1. Section 4.4 Sampling Distribution Models and the Central Limit Theorem Transition from Data Analysis and Probability to Statistics

2. Probability: n From population to sample (deduction) Statistics: n From sample to the population (induction)

3. Sampling Distributions n Population parameter: a numerical descriptive measure of a population. (for example:  p (a population proportion); the numerical value of a population parameter is usually not known) Example:  mean height of all NCSU students p=proportion of Raleigh residents who favor stricter gun control laws n Sample statistic: a numerical descriptive measure calculated from sample data. (e.g, x, s, p (sample proportion))

4. Parameters; Statistics n In real life parameters of populations are unknown and unknowable. –For example, the mean height of US adult (18+) men is unknown and unknowable n Rather than investigating the whole population, we take a sample, calculate a statistic related to the parameter of interest, and make an inference. n The sampling distribution of the statistic is the tool that tells us how close the value of the statistic is to the unknown value of the parameter.

5. DEF: Sampling Distribution n The sampling distribution of a sample statistic calculated from a sample of n measurements is the probability distribution of values taken by the statistic in all possible samples of size n taken from the same population. Based on all possible samples of size n.

6. n In some cases the sampling distribution can be determined exactly. n In other cases it must be approximated by using a computer to draw some of the possible samples of size n and drawing a histogram.

7. n If a coin is fair the probability of a head on any toss of the coin is p = 0.5. n Imagine tossing this fair coin 5 times and calculating the proportion p of the 5 tosses that result in heads (note that p = x/5, where x is the number of heads in 5 tosses). n Objective: determine the sampling distribution of p, the proportion of heads in 5 tosses of a fair coin. Sampling distribution of p, the sample proportion; an example

8. Sampling distribution of p (cont.) Step 1: The possible values of p are 0/5=0, 1/5=.2, 2/5=.4, 3/5=.6, 4/5=.8, 5/5=1 n Binomial Probabilities p(x) for n=5, p = 0.5 xp(x) 00.03125 10.15625 20.3125 30.3125 40.15625 50.03125 p0.2.4.6.81 P(p).03125.15625.3125.15625.03125 The above table is the probability distribution of p, the proportion of heads in 5 tosses of a fair coin.

9. Sampling distribution of p (cont.) n E(p) =0*.03125+ 0.2*.15625+ 0.4*.3125 +0.6*.3125+ 0.8*.15625+ 1*.03125 = 0.5 = p (the prob of heads) n Var(p) = n So SD(p) = sqrt(.05) =.2236 n NOTE THAT SD(p) = p0.2.4.6.81 P(p).03125.15625.3125.15625.03125

10. Expected Value and Standard Deviation of the Sampling Distribution of p n E(p) = p n SD(p) = where p is the “success” probability in the sampled population and n is the sample size

11. Shape of Sampling Distribution of p n The sampling distribution of p is approximately normal when the sample size n is large enough. n large enough means np ≥ 10 and n(1-p) ≥ 10

12. Shape of Sampling Distribution of p Population Distribution, p=.65 Sampling distribution of p for samples of size n

13. Example n 8% of American Caucasian male population is color blind. n Use computer to simulate random samples of size n = 1000

14. The sampling distribution model for a sample proportion p Provided that the sampled values are independent and the sample size n is large enough, the sampling distribution of p is modeled by a normal distribution with E(p) = p and standard deviation SD(p) =, that is where q = 1 – p and where n large enough means np>=10 and nq>=10 The Central Limit Theorem will be a formal statement of this fact.

15. Example: binge drinking by college students n Study by Harvard School of Public Health: 44% of college students binge drink. n 244 college students surveyed; 36% admitted to binge drinking in the past week n Assume the value 0.44 given in the study is the proportion p of college students that binge drink; that is 0.44 is the population proportion p n Compute the probability that in a sample of 244 students, 36% or less have engaged in binge drinking.

16. Example: binge drinking by college students (cont.) n Let p be the proportion in a sample of 244 that engage in binge drinking. n We want to compute n E(p) = p =.44; SD(p) = n Since np = 244*.44 = 107.36 and nq = 244*.56 = 136.64 are both greater than 10, we can model the sampling distribution of p with a normal distribution, so …

17. Example: binge drinking by college students (cont.)

18. Example: texting by college students n 2008 study : 85% of college students with cell phones use text messageing. n 1136 college students surveyed; 84% reported that they text on their cell phone. n Assume the value 0.85 given in the study is the proportion p of college students that use text messaging; that is 0.85 is the population proportion p n Compute the probability that in a sample of 1136 students, 84% or less use text messageing.

19. Example: texting by college students (cont.) n Let p be the proportion in a sample of 1136 that text message on their cell phones. n We want to compute n E(p) = p =.85; SD(p) = n Since np = 1136*.85 = 965.6 and nq = 1136*.15 = 170.4 are both greater than 10, we can model the sampling distribution of p with a normal distribution, so …

20. Example: texting by college students (cont.)

21. Another Population Parameter of Frequent Interest: the Population Mean µ n To estimate the unknown value of µ, the sample mean x is often used. n We need to examine the Sampling Distribution of the Sample Mean x (the probability distribution of all possible values of x based on a sample of size n).

