Sampling Distribution: Week 6

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1 Sampling Distribution: Week 6 Kwonsang Lee University of Pennsylvania kwonlee@wharton.upenn.edu February 27, 2015 Kwonsang Lee STAT111 February 27, / 16

2 Sampling Distribution: Sample Mean If X 1, X 2,..., X n are from the population distribution with mean µ and SD σ, then we know X = X 1+ +X n n has mean( X ) = µ, SD( X ) = σ n Further, from Central Limit Theorem (CLT), we know X approximately N(µ, σ n ) Question: When can we use this CLT about sample mean? Kwonsang Lee STAT111 February 27, / 16

3 Sampling Distribution: Sample Mean (Example) If we know both population mean and population standard deviation, then we can compute probabilities like Pr( X > k) (k: any value) Example 1. Assume that population distribution of IQ score has mean 130 and SD 30. We have a sample of size 100. What is the probability that the sample mean X is over 105? Answer: From CLT, we know mean( X ) = µ = 100 and SD( X ) = σ n = = 3; X N(100, 3). Therefore, ) Pr( X > 105) = Pr( (X > ) 3 3 = Pr(Z > 1.67) = = Note: We should know population mean µ and SD σ Kwonsang Lee STAT111 February 27, / 16

4 Sampling Distribution: Count and Proportion Y i can have two possible outcome values. Y i = 1 if Success and Y i = 0 if Failure. Y = Y Y n. Y is the total number of Success in a sample of size n. We are interested in count Y or proportion Y /n = ˆp If the population proportion p is known, what is the distribution of Y? - Binomial Distribution! Pr(Y = k) = ( n k) p k (1 p) n k for 0 k n. Kwonsang Lee STAT111 February 27, / 16

5 Sampling Distribution: Count (Simple Example) Example 2. We have a fair coin and toss it 5 times. Let s say Y is the total number of Success ( Success = getting a head ). 1 Distribution of Y Y Prob 1/32 5/32 10/32 10/32 5/32 1/32 ex) Pr(Y = 2) = ( 5 2) (1/2) 2 (1/2) 3 = 10 1/32 2 Pr(Y 1)? Pr(Y 1) = Pr(Y = 0) + Pr(Y = 1) = 1/32 + 5/32 = 6/32 Kwonsang Lee STAT111 February 27, / 16

6 Sampling Distribution Example 3. We have a fair coin and toss it 100 times. Let s say Y is the total number of heads. What is Pr(Y 30)? Naive approach: Pr(Y 30) = Pr(Y = 0)+Pr(Y = 1)+ Pr(Y = 29)+Pr(Y = 30) Much better approach: Use Normal Approximation! (CLT) Kwonsang Lee STAT111 February 27, / 16

7 Normal Approximation Sample count Y approximately follows a Normal distribution with mean(y ) = np, SD(Y ) = np(1 p) The sample proportion ˆp approximately follows a Normal distribution with p(1 p) mean(ˆp) = p, SD(ˆp) = n Kwonsang Lee STAT111 February 27, / 16

8 Normal Approximation Example Example 3. We have a fair coin and toss it 100 times. Let s say Y is the total number of heads. What is Pr(Y 30)? Answer: Here, n = 100 and p = 1/2 because a coin is fair. Since we know Y approximately follows the normal distribution N(np, np(1 p)) = N(50, 5), we can compute Pr(Y 30) = Pr( Y ) 5 5 = Pr(Z 4) 0 Kwonsang Lee STAT111 February 27, / 16

9 Question 1 Question 1. (IPS 5.25, page 319) Sheila s glucose level N(125, 10). Above 140 means having gestational diabetes. a. Single measurement X : Pr(X > 140) = Pr( X > ) = Pr(Z > 1.5) = = b. Mean of three measurements X : mean( X ) = 125, SD( X ) = 10 3 = 5.77 Pr( X > 140) = Pr( X > ) Pr(Z > 2.60) = = Kwonsang Lee STAT111 February 27, / 16

10 Question 2 Question 2. (IPS 5.51, page 345) 20% of U.S high school students stole something from a store in the past year. We take a sample of size 10. a. The distribution of those who say they have stolen something is Binomial(10, 0.2). The distribution of those who do not say they have stolen something is B(10, 0.8). b. Pr(Y 4)? Pr(Y 4) = 1 Pr(Y 3) = 1 (Pr(Y = 0) + Pr(Y = 1) + Pr(Y = 2) + Pr(Y = 3)) Kwonsang Lee STAT111 February 27, / 16

11 Question 2 (Continued) There are three ways to compute the probability. 1. Direct calculation. Pr(Y = 0) = ( ) 10 0 (0.2) 0 (0.8) 10 = Pr(Y = 1) = ( ) 10 1 (0.2) 1 (0.8) 9 = Pr(Y = 2) = ( ) 10 2 (0.2) 2 (0.8) 8 = Pr(Y = 3) = ( ) 10 3 (0.2) 3 (0.8) 7 = Therefore, Pr(Y 4) = 1 ( ) = Kwonsang Lee STAT111 February 27, / 16

12 Question 2 (Continued) 2. Use Binomial Probabilities table. Pr(Y = 0) = Pr(Y = 1) = Pr(Y = 2) = Pr(Y = 3) = Therefore, Pr(Y 4) = 1 ( ) = functions/ips6e/ips6e_table-c.pdf Kwonsang Lee STAT111 February 27, / 16

13 Question 2 (Continued) 3. Normal Approximation (Not a good idea in this case, n is too small) mean(y ) = = 2, SD(Y ) = = Pr(Y 4) = Pr( Y ) = Pr(Z 1.58) = = Kwonsang Lee STAT111 February 27, / 16

14 Question 3 Question 3. (IPS 5.69, page 347) Count of Success: Y B(900, 0.2) a. mean(y ) = = 180 and SD(Y ) = = 12. b. Sample proportion ˆp has mean(ˆp) = 0.2 and SD(ˆp) = 900 = c. Normal Approximation! (n=900) ˆp Pr(ˆp > 0.24) = Pr( > ) = Pr(Z > 3) = = Kwonsang Lee STAT111 February 27, / 16

15 Quick Review: Ch1 and Ch2 Chapter 1. Mean vs. Median; Impact of outliers on mean and median. Quartiles, IQR=Q3-Q1. Interpretation of Boxplot Chapter 2. Interpretation of Scatterplot ; Identifying positive association or negative association. Correlation formula. Regression; how to interpret regression line and how to predict using the line Equation of the least-square regression line. ŷ = b 0 + b 1 x, b 1 = r S y S x, b 0 = ȳ b 1 x Kwonsang Lee STAT111 February 27, / 16

16 Quick Review: Ch3 and Ch4 Chapter 3. Hard to summarize... Need to know keywords. Sample, Population, Bias, Randomization, Experiments, Observational studies, Simple Random Sample (SRS), Nonresponse, Nonresponse rate. Chapter 4. Disjoint and Independence events, Conditional Probability at least one Problem Formulas for mean and variance of random variable(rv). Properties of RV Standardization, z-score Kwonsang Lee STAT111 February 27, / 16

Math 180B Problem Set 3

Math 180B Problem Set 3 Problem 1. (Exercise 3.1.2) Solution. By the definition of conditional probabilities we have Pr{X 2 = 1, X 3 = 1 X 1 = 0} = Pr{X 3 = 1 X 2 = 1, X 1 = 0} Pr{X 2 = 1 X 1 = 0} = P