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.)

This question is incomplete, the complete question is;

Data on oxide thickness of semiconductors are as follows: 425, 430, 414, 419, 421, 436, 418, 412, 430, 434, 423, 426, 408, 437, 435, 428, 412, 426, 409, 439, 422, 426, 414, 416

(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.)

a) point estimate of the mean oxide thickness for all wafers in the population is 423.333

b) point estimate of the standard deviation of oxide thickness for all wafers in the population is 9.23

c) the standard error of the point estimate from part (a) is 1.88

Step-by-step explanation:

Given the data in the question;

x = 425, 430, 414, 419, 421, 436, 418, 412, 430, 434, 423, 426, 408, 437, 435, 428, 412, 426, 409, 439, 422, 426, 414, 416.  

a)

To determine the mean;

we sum the total value divided by the sample size.

mean x" =  (425 + 430 + 414 + 419 + ............... + 414 + 416) / 24

mean x'' =   10160/ 24    

mean x" = 423.3333 ≈ 423.333

Therefore; point estimate of the mean oxide thickness for all wafers in the population is 423.333

b)  standard deviation

we can determine the population standard deviation us the formula;

S.D(X) = √[ (ⁿ∑_ ( - X")²) / n ]

= √[ {(425-423.3333)² + (430-423.3333)² +......+ (416-423.3333)² } / 24 ]

= √[ 1957.333 / 24 )

S.D(X) = 9.03  

Note, the point of estimate of the variance population is biased estimate.

The unbiased point estimate of oxide thickness for all wafers in the population will be;

S.D(X) = √[ (ⁿ∑_ ( - X")²) / (n-1) ]

= √[ {(425-423.3333)² + (430-423.3333)² +......+ (416-423.3333)² } / (24-1) ]

= √[ 1957.333 / 23 )

S.D(X) = 9.225 ≈ 9.23

Therefore, point estimate of the standard deviation of oxide thickness for all wafers in the population is 9.23

c) the standard error  

SE (X") = S.D / √n

since the standard deviation is the unbiased estimate for the population, so;

SE (X") = 9.23 / √24

SE (X") = 1.884 ≈ 1.88

Therefore, the standard error of the point estimate from part (a) is 1.88