1. Chapter 4 Simple Random Sampling (SRS)

2. SRS SRS – Every sample of size n drawn from a population of size N has the same chance of being selected. Use table of random numbers (A.2) or computer software. Using the table: – Assign every sampling unit a digit – Use table of random numbers to select sample

3. Example In a population of N = 450, select a sample of size 10 using the table of random digits. – Starting digit value_______ – Ending digit value_______ – Line number started at _______ – Sample digits selected for sample:

4. Estimating population average from SRS We use (  y i /n) to estimate  ( is an unbiased estimator of  ) We use s 2 to estimate  2 (unbiased estimator) From previous, we know that V( ) =  2 /n (infinite population….or extremely large) If finite population, then V( ) = ( (N-n)/(N-1)) (  2 /n) When we replace  2 by s 2, this becomes estimated variance of y-bar = (1-(n/N))(s 2 /n)

5. Bound on the error of estimation Using 2 standard errors as our bound (think of MOE), we have 2sqrt( (1-(n/N))(s 2 /n)) When can the finite population correction (fpc) be dropped? A good rule of thumb is when (1- n/N) > 0.95 Want data to be approximately normal (sometimes transformations can be used…..the log transformation is one of the most popular transformations) Box people example Problem 4.16 (and put a bound on it)

6. Estimating population total using SRS Since a SRS assumes all observations have an equally likely chance to be selected, we set  i to be  i = n/N) We use  -hat to estimate  ( =  y i /  i =N*y-bar is an unbiased estimator of  ) Therefore, for finite population, V(  ) = N 2 ( (N- n)/(N-1)) (  2 /n) When we replace  2 by s 2, this becomes estimated variance of  = N 2 (1-(n/N))(s 2 /n)

7. Bound on the error of estimation Using 2 standard errors as our bound (think of MOE), we have 2sqrt( N 2 (1-(n/N))(s 2 /n)) Normality is still important here!! (transform if necessary….i.e. small sample size and skewed data) Problem 4.17

8. Selecting Sample Size for  Use the variance of y-bar, which is V(y-bar) = ( (N-n)/(N-1)) (  2 /n) Set B = 2sqrt(V(y-bar)), which is B = 2sqrt(( (N-n)/(N-1)) (  2 /n) ) and solve for n ….which yields n = (N  2 )/((N-1)D+  2 ) where D=B 2 /4 Since  2 is usually not known, estimate it with s 2 (or s is approximately range/4)

9. Selecting Sample Size for  Set B = 2sqrt(N 2 V(y-bar)), which is B = 2sqrt(N 2 ( (N-n)/(N-1)) (  2 /n) ) and solve for n ….which yields n = (N  2 )/((N-1)D+  2 ) where D=B 2 /(4N 2 ) Since  2 is usually not known, estimate it with s 2 (or s is approximately range/4)

10. Examples 4.13, 4.23, 4.24, 4.27, 4.28

11. 4.5 Estimation of a Population Proportion Define y i as 0 (if unit does not have quantity of interest) and y i =1 (if unit does have quantity of interest) Then p-hat =  y i /n p-hat is an unbiased estimator of p Estimated variance of p-hat (for infinite sample sizes) is p-hat*q-hat/n Estimated variance of p-hat (for finite sample sizes) is (1-n/N)(p-hat*q-hat)/(n-1), where q-hat= 1-p-hat Bound = 2*sqrt(Estimated variance of p-hat) Problem 4.14

12. To estimate sample size n = Npq/( (N-1)D + pq ) where D = B 2 /4 If p is unknown, then we use p = 0.5 Normality is important here!! Problem 4.15 Question: All the bounds that we have looked at so far assumes what level of confidence?

13. 4.6 Comparing Estimates Comparing two means, or two totals or two proportions: Quantity of interest is  hat1 -  hat2 Variance of quantity of interest is V(  hat1 ) + V(  hat2 ) – 2cov(  hat1,  hat2 ) ********NOTE: We will NOT be using finite population correction factor in this section!! If statistics come from two independent samples, then cov(  hat1,  hat2 ) = 0 Problem 4.18

14. Examples A question asked to high school students was if they lied to a teacher at least one during the past year. The information is presented below MaleFemale Lied at least once Yes322810295 No96594620 Find the estimated difference in proportion for those who lied at least once to the teacher during the past year by gender. Place a bound on this estimated difference.* *Source: Moore, McCabe and Craig

15. Multinomial example If statistics are from a multinomial distribution, then cov(  hat1,  hat2 ) = (-p 1 p 2 /n) In a class with 30 students, the table below illustrates the breakdown of class: Freshmen10 Sophomore5 Junior7 Senior8 Estimate the difference in percent Freshmen and percent Junior and place a bound on this difference.