Inductive Statistics Dr. Ning DING [email_address] I.007 IBS, Hanze You’d better use the full-screen mode to view this PPT...
Table of Contents Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Basics-Independent σ  is known: σ  is unknown: H 0 H 1 n <30 &  σ  is unkn...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice Ch 9 No. 9-9 P.466 Step 1: Formulate hypotheses 9-9 Step 2: Find ...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Review: Chapter 9: Testing Hypothe...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Will the partic...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Step 2: Calcula...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Step 3: Find th...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice Ch 9 No. 9-15  P.474 Step 4: Visalize and Calculate the t values ...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Example: Ch 9 Example P.476 You are testing whether the two dru...
Step 4: Visualize and get the z values -1.96  +1.96  Accept H 0  No significant difference Chapter 9 Testing Hypotheses:...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Step 1: Formulate Hypotheses Step 3: Calculate the Standard Err...
Step 4: Visualize and get the z values One-tailed  α =0.15   P=0.35    z= -1.04  Accept H 0  No significant difference...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice 9-20 Ch 9 No. 9-20  P.483 Step 1: Formulate Hypotheses Step 3: Ca...
Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice 9-21 Step 1: Formulate Hypotheses Step 3: Calculate the Standard ...
Chapter 11  Chi-Square Test-Basics Contingency Table (Cross break table) Rows * Columns == 2*2 table Review: Chapter 9: Te...
Chapter 11  Chi-Square Test-Basics Contingency Table (Cross break table) Rows * Columns == 2*2 table Expected and Observed...
Chapter 11  Chi-Square Test-Basics Contingency Table (Cross break table) 1*1=1 2*2=4 Degree of freedom= (row-1)*(column-1)...
Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~De...
Chapter 11  Chi-Square Test–Calculate  χ 2  for 2-row Table Step 2: Calculate the  χ 2 2.7644 Ch 11 Example P.570 Review: ...
Step 3: Find the critical  χ 2 Chapter 11  Chi-Square Test–Calculate  χ 2  for 2-row Table α  = 0.10  2*4   df= 3 χ 2 =6....
Chapter 11  Chi-Square Test–Calculate  χ 2  for 3-row Table For a national health insurance program, you believes that len...
Chapter 11  Chi-Square Test–Calculate  χ 2  for 3-row Table For a national health insurance program, you believes that len...
Chapter 11  Chi-Square Test–Calculate  χ 2  for 3-row Table Ch 11 Example P.575 Step 3: Calculate the Chi-Square Example: ...
Chapter 11  Chi-Square Test–Calculate  χ 2  for 3-row Table Ch 11 Example P.575 Step 3: Calculate the Chi-Square Example: ...
Chapter 11  Chi-Square Test–Calculate  χ 2  for 3-row Table Ch 11 Example P.575 Step 4: Find the critical Chi-Square For a...
Chapter 11  Chi-Square Test–Calculate  χ 2  for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 1: Formulate th...
Chapter 11  Chi-Square Test–Calculate  χ 2  for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 1: Formulate th...
Chapter 11  Chi-Square Test–Calculate  χ 2  for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 2: Calculate th...
Chapter 11  Chi-Square Test–Calculate  χ 2  for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 3: Calculate th...
Table of Contents Review: Chapter 8 Testing Hypothesis Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independen...
The Normal Distribution SPSS Tips The data can be downloaded from: Blackboard – Inductive Statsitics STA2—SPSS-- Week 6 Ch...
The Normal Distribution SPSS Tips APPLE shop in Groningen wants to know whether our IBS students have an intention to buy ...
The Normal Distribution SPSS Tips Step 1: Analyze   Descriptive Statistics   Crosstabs…
The Normal Distribution SPSS Tips Step 2: Move the variable to “Rows” and “Columns” respectively.  Click on “Statistics” a...
The Normal Distribution SPSS Tips Step 3:Click on “Cells” and choose: Observed and Expected.
The Normal Distribution SPSS Tips Now you get the output! But how to interpret it?
The Normal Distribution SPSS Tips <ul><li>In our example, the cross break table (Table 1) has shown the observed and expec...
The Normal Distribution SPSS Tips Now you get the output! But how to interpret it? important
The Normal Distribution SPSS Tips <ul><li>We can reject at the .05 level the null hypothesis that population proportions a...
