1. South Dakota School of Mines & Technology Estimation Industrial Engineering

2. Estimation Interval Estimates Industrial Engineering

3. 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

4. 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 - =

5. 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 - =

6. Interval Estimates n x ) ( z P s m - £ = n ) ( z x P s m £ - = x ) ( 12 / a s m z n P - = n ) ( 2 / z x P s m a - = n x ) ( 2 / z P s m a - =

7. Interval Estimates n x ) ( z P s m - £ = x ) ( n z P s m - ³ + = 1 - a2 / z P s m a - =
1 - a x ) ( 2 / n z P s m a - + =

8. Interval Estimates s x ± z n x ) ( n z P s m - ³ + = 1 - a2 / n z P s m a - + =
1 - a
In words, we are (1 - a)% confident that the
true mean lies within the interval s x z a / 2 n

9. 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

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

11. 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

12. 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

13. 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, , ,628.9
51.5
1, , ,647.5

14. 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.

15. 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

16. 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