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Statistics 2Dr. Ning DINGIBS I.007n.ding@pl.hanze.nl
Table of ContentReview of Chapter 5Sampling DistributionSPSS
Quiz AnswerIn 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?QuizReviewSampling DistributionSPSSz = (820-1026)/209 =-0.99P=0.3389P=0.5-0.3389=0.1611P=0.5+0.3389=0.8389
Quiz AnswerIt 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?QuizReviewSampling DistributionSPSSn=20, p=0.45  P=0.1623n=20, p=0.10 P=0.2702n=20, p=0.20 P=0.0115+0.0576+0.1369=0.2060
ReviewBionomialDistributionPoissonDistributionNormalDistributionQuizReviewSampling DistributionSPSSDiscretewithin a rangecontinuousDiscretewith meann!r ! (n-r) !Probability of r successes in n trials =λ: meanprqn-rA-3A-13A-1
Chapter 5, No. 5-37 P.270ReviewQuizReviewSampling DistributionSPSS1. Visualization is veryuseful !5-37 Normal Distribution, μ=6.4  σ=2.7(a).  P(4<X<5)=?Z1=(4-6.4)/2.7=-0.52P0.52=0.1985Z2=(5-6.4)/2.7=-0.89P0.89=0.3133P(4<X<5)=0.3133-0.1985=0.1148
Chapter 5, No. 5-38 P.270ReviewQuizReviewSampling DistributionSPSS2. FromBinomial Distribution to NormalApproximation5-38 NormalapproximationN=50   p=0.25(a). P(x>10)=?
ReviewBionomialDistributionNormalDistributionQuizReviewSampling DistributionSPSSDiscretewithin a rangecontinuousStep 1: Is itallowed?When n*p>=5 AND n*q>=5Step 2: μ =n*pσ =√n*p*qStep 3: Find CCF (ContinuityCorrection Factor)
ReviewBionomialDistributionNormalDistributionQuizReviewSampling DistributionSPSSDiscretewithin a rangecontinuousStep 3: Find CCF (ContinuityCorrection Factor)X<3X=<3X>3X>=3Discrete:  1         2         3         4         5         6         7        8CCF X>3.5CCF X>2.5CCF X<2.5CCF X<3.5Continuous:  1         2         3         4         5         6         7        8
Chapter 5, No. 5-38 P.270ReviewQuizReviewSampling DistributionSPSS2. FromBinomial Distribution to NormalApproximation5-38 NormalapproximationN=50   p=0.25(a). P(x>10)=?Step 1: n*p=50*0.25=12.5   >5AND n*q=50*0.75=37.5     >5Yes, it is allowed. Step 2: n*p=50*0.25=12.5   =μσ=√n*p*q= √50*0.25*0.75=3.062Step 3: CCF X=10.5P(x>10.5)=?z=(10.5-12.5)/3.062 =0.65
Chapter 5, No. 5-38 P.270ReviewQuizReviewSampling DistributionSPSS2. FromBinomial Distribution to NormalApproximation5-38 NormalapproximationN=50   p=0.25(a). P(x>10)=?Step 3: CCF X=10.5P(x>10.5)=?z=(10.5-12.5)/3.062 =0.65Step 4: VisualizationP0.65=0.2422P(x>10)=0.2422+0.5=0.7422
Chapter 5, No. 5-39 P.270ReviewQuizReviewSampling DistributionSPSS3. 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%VisualizationPz=0.50-0.14=0.36   z=1.08(21-μ)/5.0=1.08μ=15.6Pz=0.5-0.04=0.46 z=1.75(X1-15.6)/5.0=-1.75X1=6.85
Chapter 5, No. 5-41 P.270Review5-41 μ=100, P(X<115)=0.9(a). σ=?QuizReviewSampling DistributionSPSSVisualizationz =1.28z =1.28=(115-100)/σσ=11.72
Chapter 5, No. 5-41 P.270Review5-41 μ=100, P(X<115)=0.9(a). σ=11.72(b). Stock at least ___?QuizReviewSampling DistributionSPSSVisualization(X-100)11.72=1.645   z =1.645  X=119.28Therefore, we’dbetter stock at least 120 tubes.