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Lesson 06 chapter 9 two samples test and Chapter 11 chi square test

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  • Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don&apos;t qualify for favorable rates, then Y , which equals 10 −  X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can&apos;t use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite  P ( X  ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it&apos;s just a matter of looking up the probability in the right place on our cumulative binomial table. To find  P ( Y  ≤ 6), we:  Find  n  = 10  in the first column on the left. Find the column containing  p  = 0.30 . Find the  6  in the second column on the left, since we want to find  F (6) =  P ( Y  ≤ 6). Now, all we need to do is read the probability value where the  p  = 0.30 column and the ( n  = 10, y  = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that  P ( Y  ≤ 6) =  P ( X  ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
  • Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don&apos;t qualify for favorable rates, then Y , which equals 10 −  X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can&apos;t use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite  P ( X  ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it&apos;s just a matter of looking up the probability in the right place on our cumulative binomial table. To find  P ( Y  ≤ 6), we:  Find  n  = 10  in the first column on the left. Find the column containing  p  = 0.30 . Find the  6  in the second column on the left, since we want to find  F (6) =  P ( Y  ≤ 6). Now, all we need to do is read the probability value where the  p  = 0.30 column and the ( n  = 10, y  = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that  P ( Y  ≤ 6) =  P ( X  ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
  • Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don&apos;t qualify for favorable rates, then Y , which equals 10 −  X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can&apos;t use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite  P ( X  ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it&apos;s just a matter of looking up the probability in the right place on our cumulative binomial table. To find  P ( Y  ≤ 6), we:  Find  n  = 10  in the first column on the left. Find the column containing  p  = 0.30 . Find the  6  in the second column on the left, since we want to find  F (6) =  P ( Y  ≤ 6). Now, all we need to do is read the probability value where the  p  = 0.30 column and the ( n  = 10, y  = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that  P ( Y  ≤ 6) =  P ( X  ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
  • Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don&apos;t qualify for favorable rates, then Y , which equals 10 −  X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can&apos;t use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite  P ( X  ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it&apos;s just a matter of looking up the probability in the right place on our cumulative binomial table. To find  P ( Y  ≤ 6), we:  Find  n  = 10  in the first column on the left. Find the column containing  p  = 0.30 . Find the  6  in the second column on the left, since we want to find  F (6) =  P ( Y  ≤ 6). Now, all we need to do is read the probability value where the  p  = 0.30 column and the ( n  = 10, y  = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that  P ( Y  ≤ 6) =  P ( X  ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
  • Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don&apos;t qualify for favorable rates, then Y , which equals 10 −  X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can&apos;t use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite  P ( X  ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it&apos;s just a matter of looking up the probability in the right place on our cumulative binomial table. To find  P ( Y  ≤ 6), we:  Find  n  = 10  in the first column on the left. Find the column containing  p  = 0.30 . Find the  6  in the second column on the left, since we want to find  F (6) =  P ( Y  ≤ 6). Now, all we need to do is read the probability value where the  p  = 0.30 column and the ( n  = 10, y  = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that  P ( Y  ≤ 6) =  P ( X  ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
  • Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don&apos;t qualify for favorable rates, then Y , which equals 10 −  X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can&apos;t use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite  P ( X  ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it&apos;s just a matter of looking up the probability in the right place on our cumulative binomial table. To find  P ( Y  ≤ 6), we:  Find  n  = 10  in the first column on the left. Find the column containing  p  = 0.30 . Find the  6  in the second column on the left, since we want to find  F (6) =  P ( Y  ≤ 6). Now, all we need to do is read the probability value where the  p  = 0.30 column and the ( n  = 10, y  = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that  P ( Y  ≤ 6) =  P ( X  ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
  • Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don&apos;t qualify for favorable rates, then Y , which equals 10 −  X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can&apos;t use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite  P ( X  ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it&apos;s just a matter of looking up the probability in the right place on our cumulative binomial table. To find  P ( Y  ≤ 6), we:  Find  n  = 10  in the first column on the left. Find the column containing  p  = 0.30 . Find the  6  in the second column on the left, since we want to find  F (6) =  P ( Y  ≤ 6). Now, all we need to do is read the probability value where the  p  = 0.30 column and the ( n  = 10, y  = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that  P ( Y  ≤ 6) =  P ( X  ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
  • Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don&apos;t qualify for favorable rates, then Y , which equals 10 −  X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can&apos;t use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite  P ( X  ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it&apos;s just a matter of looking up the probability in the right place on our cumulative binomial table. To find  P ( Y  ≤ 6), we:  Find  n  = 10  in the first column on the left. Find the column containing  p  = 0.30 . Find the  6  in the second column on the left, since we want to find  F (6) =  P ( Y  ≤ 6). Now, all we need to do is read the probability value where the  p  = 0.30 column and the ( n  = 10, y  = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that  P ( Y  ≤ 6) =  P ( X  ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
  • Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don&apos;t qualify for favorable rates, then Y , which equals 10 −  X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can&apos;t use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite  P ( X  ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it&apos;s just a matter of looking up the probability in the right place on our cumulative binomial table. To find  P ( Y  ≤ 6), we:  Find  n  = 10  in the first column on the left. Find the column containing  p  = 0.30 . Find the  6  in the second column on the left, since we want to find  F (6) =  P ( Y  ≤ 6). Now, all we need to do is read the probability value where the  p  = 0.30 column and the ( n  = 10, y  = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that  P ( Y  ≤ 6) =  P ( X  ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
  • Lesson 06 chapter 9 two samples test and Chapter 11 chi square test

    1. 1. Inductive Statistics Dr. Ning DING [email_address] I.007 IBS, Hanze You’d better use the full-screen mode to view this PPT file.
    2. 2. Table of Contents Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    3. 3. Chapter 9 Testing Hypotheses: Two-Sample Tests: Basics-Independent σ is known: σ is unknown: H 0 H 1 n <30 & σ is unknown Two-tailed test One-tailed test
    4. 4. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice Ch 9 No. 9-9 P.466 Step 1: Formulate hypotheses 9-9 Step 2: Find the Pooled Estimate of σ 2 Step 3: Calculate the standard error Step 4: Visualize and Find the t scores One-tailed Test df = 16 area=0.10 t=1.746 Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Gender Mean Standard Deviation Sample size Female 12.8 1.0667 10 Male 11.625 1.4107 8
    5. 5. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    6. 6. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Will the participant lose more than 17 pounds after the weight-reducing program? The survey data is: Step 1: Formulate Hypotheses One-tailed Test Example: Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    7. 7. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Step 2: Calculate the estimated standard deviation of the population difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    8. 8. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Step 3: Find the Standard Error of the population difference Step 4: Calculate the t value Step 5: Visualize and get the t values df = 10-1=9 area = 0.10 t=1.833 One-tailed Test  reject H 0  significant difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    9. 9. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice Ch 9 No. 9-15 P.474 Step 4: Visalize and Calculate the t values t=1.895 9-15 Step 3: Find the Standard Error of the population difference Step 1: Formulate Hypotheses Step 2: Calculate the estimated standard deviation of the population difference df=7 area=0.10 reject H 0 sig difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row One-tailed Test
