1. EML4550 2007 1 EML 4550: Engineering Design Methods Probability and Statistics in Engineering Design Class Notes Probability & Statistics for Engineers & Scientists, by Walpole, Myers, Myers and Ye

2. EML4550 -- 2007 Sampling Distributions nEstimation of population (infinity or very large sample) based on sample (limited ensemble) information nSampling distribution of mean Note: we are going to specify the population mean as  and the population standard deviation as  nCentral limit theorem:

3. EML4550 -- 2007 Example nAn engineering process is producing ball bearings that have an average diameter of 1 cm and a standard deviation of 0.05 cm. Find the probability that a random sample of 64 ball bearings will have an average diameter of less than 0.99 cm. With a confidence level of 99%, what do you expect the range of the average of a random sample of 36 ball bearings should be?

4. Example What is the probability that an elevator will overload when 12 people (randomly selected) ride at the same time? Given average weight of the population is  =170 lbs, standard deviation is  =30 lbs and the capacity of the elevator is rated at 2400 lbs. EML4550 -- 2007

5. Sampling Distribution of s 2  Chi-squared distribution The probability (  ) that the sample produces a  2 value greater than specific   2 value.

6. EML4550 -- 2007 Chi-square table As indicated in the table, it is significant if the  2 is large for given degree of freedom. For example, with a two DOF only 5% probability (1 out of 20 chance) that  2 is greater than 5.99. Therefore, it is likely that certain specifications are not correct resulting a value > 5.99. For values that give probability of greater than 10% probability, one can usually accept the hypothesis (without strong contradiction.)

7. EML4550 -- 2007 Example: check hypothesis nA manufacturer company claims that it produces hard drives that have an average life of 10 years with a standard deviation of 1 year. If six of these drives examined have lifetime of 9.8, 10.5, 12, 8.8, 11.5, 10.2 years, can we believe the company’s claim based on these samples? Assume normally distributed population. Check the chi-square table with 5 degrees of freedom (lost one DOF obtaining the average.) One expects most (90%, between 5 and 95%) of the  2 value will fall between 1.14 and 11.07, therefore, it is likely that the claim by the company is correct (at least one does not have enough evidence to dispute.)

8. EML4550 -- 2007 Example: estimate variance nA sample set of steel rods are measured to have the following lengths: 46.4, 45.8, 46.1, 46.9, 45.7, 46.0, 45.9 cm. Assuming the entire stock has a normal distribution, estimate the mean and variance of the entire stock given the confidence level of 90%.

9. EML4550 -- 2007 Student t-distribution Central limit theorem is good given that the standard deviation  of a population is known. However, knowledge about  usually is not available. Therefore, an estimation of  must be used for the estimation of the sample average . A new variable is defined to handle this situation nIf the sample size is large, one would expect that s ~  and the T-variable follows closely to a standard normal distribution as defined in the central limit theorem. The t-test is important when the sample size is small (n<30).

10. EML4550 -- 2007 t-distribution

11. EML4550 -- 2007 Example: estimation based on small sample size nThe manufacturer collect 8 samples of the years of failure for a newly designed advanced engine; they are: 15.4, 14.7, 18.1, 16.5, 17.2, 13.5, 15.8, 18.0. It is believed that the lifetimes of all engines are normally distributed with an unknown standard deviation. Therefore, we can not apply central limit theorem to estimate the average lifetime of this engine. Instead, t-distribution is used instead. For a 90% confidence level, estimate the average lifetime of the given engine. Choose one-tail 0.05 or two-tails 0.10 (1-0.9=0.1) as the starting point. The DOF is 8-1=7