Estimation Interval Estimates Industrial Engineering Estimation Interval Estimates Industrial Engineering

Interval Estimates  Suppose our light bulbs have some underlying distribution f(x) with finite mean  and variance  2. Regardless of the distribution, recall from that central limit theorem that X n ),(N  

Interval Estimates u Recall that for a standard normal distribution,  z  /2  z  /2  /2 1 -  )(1 2/2/   zZzP 

Interval Estimates But, so, Then, X n ),(N   )1,0(N n X Z      x )(1 2/2/     z n zP   

Interval Estimates x )(1 2/2/     z n zP    n )( 2/2/ n zxzP      n xx n )( 2/2/ zzP     

1 -  n xx n )( 2/2/ zzP      xx)( 2/2/ n z n zP     

x Interval Estimates 1 -  xx)( 2/2/ n z n zP      In words, we are (1 - a)% confident that the true mean lies within the interval n z   2/ 

Example  Suppose we know that the variance of the bulbs is given by  2 = 10,000. A sample of 25 bulbs yields a sample mean of 1,596. Then a 90% confidence interval is given by ,1  ,1 

Example or 1,563.1 <  < 1, , ,596 1,

Example or 1,563.1 <  < 1, , ,596 1, is called the precision (E) of the interval and is given by n zE   2/ 

Example u Suppose we repeat this process 4 times and get 4 sample means of 1596, 1578, 1612, and Computing confidence intervals then gives 1,612 1,596 1,578 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  and 10 would not. 1,612 1,596 1,578 1,584

A Word on Confidence Int. u Suppose instead of a 90% confidence, we wish to be 99% confident the mean is in the interval. Then ,1  ,1 

A Word on Confidence Int. u That is, all we have done is increase the interval so that we are more confident that the true mean is in the interval. 1, ,596 1, , ,596 1, % Confidence 99% Confidence

Sample Sizes u 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 u Recall the precision is given by Solving for n gives n zE   2/  2 2/        E z n  

Sample Sizes u We wish to determine the sample size required in order to estimate the true mean within + 20 hours with 90% confidence )100(         n

Sample Sizes u We wish to determine the sample size required in order to estimate the true mean within + 20 hours with 90% confidence )100(         n Greater precision requires a larger sample size.

South Dakota School of Mines & Technology Estimation Industrial Engineering

Estimation Interval Estimates (  unknown) Industrial Engineering Estimation Interval Estimates (  unknown) Industrial Engineering

Confidence Intervals  unknown u Suppose we do not know the true variance of the population, but we can estimate it with the sample variance. x      n nx s n i i

Confidence Intervals (  unknown) u 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  2 with s 2 and compute confidence interval as before. x      n nx s n i i

Confidence Intervals (  unknown) u For small samples we need to replace the standard normal, N(0,1), with the t- distribution. Specifically, n 1     n t s x t 

Confidence Interval (  unknown)  t n-1,  /2  t n-1,  /2  /2 1 -  t n-1 n 1     n t s x t 

Confidence Interval (  unknown)  t n-1,  /2  t n-1,  /2  /2 1 -  t n-1 n 1     n t s x t  Assumption: x is normally distributed

Confidence Interval (  unknown) )(1 2/,12/,1        nn t n s x tP  t n-1,  /2  t n-1,  /2  /2 1 -  t n-1

Confidence Interval (  unknown) )(1 2/,12/,1        nn t n s x tP n x s t n 2/,1   Miracle 17b occurs

Example u 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. n x s t n 2/,1   ,1 

Example u 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. n x s t n 2/,1   ,1  1,

Example u 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). 1, ,596 1, , ,596 1,  known  unknown

A Final Word N(0,1) t 10

A Final Word N(0,1) t 20

A Final Word N(0,1) t 30

A Final Word u 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. 2/2/,1  zt n  