    10. 10. Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Example: Ch 9 Example P.476 You are testing whether the two drugs cause different blood-pressure levels. The data is as below: Step 1: Formulate Hypotheses Step 3: Calculate the Standard Error of Proportion Difference Step 2: Calculate the Estimated Proportion Difference Two-tailed Test Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    11. 11. Step 4: Visualize and get the z values -1.96 +1.96  Accept H 0  No significant difference Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Ch 9 Example P.476 Two-tailed Test Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    12. 12. Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Step 1: Formulate Hypotheses Step 3: Calculate the Standard Error of Proportion Difference Step 2: Calculate the Estimated Proportion Difference One-tailed Test Example: Ch 9 Example P.480 You are testing whether personal-appearance method causes fewer tax mistakes than mail method. The data is as below: Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    13. 13. Step 4: Visualize and get the z values One-tailed α =0.15  P=0.35  z= -1.04  Accept H 0  No significant difference Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Ch 9 Example P.480 One-tailed Test -1.04 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    14. 14. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice 9-20 Ch 9 No. 9-20 P.483 Step 1: Formulate Hypotheses Step 3: Calculate the Standard Error of Proportion Difference Step 2: Calculate the Estimated Proportion Difference Step 4: Visualize and get the z values z= -1.28  reject H 0  sig difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    15. 15. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice 9-21 Step 1: Formulate Hypotheses Step 3: Calculate the Standard Error of Proportion Difference Step 2: Calculate the Estimated Proportion Difference Ch 9 No. 9-21 P.483 Step 4: Visualize and get the z values  accept H 0  no sig difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    16. 16. Chapter 11 Chi-Square Test-Basics Contingency Table (Cross break table) Rows * Columns == 2*2 table Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row   Male Female Total Junior high school 40 60 100 Senior high school 60 40 100 Total 100 100 200
    17. 17. Chapter 11 Chi-Square Test-Basics Contingency Table (Cross break table) Rows * Columns == 2*2 table Expected and Observed Values Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row   Male Female Total Junior high school 40 60 100 Senior high school 60 40 100 Total 100 100 200   Male Female Total Junior high school 40 (50) 60 (50) 100 Senior high school 60 (50) 40 (50) 100 Total 100 100 200
    18. 18. Chapter 11 Chi-Square Test-Basics Contingency Table (Cross break table) 1*1=1 2*2=4 Degree of freedom= (row-1)*(column-1) 2*2 table 3*3 table Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row   Male Female Total Junior high school 40 60 100 Senior high school 60 40 100 Total 100 100 200   Social Science Nature Science Sports Junior high school 40 60 20 Senior high school 60 40 40 University 50 30 120
    19. 19. Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Chapter 11 Chi-Square Test–Calculate χ 2 for 2-row Table Step 1: Calculate the expected values Ch 11 Example P.570 Example: Our employees’ attitude toward job-performance reviews. There are two review methods, the present one or the new one. Is the attitude dependent on geography? The survey looks like below:
    20. 20. Chapter 11 Chi-Square Test–Calculate χ 2 for 2-row Table Step 2: Calculate the χ 2 2.7644 Ch 11 Example P.570 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    21. 21. Step 3: Find the critical χ 2 Chapter 11 Chi-Square Test–Calculate χ 2 for 2-row Table α = 0.10 2*4  df= 3 χ 2 =6.251 10% χ 2 =2.7644 <ul><li>Accept H 0 </li></ul><ul><li>No sig difference among groups </li></ul>Ch 11 Example P.570 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Acceptance Region
    22. 22. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: Ch 11 Example P.575 Step 1: Formulate the hypotheses H 0 : length of stay and insurance types are independent H 1 : length of stay depends on insurance types α=0.01 Example: Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row   Days in Hospital Cost Cover <5 5-10 >10 Total <25% 40 75 65 180 25-50% 30 45 75 150 >50% 40 100 190 330 Total 110 220 330 660
    23. 23. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: Example: Ch 11 Example P.575 Step 2: Calculate the Expected Frequency For Any Cell RT CT (30) (60) Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row   Days in Hospital Cost Cover <5 5-10 >10 Total <25% 40 75 65 180 25-50% 30 45 75 150 >50% 40 100 190 330 Total 110 220 330 660