The Normal DistributionMolly 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 DiscussionQuizReviewSampling DistributionSPSS
Self-PracticeChapter 5, No. 5-37 P.270QuizReviewSampling DistributionSPSSGiven that a random variable, X, has a normal distribution with mean 6.4 and standard deviation 2.7, findQ1. P (4.0<x<5.0)Q2. P(x>20)Q3. P(x<7.2)Q4. P((x<3.0) or (x>9.0))
Chapter 5, No. 5-37 P.270QuizReviewSampling DistributionSPSSGiven that a random variable, X, has a normal distribution with mean 6.4 and standard deviation 2.7, findQ1. 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
Sampling and Sampling DistributionQuizReviewSampling DistributionSPSSPopulationSample= all items chosen for study= a portion chosen from the populationParameterStatistic Greek or capital letters Lowercase Roman letters
Standard ErrorPopulation Range=80~240QuizReviewSampling DistributionSPSSµ = 100σ = 25=95Sample Range=90~120=106=101Standard Error of meanStandard Deviation of population____
Standard ErrorQuizReviewSampling DistributionSPSSSample size Dispersion of meansStandard Error
The Normal DistributionSPSS Tip: Descriptive StatisticsThe data can be downloaded from:Blackboard – Inductive Statsitics STA2—SPSS--Week 2 Creating Graphs.sav
SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingType data into SPSS	- Gender	- Age	- Satisfaction with STA21                     2                         3                       4                  5Hate it -------- Not like it --------- So so------------Like it-----Addicted to it
SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingType data into SPSS	- Gender	- Age	- Satisfactionwith STA22. Click on “Values” to define variablese.g. 0=male; 1=female1                     2                         3                       4                  5Hate it -------- Not like it --------- So so------------Like it-----Addicted to it1. Use the Variable View Tab
SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingType data into SPSS	- Gender	- Age	- Satisfaction with STA24. Input data intoeach column3. Use the Data View Tab
SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingType data into SPSS	- Gender	- Age	- Satisfaction with STA25. Don’tforget to choose the measurement level. * . Youcanedit the variables again.
6. Analyze—descriptivestatistics--DescriptiveSPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about Sampling
SPSS Tip: Descriptive StatisticsNowyougetit!QuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about Sampling
7. Analyze—descriptivestatistics--FrequenciesSPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about Sampling
8. Choose the variableyouneed to analyzeSPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about Sampling9. Click “Statistcs” to choose more
SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingNowyougetit!
SummaryReview of Chapter 5Sampling DistributionSPSS

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Lesson 02 class practices

  • 1. Statistics 2Dr. Ning DINGIBS I.007n.ding@pl.hanze.nl
  • 2. Table of ContentReview of Chapter 5Sampling DistributionSPSS
  • 3. Quiz AnswerIn 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?QuizReviewSampling DistributionSPSSz = (820-1026)/209 =-0.99P=0.3389P=0.5-0.3389=0.1611P=0.5+0.3389=0.8389
  • 4. Quiz AnswerIt 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?QuizReviewSampling DistributionSPSSn=20, p=0.45 P=0.1623n=20, p=0.10 P=0.2702n=20, p=0.20 P=0.0115+0.0576+0.1369=0.2060
  • 5. ReviewBionomialDistributionPoissonDistributionNormalDistributionQuizReviewSampling DistributionSPSSDiscretewithin a rangecontinuousDiscretewith meann!r ! (n-r) !Probability of r successes in n trials =λ: meanprqn-rA-3A-13A-1
  • 6. Chapter 5, No. 5-37 P.270ReviewQuizReviewSampling DistributionSPSS1. Visualization is veryuseful !5-37 Normal Distribution, μ=6.4 σ=2.7(a). P(4<X<5)=?Z1=(4-6.4)/2.7=-0.52P0.52=0.1985Z2=(5-6.4)/2.7=-0.89P0.89=0.3133P(4<X<5)=0.3133-0.1985=0.1148
  • 7. Chapter 5, No. 5-38 P.270ReviewQuizReviewSampling DistributionSPSS2. FromBinomial Distribution to NormalApproximation5-38 NormalapproximationN=50 p=0.25(a). P(x>10)=?
