ENGR 224/STAT 224 Probability and Statistics Lecture 17

1. ENGR 224/STAT 224 Probability and StatisticsLecture 17 1

2. Sampling Distributions • We previously considered distributions of individual scores, {x1, x2, x3, … xn} • Because sample statistics are calculated from sample data, and each sample is potentially different, it too is a random variable. • Prior to the data being collected there is uncertainty about the value of the statistic. That is, it has a distribution. • Once the data is obtained, we can evaluate the observed value of the statistic. • To determine the distribution of the statistic, we must not only consider the sample we did observe, but the possibility of other samples we could have observed. • We shall do some simulations.

3. Definition: Sampling Distribution of Sample Statistics Evaluating the distribution of a statistic calculated from a sample with arbitrary joint distributions is very difficult, so one simplifies the situation by assuming that the sample is a random sample from the same distribution. The sampling distribution of Sample Statistics (E.g., ) is the distribution of the Sample Statistics obtained when we repeatedly draw random samples of the same size, n, from the same population.

4. Simulation in R (Central Limit Theorem) # This is a simulation in which we take 10000 samples from a normal population par(mfrow=c(2,2)) # This allows us to view 4 graphs at once mu = 10 # We can select a mean sigma = 5 # We can select the standard deviation sample_sizes = c(1,5,50,500) # We can choose the sample sizes results=c() # initializes the vector to hold the sample means for (n in sample_sizes){ for(i in 1:10000){ # we are going to take 10000 samples X=rnorm(n,mu,sigma) # of size n from a normal population with a mean mu and st dev sigma results[i]=mean(X) # Here we calculate the sample mean } hist(results,probability=T,new=T) # Here we plot the histogram curve(dnorm(x,mu,sigma/sqrt(n)),add=T) # Here we plot the normal distribution }

5. Simulation in R (Central Limit Theorem) # This is a simulation in which we take 10000 samples from a exponential population par(mfrow=c(2,2)) # This allows us to view 4 graphs at once lambda = 5 # We can select a rate parameter sample_sizes = c(1,5,50,500) # We can choose the sample sizes mu=1/lambda # We calculate the Mean sigma=1/lambda # We calculate the Standard Deviation results=c() for (n in sample_sizes){ for(i in 1:10000){ # we are going to take 10000 samples X=rexp(n,lambda) # of size n from a normal population with a mean mu and st dev sigma results[i]=mean(X) # Here we calculate the sample mean } hist(results,probability=T,new=T) # Here we plot the histogram curve(dnorm(x,mu,sigma/sqrt(n)),add=T) # Here we plot the normal distribution }

6. Central Limit Theorem • Given • Let X1, X2, …Xn, be a random sample drawn from a distribution (which may or may not be normal) with mean m and standard deviation s, • Then • The distribution of the sample means, will as the sample size increases (sufficiently large n) approach a normal distribution. (If population is Normal, then true for any n) • The mean of the sample means will be the population mean m. • The standard deviation of the sample means will be

7. Notation • Let samples of size n be selected from a population with mean m and standard deviation s, • The mean of the sample means is denoted as • Therefore the CLT says that • The standard deviation of the sample means is denoted as • According to the CLT, we have thatfor large populations.

8. Application of the CLT Therefore, the distribution of the sample mean X of a random sample drawn from practically any population with mean m and standard deviation s can be approximated by a normal distribution with mean m , and standard deviation provided the sample is large (If population is Normal then true for any n)

9. Finding the probability that a sample mean is between a and b

10. Example: In human engineering and product design, it is often important to consider the weights of peoples so that airplanes or elevators are not overloaded, chairs don’t break and other unpleasant happenings don’t occur. Assume that the population of men and women has normally distributed weights with a mean of 173 pounds and a standard deviation of 30 pounds. • A) If 1 person is randomly selected, find the probability that their weight is greater than 180 pounds. • B) If 36 different people are randomly selected from this population find the probability that their mean weight is greater than 180 pounds. Solutions: A) .4090 B) .0808

11. Example: To avoid false advertisement suits, a beverage bottler must make reasonably certain that 500 ml bottles actually contain 500 ml. To infer whether a bottling machine is working satisfactorily, the bottler randomly samples 10 bottles per hour and measures the amount of beverage in each bottle. The mean of the 10 measurements is used to determine whether to readjust the amount of beverage delivered to each bottle. Records show that the amount of fill per bottle is normally distributed with a standard deviation of 1.5 ml and a mean fill of 501 ml. What is the probability that the sample mean of 10 test bottles is less than 500 ml. Solution: P( x  500) = 0.0174

12. Overview Distributions of Statistics Central Limit Theorem Sections 5.3. 5.4 of text 12

13. Homework Reread 5.3 and 5.4 Read 5.5, 6.1, 6.2 (skim we are not going to do all detail) Start Preparing for the Midterm Exam (Chaps 1 – 5 ) 13