    24. 24. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table Ch 11 Example P.575 Step 3: Calculate the Chi-Square Example: For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: 3.33 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row   Days in Hospital   Cost Cover <5 5-10 >10 Total <25% 40 (30) 75 (60) 65 (90) 180 25-50% 30 (25) 45 (50) 75 (75) 150 >50% 40 (55) 100 (110) 190 (165) 330 Total 110 220 330 660
    25. 25. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table Ch 11 Example P.575 Step 3: Calculate the Chi-Square Example: For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: 3.33 3.75 6.94 1.00 4.09 0.50 0.91 0.00 3.79 =3.33+3.75+6.94+1.00+0.50+0.00+4.09+0.91+3.79 =24.32 Chi-Square χ 2 =24.32 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row   Days in Hospital   Cost Cover <5 5-10 >10 Total <25% 40 (30) 75 (60) 65 (90) 180 25-50% 30 (25) 45 (50) 75 (75) 150 >50% 40 (55) 100 (110) 190 (165) 330 Total 110 220 330 660
    26. 26. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table Ch 11 Example P.575 Step 4: Find the critical Chi-Square For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: Chi-Square χ 2 =24.32 Example: 1% α = 0.01 3*3  df= 4 χ 2 =13.277 <ul><li>Reject H 0 </li></ul><ul><li>dependent on each other </li></ul>Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row   Days in Hospital Cost Cover <5 5-10 >10 Total <25% 40 75 65 180 25-50% 30 45 75 150 >50% 40 100 190 330 Total 110 220 330 660 Acceptance Region
    27. 27. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 1: Formulate the hypotheses Step 2: Calculate the Expected Frequency For Any Cell Step 3: Calculate the Chi-Square Step 4: Find the critical Chi-Square Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Economy Weekly Chip Sales High Medium Low Total At peak 20 7 3 30 At Through 30 40 30 100 Rising 20 8 2 30 Falling 30 5 5 40 Total 100 60 40 200
    28. 28. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 1: Formulate the hypotheses H 0 : Sales and economy are independent H 1 : sales depends on economy α=0.10 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Economy Weekly Chip Sales High Medium Low Total At peak 20 7 3 30 At Through 30 40 30 100 Rising 20 8 2 30 Falling 30 5 5 40 Total 100 60 40 200
    29. 29. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 2: Calculate the Expected Frequency For Any Cell Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Economy Weekly Chip Sales High Medium Low Total At peak 20 7 3 30 At Through 30 40 30 100 Rising 20 8 2 30 Falling 30 5 5 40 Total 100 60 40 200 Economy Weekly Chip Sales High Medium Low Total At peak 20 (15) 7 (9) 3 (6) 30 At Through 30 (50) 40 (30) 30 (20) 100 Rising 20 (15) 8 (9) 2 (6) 30 Falling 30 (20) 5 (12) 5 (8) 40 Total 100 60 40 200
    30. 30. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 3: Calculate the Chi-Square = 34.60 10% α = 0.10 4*3  df= 6 χ 2 =10.645 Chi-Square χ 2 =34.60 <ul><li>Reject H 0 </li></ul><ul><li>dependent on each other </li></ul>Step 4: Find the critical Chi-Square Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Acceptance Region Economy Weekly Chip Sales High Medium Low Total At peak 20 (15) 7 (9) 3 (6) 30 At Through 30 (50) 40 (30) 30 (20) 100 Rising 20 (15) 8 (9) 2 (6) 30 Falling 30 (20) 5 (12) 5 (8) 40 Total 100 60 40 200 Economy Weekly Chip Sales High Medium Low Total At peak 1.67 0.44 1.50 30 At Through 8.00 3.33 5.00 100 Rising 1.67 0.11 2.67 30 Falling 5.00 4.08 1.13 40 Total 100 60 40 200
    31. 31. Table of Contents Review: Chapter 8 Testing Hypothesis Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
    32. 32. The Normal Distribution SPSS Tips The data can be downloaded from: Blackboard – Inductive Statsitics STA2—SPSS-- Week 6 Chi-Square.sav
    33. 33. The Normal Distribution SPSS Tips APPLE shop in Groningen wants to know whether our IBS students have an intention to buy an iPad and whether their interest depends on their nationalities. They have interviewed 64 Year TWO students and the data can be downloaded from Blackboard—STA2—SPSS –Chi square.sav.
    34. 34. The Normal Distribution SPSS Tips Step 1: Analyze  Descriptive Statistics  Crosstabs…
    35. 35. The Normal Distribution SPSS Tips Step 2: Move the variable to “Rows” and “Columns” respectively. Click on “Statistics” and choose: Chi-square.
    36. 36. The Normal Distribution SPSS Tips Step 3:Click on “Cells” and choose: Observed and Expected.
    37. 37. The Normal Distribution SPSS Tips Now you get the output! But how to interpret it?
    38. 38. The Normal Distribution SPSS Tips <ul><li>In our example, the cross break table (Table 1) has shown the observed and expected values of the frequency concerning those who are willing to buy an iPad. </li></ul>Table 1: Nationality and Intention to buy an iPad
    39. 39. The Normal Distribution SPSS Tips Now you get the output! But how to interpret it? important
    40. 40. The Normal Distribution SPSS Tips <ul><li>We can reject at the .05 level the null hypothesis that population proportions are equal across the three categories, X 2 ( 3 , N =64) = 0.645 , p =.886 . </li></ul>

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