As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective

(a) What population is under consideration in the data set?

The answer: The population under consideration is the quantity of chips produced during a week of production = 212 chips.

(b) What parameter is being estimated?

The answer: The fraction of computer chips that had defects 27.

(c) What is the point estimate for the parameter?

The answer: the point estimate is around 12.7%

n <- 212
x <- 27
p_hat <- x / n
p_hat

## [1] 0.1273585

(d) What is the name of the statistic can we use to measure the uncertainty of the point estimate?

The answer: Standard error (SE).

(e) Compute the value from part (d) for this context.

The answer: Standard error (SE) around 2.3%.

\(SE = \sqrt[]{\frac {p(1-p)}{n}}\)

SE <- sqrt(p_hat * ( 1 - p_hat) / n)
SE

## [1] 0.02289623

(f) The historical rate of defects is 10%. Should the engineer be surprised by the observed rate of defects during the current week?

The answer: Difference between p_hat and historical rage of defects is higher than the SE. No, the engineer should not be surprised.

p_hat - 0.10

## [1] 0.02735849

SE

## [1] 0.02289623

(g) Suppose the true population value was found to be 10%. If we use this proportion to recompute the value in part (e) using p = 0.1 instead of pˆ, does the resulting value change much?

The answer: SE = 0.021 using p = 0.1 instead of p, which is not very different, and thats typical when the SE is calculated using similar proportion or little different.

\(SE = \sqrt[]{\frac {p(1-p)}{n}}\)

n <- 212
p <- 0.1
SE <- sqrt(p * ( 1 - p) / n)
SE

## [1] 0.02060408

Thanks