Estimation Interval Estimates Industrial Engineering

1. Estimation Interval EstimatesIndustrial Engineering

2. s » m X N ( , ) n Interval Estimates • 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

3. a/2 a/2 1 - a za/2 za/2 0 - a = - £ £ 1 P ( z Z z ) a a / 2 / 2 Interval Estimates • Recall that for a standard normal distribution,

4. - m X = » Z N ( 0 , 1 ) s n s » m X N ( , ) n - m x - a = - £ £ 1 P ( z z ) s a a / 2 / 2 n Interval Estimates But, so, Then,

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

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

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

8. 100 ± 1 , 596 1 . 645 25 ± 1 , 596 32 . 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

9. Example or 1,563.1 < m < 1,628.9 32.9 1,563.1 1,596 1,628.9

10. s = E z a / 2 n 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

11. 1,596 1,578 1,612 1,584 Example • Suppose we repeat this process 4 times and get 4 sample means of 1596, 1578, 1612, and 1584. Computing confidence intervals then gives

12. 1,596 1,578 1,612 1,584 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.

13. 100 ± 1 , 596 2 . 575 25 ± 1 , 596 51 . 5 A Word on Confidence Int. • Suppose instead of a 90% confidence, we wish to be 99% confident the mean is in the interval. Then

14. 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

15. 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.

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

17. 2 æ ö 1 . 645 ( 100 ) = ç ÷ n 20 è ø = = 67 . 65 68 Sample Sizes • We wish to determine the sample size required in order to estimate the true mean within + 20 hours with 90% confidence.

18. 2 æ ö 1 . 645 ( 100 ) = ç ÷ n 20 è ø = = 67 . 65 68 Sample Sizes • We wish to determine the sample size required in order to estimate the true mean within + 20 hours with 90% confidence. Greater precision requires a larger sample size.

20. South Dakota School of Mines & TechnologyEstimationIndustrial Engineering

21. Estimation Interval Estimates(s unknown)Industrial Engineering

22. n å 2 - 2 x n x i = 2 = i 1 s - n 1 Confidence Intervals s unknown • Suppose we do not know the true variance of the population, but we can estimate it with the sample variance.

23. n å 2 - 2 x n x i = 2 = i 1 s - n 1 Confidence Intervals (s unknown) • Suppose we do not know the true variance of the population, but we can estimate it with the sample variance. For large samples (>30), replace s2 with s2 and compute confidence interval as before.

24. - m x = » t t - 1 n s n Confidence Intervals (s unknown) • For small samples we need to replace the standard normal, N(0,1) , with the t-distribution. Specifically,

25. tn-1 - m a/2 a/2 x = » t t 1 - a - 1 n s n tn-1,a/2 tn-1,a/2 0 Confidence Interval (s unknown)

26. tn-1 - m a/2 a/2 x = » t t 1 - a - 1 n s n tn-1,a/2 tn-1,a/2 0 Confidence Interval (s unknown) Assumption: x is normally distributed

27. tn-1 a/2 a/2 1 - a tn-1,a/2 tn-1,a/2 0 - m x - a = - £ £ 1 P ( t t ) - a - a 1 , / 2 1 , / 2 n n s n Confidence Interval (s unknown)

28. - m x s - a = - £ £ 1 P ( t t ) ± x t - a - a 1 , / 2 1 , / 2 n n s - a 1 , / 2 n n n Confidence Interval (s unknown) Miracle 17b occurs

29. 100 ± 1 , 596 1 . 711 s 25 ± x t - a 1 , / 2 n n Example • Suppose in our light bulb example, we wish to estimate an interval for the mean with 90% confidence. A sample of 25 bulbs yields a sample mean of 1,596 and a sample variance of 10,000.

30. 100 ± 1 , 596 1 . 711 s 25 ± x t - a 1 , / 2 n n Example • Suppose in our light bulb example, we wish to estimate an interval for the mean with 90% confidence. A sample of 25 bulbs yields a sample mean of 1,596 and a sample variance of 10,000. 1,596 + 34.2

31. 32.9 1,563.1 1,596 1,628.9 34.2 1,561.8 1,596 1,630.2 Example • Note that lack of knowledge of s gives a slightly bigger confidence interval (we know less, therefore we feel less confident about the same size interval). s known s unknown

32. -4.00 -2.00 0.00 2.00 4.00 A Final Word N(0,1) t10

33. -4.00 -2.00 0.00 2.00 4.00 A Final Word N(0,1) t20

34. -4.00 -2.00 0.00 2.00 4.00 A Final Word N(0,1) t30

35. ® t z - a a n 1 , / 2 / 2 A Final Word • Note that on the t-distribution chart, as n becomes larger, hence, for larger samples (n > 30) we can replace the t-distribution with the standard normal.