22. Example n Professor Stickler has a large statistics class of over 300 students. He asked them the ages of their cars and obtained the following probability distribution : x2345678x2345678 p(x)1/141/142/142/142/143/143/14 n SRS n=2 is to be drawn from pop. n Find the sampling distribution of the sample mean x for samples of size n = 2.

23. Solution n 7 possible ages (ages 2 through 8) n Total of 7 2 =49 possible samples of size 2 n All 49 possible samples with the corresponding sample mean are on p. 47.

24. Solution (cont.) n Probability distribution of x: x 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 p(x) 1/196 2/196 5/196 8/196 12/196 18/196 24/196 26/196 28/196 24/196 21/196 18/196 1/196 n This is the sampling distribution of x because it specifies the probability associated with each possible value of x n From the sampling distribution above P(4  x  6) = p(4)+p(4.5)+p(5)+p(5.5)+p(6) = 12/196 + 18/196 + 24/196 + 26/196 + 28/196 = 108/196

25. Expected Value and Standard Deviation of the Sampling Distribution of x

26. Example (cont.) n Population probability dist. x 2 3 4 5 6 7 8 p(x)1/141/142/142/142/143/143/14 n Sampling dist. of x x 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 p(x) 1/196 2/196 5/196 8/196 12/196 18/196 24/196 26/196 28/196 24/196 21/196 18/196 1/196

27. Population probability dist. x 2 3 4 5 6 7 8 p(x)1/141/142/142/142/143/143/14 Sampling dist. of x x 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 p(x) 1/196 2/196 5/196 8/196 12/196 18/196 24/196 26/196 28/196 24/196 21/196 18/196 1/196 Population mean E(X)=  = 5.714 E(X)=2(1/14)+3(1/14)+4(2/14)+ … +8(3/14)=5.714 E(X)=2(1/196)+2.5(2/196)+3(5/196)+3.5(8/196)+4(12/196)+4.5(18/196)+5(24/196) +5.5(26/196)+6(28/196)+6.5(24/196)+7(21/196)+7.5(18/196)+8(1/196) = 5.714 Mean of sampling distribution of x: E(X) = 5.714

28. Example (cont.) SD(X)=SD(X)/  2 =  /  2

29. IMPORTANT

30. Sampling Distribution of the Sample Mean X: Example n An example –A fair 6-sided die is thrown; let X represent the number of dots showing on the upper face. –The probability distribution of X is x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 Population mean  :  = E(X) = 1(1/6) +2(1/6) + 3(1/6) +……… = 3.5. Population variance  2  2 =V(X) = (1-3.5) 2 (1/6)+ (2-3.5) 2 (1/6)+ ……… ………. = 2.92

31. Suppose we want to estimate  from the mean of a sample of size n = 2. n What is the sampling distribution of in this situation?

32. 1 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6/36 5/36 4/36 3/36 2/36 1/36 E( ) =1.0(1/36)+ 1.5(2/36)+….=3.5 V(X) = (1.0-3.5) 2 (1/36)+ (1.5-3.5) 2 (2/36)... = 1.46

33. 1 1 1 6 6 6 Notice that is smaller than Var(X). The larger the sample size the smaller is. Therefore, tends to fall closer to , as the sample size increases.

34. The variance of the sample mean is smaller than the variance of the population. 123 Also, Expected value of the population = (1 + 2 + 3)/3 = 2 Mean = 1.5Mean = 2.5Mean = 2. Population 1.5 2.5 2 2 2 2 2 2 2 2 2 2 2 Expected value of the sample mean = (1.5 + 2 + 2.5)/3 = 2 Compare the variability of the population to the variability of the sample mean. Let us take samples of two observations

35. Properties of the Sampling Distribution of x

36. BUS 350 - Topic 6.16.1 -14 The central tendency is down the center Unbiased Handout 6.1, Page 1 µ l Confidence l Precision

39. Consequences

40. A Billion Dollar Mistake n “Conventional” wisdom: smaller schools better than larger schools n Late 90’s, Gates Foundation, Annenberg Foundation, Carnegie Foundation n Among the 50 top-scoring Pennsylvania elementary schools 6 (12%) were from the smallest 3% of the schools n But …, they didn’t notice … n Among the 50 lowest-scoring Pennsylvania elementary schools 9 (18%) were from the smallest 3% of the schools

41. A Billion Dollar Mistake (cont.) n Smaller schools have (by definition) smaller n’s. n When n is small, SD(x) = is larger n That is, the sampling distributions of small school mean scores have larger SD’s n http://www.forbes.com/2008/11/18/gate s-foundation-schools-oped- cx_dr_1119ravitch.html http://www.forbes.com/2008/11/18/gate s-foundation-schools-oped- cx_dr_1119ravitch.html

42. We Know More! n We know 2 parameters of the sampling distribution of x :