  • 8. ReviewBionomialDistributionNormalDistributionQuizReviewSampling DistributionSPSSDiscretewithin a rangecontinuousStep 1: Is itallowed?When n*p>=5 AND n*q>=5Step 2: μ =n*pσ =√n*p*qStep 3: Find CCF (ContinuityCorrection Factor)
  • 9. ReviewBionomialDistributionNormalDistributionQuizReviewSampling DistributionSPSSDiscretewithin a rangecontinuousStep 3: Find CCF (ContinuityCorrection Factor)X<3X=<3X>3X>=3Discrete: 1 2 3 4 5 6 7 8CCF X>3.5CCF X>2.5CCF X<2.5CCF X<3.5Continuous: 1 2 3 4 5 6 7 8
  • 10. Chapter 5, No. 5-38 P.270ReviewQuizReviewSampling DistributionSPSS2. FromBinomial Distribution to NormalApproximation5-38 NormalapproximationN=50 p=0.25(a). P(x>10)=?Step 1: n*p=50*0.25=12.5 >5AND n*q=50*0.75=37.5 >5Yes, it is allowed. Step 2: n*p=50*0.25=12.5 =μσ=√n*p*q= √50*0.25*0.75=3.062Step 3: CCF X=10.5P(x>10.5)=?z=(10.5-12.5)/3.062 =0.65
  • 11. Chapter 5, No. 5-38 P.270ReviewQuizReviewSampling DistributionSPSS2. FromBinomial Distribution to NormalApproximation5-38 NormalapproximationN=50 p=0.25(a). P(x>10)=?Step 3: CCF X=10.5P(x>10.5)=?z=(10.5-12.5)/3.062 =0.65Step 4: VisualizationP0.65=0.2422P(x>10)=0.2422+0.5=0.7422
  • 12. Chapter 5, No. 5-39 P.270ReviewQuizReviewSampling DistributionSPSS3. 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%VisualizationPz=0.50-0.14=0.36  z=1.08(21-μ)/5.0=1.08μ=15.6Pz=0.5-0.04=0.46 z=1.75(X1-15.6)/5.0=-1.75X1=6.85
  • 13. Chapter 5, No. 5-41 P.270Review5-41 μ=100, P(X<115)=0.9(a). σ=?QuizReviewSampling DistributionSPSSVisualizationz =1.28z =1.28=(115-100)/σσ=11.72
  • 14. Chapter 5, No. 5-41 P.270Review5-41 μ=100, P(X<115)=0.9(a). σ=11.72(b). Stock at least ___?QuizReviewSampling DistributionSPSSVisualization(X-100)11.72=1.645  z =1.645  X=119.28Therefore, we’dbetter stock at least 120 tubes.
  • 15. The Normal DistributionMolly 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 DiscussionQuizReviewSampling DistributionSPSS
  • 16. Self-PracticeChapter 5, No. 5-37 P.270QuizReviewSampling DistributionSPSSGiven that a random variable, X, has a normal distribution with mean 6.4 and standard deviation 2.7, findQ1. 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.270QuizReviewSampling DistributionSPSSGiven that a random variable, X, has a normal distribution with mean 6.4 and standard deviation 2.7, findQ1. 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 DistributionQuizReviewSampling DistributionSPSSPopulationSample= all items chosen for study= a portion chosen from the populationParameterStatistic Greek or capital letters Lowercase Roman letters
  • 19. Standard ErrorPopulation Range=80~240QuizReviewSampling DistributionSPSSµ = 100σ = 25=95Sample Range=90~120=106=101Standard Error of meanStandard Deviation of population____
  • 20. Standard ErrorQuizReviewSampling DistributionSPSSSample size Dispersion of meansStandard Error
  • 21. The Normal DistributionSPSS Tip: Descriptive StatisticsThe data can be downloaded from:Blackboard – Inductive Statsitics STA2—SPSS--Week 2 Creating Graphs.sav
  • 22. SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingType data into SPSS - Gender - Age - Satisfaction with STA21 2 3 4 5Hate it -------- Not like it --------- So so------------Like it-----Addicted to it
  • 23. SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingType data into SPSS - Gender - Age - Satisfactionwith STA22. Click on “Values” to define variablese.g. 0=male; 1=female1 2 3 4 5Hate it -------- Not like it --------- So so------------Like it-----Addicted to it1. Use the Variable View Tab
  • 24. SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingType data into SPSS - Gender - Age - Satisfaction with STA24. Input data intoeach column3. Use the Data View Tab
  • 25. SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingType data into SPSS - Gender - Age - Satisfaction with STA25. Don’tforget to choose the measurement level. * . Youcanedit the variables again.
  • 26. 6. Analyze—descriptivestatistics--DescriptiveSPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about Sampling
  • 27. SPSS Tip: Descriptive StatisticsNowyougetit!QuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about Sampling
  • 28. 7. Analyze—descriptivestatistics--FrequenciesSPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about Sampling
  • 29. 8. Choose the variableyouneed to analyzeSPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about Sampling9. Click “Statistcs” to choose more
  • 30. SPSS Tip: Descriptive StatisticsQuizSamplingRandom SamplingExperiment DesignSampling DistributionMore about SamplingNowyougetit!
  • 31. SummaryReview of Chapter 5Sampling DistributionSPSS