430   431   432   434   436   437   439   

As you can see, we do not have just one middle number but we have a pair of middle numbers, so the median is the average of these two numbers:

Median=

e. Let us first arrange the data

408
409
412
412
413
416
417
417
417
421
422
423
424
426
426
426
426
430
431
432
434
436
437
439

So out of 24, we see that 6 are above 430, so