South Dakota School of Mines & Technology Estimation Industrial Engineering

Estimation Interval Estimates Industrial Engineering

Interval Estimates X n ) , ( N s m » Suppose our light bulbs have some underlying distribution f(x) with finite mean m and variance s2. Regardless of the distribution, recall from that central limit theorem that X n ) , ( N s m »

Interval Estimates ) ( 1 z Z P £ - = Recall that for a standard normal distribution, za/2 za/2 a/2 1 - a ) ( 1 2 / a z Z P £ - =

Interval Estimates X n ) , ( N s m » x ) ( 1 s m z n P £ - = But, so, Then, ) 1 , ( N n X Z » - = s m x ) ( 1 2 / a s m z n P £ - =

Interval Estimates n x ) ( z P s m - £ = n ) ( z x P s m £ - = x ) ( 1 2 / a s m z n P £ - = n ) ( 2 / z x P s m a £ - = n x ) ( 2 / z P s m a - £ =

Interval Estimates n x ) ( z P s m - £ = x ) ( n z P s m - ³ + = 1 - a 2 / z P s m a - £ = 1 - a x ) ( 2 / n z P s m a - ³ + =

Interval Estimates s x ± z n x ) ( n z P s m - ³ + = 1 - a 2 / n z P s m a - ³ + = 1 - a In words, we are (1 - a)% confident that the true mean lies within the interval s x ± z a / 2 n

Example Suppose we know that the variance of the bulbs is given by s2 = 10,000. A sample of 25 bulbs yields a sample mean of 1,596. Then a 90% confidence interval is given by 25 100 645 . 1 596 , ± 9 . 32 596 , 1 ±

Example or 1,563.1 < m < 1,628.9 32.9 1,563.1 1,596 1,628.9 32.9 is called the precision (E) of the interval and is given by n z E s a 2 / =

Interpretation Either the mean is in the confidence interval or it is not. A 90% confidence interval says that if we construct 100 intervals, we would expect 90 to contain the true mean m and 10 would not. 1,612 1,596 1,578 1,584

A Word on Confidence Int. Suppose instead of a 90% confidence, we wish to be 99% confident the mean is in the interval. Then 25 100 575 . 2 596 , 1 ± 5 . 51 596 , 1 ±

A Word on Confidence Int. That is, all we have done is increase the interval so that we are more confident that the true mean is in the interval. 32.9 90% Confidence 99% Confidence 1,563.1 1,596 1,628.9 51.5 1,544.5 1,596 1,647.5

Sample Sizes Suppose we wish to compute a sample size required in order to have a specified precision. In this case, suppose we wish to determine the sample size required in order to estimate the true mean within + 20 hours.

Sample Sizes n z E s = ÷ ø ö ç è æ = E z n s Recall the precision is given by Solving for n gives n z E s a 2 / = 2 / ÷ ø ö ç è æ = E z n s a

Sample Sizes We wish to determine the sample size required in order to estimate the true mean within + 20 hours with 90% confidence. 68 65 . 67 20 ) 100 ( 645 1 2 = ÷ ø ö ç è æ n