3. Quiz Answer In 2003, 1.4 millionstudentstook the SATs. The mean of SAT score is 1026 and the standarddeviation is 209. Whatproportion of all students had SAT scores of lessthan820? b. Whatproportion of all students had SAT scores of more than820? Quiz Review Sampling Distribution SPSS z = (820-1026)/209 =-0.99 P=0.3389 P=0.5-0.3389=0.1611 P=0.5+0.3389=0.8389
4. Quiz Answer It was foundthat the probability of Americansspendingtheirleisure time fortwo weeks is 0.45; one week is 0.10, and 0.20 forthreeor more weeks. Suppose we have surveyed 20 Americans, What is the probabilitythat 8 have 2 weeks of vacation of time? What is the probabilitythatonlyone American has one week of vacation time? What is the probabilitythat at most 2 of the workers have threeor more weeks of vacation time? Quiz Review Sampling Distribution SPSS n=20, p=0.45 P=0.1623 n=20, p=0.10 P=0.2702 n=20, p=0.20 P=0.0115+0.0576+0.1369=0.2060
5. Review Bionomial Distribution Poisson Distribution Normal Distribution Quiz Review Sampling Distribution SPSS Discrete within a range continuous Discrete with mean n! r ! (n-r) ! Probability of r successes in n trials = λ: mean prqn-r A-3 A-13 A-1
7. Chapter 5, No. 5-38 P.270 Review Quiz Review Sampling Distribution SPSS 2. FromBinomial Distribution to NormalApproximation 5-38 Normalapproximation N=50 p=0.25 (a). P(x>10)=?
8. Review Bionomial Distribution Normal Distribution Quiz Review Sampling Distribution SPSS Discrete within a range continuous Step 1: Is itallowed? When n*p>=5 AND n*q>=5 Step 2: μ =n*p σ =√n*p*q Step 3: Find CCF (ContinuityCorrection Factor)
9. Review Bionomial Distribution Normal Distribution Quiz Review Sampling Distribution SPSS Discrete within a range continuous Step 3: Find CCF (ContinuityCorrection Factor) X<3 X=<3 X>3 X>=3 Discrete: 1 2 3 4 5 6 7 8 CCF X>3.5 CCF X>2.5 CCF X<2.5 CCF X<3.5 Continuous: 1 2 3 4 5 6 7 8
10. Chapter 5, No. 5-38 P.270 Review Quiz Review Sampling Distribution SPSS 2. FromBinomial Distribution to NormalApproximation 5-38 Normalapproximation N=50 p=0.25 (a). P(x>10)=? Step 1: n*p=50*0.25=12.5 >5 AND n*q=50*0.75=37.5 >5 Yes, it is allowed. Step 2: n*p=50*0.25=12.5 =μ σ=√n*p*q= √50*0.25*0.75=3.062 Step 3: CCF X=10.5 P(x>10.5)=? z=(10.5-12.5)/3.062 =0.65
12. Chapter 5, No. 5-39 P.270 Review Quiz Review Sampling Distribution SPSS 3. If P is known, butcannotbefound in the table, it is allowed to use the closestone. 5-39 Normal Distribution σ=5.0 X=21 P=0.14 (a). μ=? (b). X1? P(X<X1)= 4% Visualization Pz=0.50-0.14=0.36 z=1.08 (21-μ)/5.0=1.08 μ=15.6 Pz=0.5-0.04=0.46 z=1.75 (X1-15.6)/5.0=-1.75 X1=6.85
13. Chapter 5, No. 5-41 P.270 Review 5-41 μ=100, P(X<115)=0.9 (a). σ=? Quiz Review Sampling Distribution SPSS Visualization z =1.28 z =1.28=(115-100)/σσ=11.72
14. Chapter 5, No. 5-41 P.270 Review 5-41 μ=100, P(X<115)=0.9 (a). σ=11.72 (b). Stock at least ___? Quiz Review Sampling Distribution SPSS Visualization (X-100)11.72=1.645 z =1.645 X=119.28 Therefore, we’dbetter stock at least 120 tubes.
15. The Normal Distribution Molly earned a score of 940 on a national achievement test. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Molly? (A) 0.10 (B) 0.18 (C) 0.50 (D) 0.82 (E) 0.90 Group Discussion Quiz Review Sampling Distribution SPSS
16. Self-Practice Chapter 5, No. 5-37 P.270 Quiz Review Sampling Distribution SPSS Given that a random variable, X, has a normal distribution with mean 6.4 and standard deviation 2.7, find Q1. P (4.0<x<5.0) Q2. P(x>20) Q3. P(x<7.2) Q4. P((x<3.0) or (x>9.0))
17. Chapter 5, No. 5-37 P.270 Quiz Review Sampling Distribution SPSS Given that a random variable, X, has a normal distribution with mean 6.4 and standard deviation 2.7, find Q1. P (4.0<x<5.0) Q2. P(x>20) Q3. P(x<7.2) Q4. P((x<3.0) or (x>9.0)) Self-Practice
18. Sampling and Sampling Distribution Quiz Review Sampling Distribution SPSS Population Sample = all items chosen for study = a portion chosen from the population Parameter Statistic Greek or capital letters Lowercase Roman letters
19. Standard Error Population Range=80~240 Quiz Review Sampling Distribution SPSS µ = 100 σ = 25 =95 Sample Range=90~120 =106 =101 Standard Error of mean Standard Deviation of population ____
20. Standard Error Quiz Review Sampling Distribution SPSS Sample size Dispersion of means Standard Error
21. The Normal Distribution SPSS Tip: Descriptive Statistics The data can be downloaded from: Blackboard – Inductive Statsitics STA2—SPSS--Week 2 Creating Graphs.sav
22. SPSS Tip: Descriptive Statistics Quiz Sampling Random Sampling Experiment Design Sampling Distribution More about Sampling Type data into SPSS - Gender - Age - Satisfaction with STA2 1 2 3 4 5 Hate it -------- Not like it --------- So so------------Like it-----Addicted to it
23. SPSS Tip: Descriptive Statistics Quiz Sampling Random Sampling Experiment Design Sampling Distribution More about Sampling Type data into SPSS - Gender - Age - Satisfactionwith STA2 2. Click on “Values” to define variables e.g. 0=male; 1=female 1 2 3 4 5 Hate it -------- Not like it --------- So so------------Like it-----Addicted to it 1. Use the Variable View Tab
24. SPSS Tip: Descriptive Statistics Quiz Sampling Random Sampling Experiment Design Sampling Distribution More about Sampling Type data into SPSS - Gender - Age - Satisfaction with STA2 4. Input data intoeach column 3. Use the Data View Tab
25. SPSS Tip: Descriptive Statistics Quiz Sampling Random Sampling Experiment Design Sampling Distribution More about Sampling Type data into SPSS - Gender - Age - Satisfaction with STA2 5. Don’tforget to choose the measurement level. * . Youcanedit the variables again.
29. 8. Choose the variableyouneed to analyze SPSS Tip: Descriptive Statistics Quiz Sampling Random Sampling Experiment Design Sampling Distribution More about Sampling 9. Click “Statistcs” to choose more
30. SPSS Tip: Descriptive Statistics Quiz Sampling Random Sampling Experiment Design Sampling Distribution More about Sampling Nowyougetit!