WARM UP Pick the 4 papers from back of room and then check your answer to 12 (d-i)

Slides:



Advertisements
Similar presentations
Chi-square test Chi-square test or  2 test. Chi-square test countsUsed to test the counts of categorical data ThreeThree types –Goodness of fit (univariate)
Advertisements

AP Statistics Tuesday, 15 April 2014 OBJECTIVE TSW (1) identify the conditions to use a chi-square test; (2) examine the chi-square test for independence;
By Josh Spiezle, Emy Chinen, Emily Lopez, Reid Beloff.
Chi-Squared Hypothesis Testing Using One-Way and Two-Way Frequency Tables of Categorical Variables.
CHAPTER 23: Two Categorical Variables: The Chi-Square Test
Chapter 11 Inference for Distributions of Categorical Data
Chapter 13: Inference for Distributions of Categorical Data
AP Statistics Thursday, 24 April 2014 OBJECTIVE TSW review for tomorrow’s Chi-Square Inference test. DUAL CREDIT FINAL: NEXT WEEK –Everyone will take this.
Copyright ©2011 Brooks/Cole, Cengage Learning More about Inference for Categorical Variables Chapter 15 1.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. More About Categorical Variables Chapter 15.
1 Chi-Squared Distributions Inference for Categorical Data and Multiple Groups.
CHAPTER 11 Inference for Distributions of Categorical Data
Stat 217 – Day 27 Chi-square tests (Topic 25). The Plan Exam 2 returned at end of class today  Mean.80 (36/45)  Solutions with commentary online  Discuss.
Chi-square Goodness of Fit Test
1 Chapter 20 Two Categorical Variables: The Chi-Square Test.
Presentation 12 Chi-Square test.
Chapter 13 Chi-Square Tests. The chi-square test for Goodness of Fit allows us to determine whether a specified population distribution seems valid. The.
Chi-Square Distributions
Analysis of Count Data Chapter 26
AP Statistics Section 14.2 B. We can use the chi-square test of association/independence to test the null hypothesis when you have a two-way table from.
Copyright © 2013 Pearson Education, Inc. All rights reserved Chapter 10 Inferring Population Means.
AP Statistics Chapter 26 Notes
By: Jackie, Molly & Franny Hey What’s up? What’s your Favorite Color? TEXT REACTION.
Chi-square test or c2 test
The Practice of Statistics Third Edition Chapter 13: Comparing Two Population Parameters Copyright © 2008 by W. H. Freeman & Company Daniel S. Yates.
Warm-up 10.3 Chi-Square test of Independence
Chi-square test Chi-square test or  2 test Notes: Page Goodness of Fit 2.Independence 3.Homogeneity.
Chapter 26 Chi-Square Testing
Chi-Square Procedures Chi-Square Test for Goodness of Fit, Independence of Variables, and Homogeneity of Proportions.
Chi-Square Distributions. Recap Analyze data and test hypothesis Type of test depends on: Data available Question we need to answer What do we use to.
Chapter 11 The Chi-Square Test of Association/Independence Target Goal: I can perform a chi-square test for association/independence to determine whether.
13.2 Chi-Square Test for Homogeneity & Independence AP Statistics.
+ Chi Square Test Homogeneity or Independence( Association)
Chapter 14: Chi-Square Procedures – Test for Goodness of Fit.
Chapter 11 Chi- Square Test for Homogeneity Target Goal: I can use a chi-square test to compare 3 or more proportions. I can use a chi-square test for.
Statistical Significance for a two-way table Inference for a two-way table We often gather data and arrange them in a two-way table to see if two categorical.
Inference for Distributions of Categorical Variables (C26 BVD)
Test of Independence Lecture 43 Section 14.5 Mon, Apr 23, 2007.
1 Chapter 10. Section 10.1 and 10.2 Triola, Elementary Statistics, Eighth Edition. Copyright Addison Wesley Longman M ARIO F. T RIOLA E IGHTH E DITION.
Slide 1 Copyright © 2004 Pearson Education, Inc..
AP STATISTICS LESSON (DAY 1) INFERENCE FOR TWO – WAY TABLES.
AGENDA:. AP STAT Ch. 14.: X 2 Tests Goodness of Fit Homogeniety Independence EQ: What are expected values and how are they used to calculate Chi-Square?
Section 12.2: Tests for Homogeneity and Independence in a Two-Way Table.
Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Chapter 11 Analyzing the Association Between Categorical Variables Section 11.2 Testing Categorical.
Lecture PowerPoint Slides Basic Practice of Statistics 7 th Edition.
Chi Square Tests PhD Özgür Tosun. IMPORTANCE OF EVIDENCE BASED MEDICINE.
By.  Are the proportions of colors of each M&M stated by the M&M company true proportions?
+ Section 11.1 Chi-Square Goodness-of-Fit Tests. + Introduction In the previous chapter, we discussed inference procedures for comparing the proportion.
WARM – UP: The Math club and the Spanish club traditionally are composed of a similar distribution of class level. A random sample of this year’s math.
Section 13.2 Chi-Squared Test of Independence/Association.
The Practice of Statistics, 5th Edition Starnes, Tabor, Yates, Moore Bedford Freeman Worth Publishers CHAPTER 11 Inference for Distributions of Categorical.
Chi-Squared Test of Homogeneity Are different populations the same across some characteristic?
Chapter 11: Categorical Data n Chi-square goodness of fit test allows us to examine a single distribution of a categorical variable in a population. n.
13.2 Inference for Two Way Tables.  Analyze Two Way Tables Using Chi-Squared Test for Homogeneity and Independence.
AP Statistics Tuesday, 05 April 2016 OBJECTIVE TSW (1) identify the conditions to use a chi-square test; (2) examine the chi- square test for independence;
12.2 Tests for Homogeneity and Independence in a two-way table Wednesday, June 22, 2016.
Goodness-of-Fit and Contingency Tables Chapter 11.
Chi Square Procedures Chapter 14. Chi-Square Goodness-of-Fit Tests Section 14.1.
AP Stats Check In Where we’ve been… Chapter 7…Chapter 8… Where we are going… Significance Tests!! –Ch 9 Tests about a population proportion –Ch 9Tests.
Chapter 12 Lesson 12.2b Comparing Two Populations or Treatments 12.2: Test for Homogeneity and Independence in a Two-way Table.
 Check the Random, Large Sample Size and Independent conditions before performing a chi-square test  Use a chi-square test for homogeneity to determine.
Chi-Square hypothesis testing
X2 = X2 Test of Independence P-Value =
AP Stats Check In Where we’ve been… Chapter 7…Chapter 8…
Does ones regional location have an affect on their Political affiliation? To begin to investigate this situation data from 177 random voters was analyzed.
AP Stats Check In Where we’ve been… Chapter 7…Chapter 8…
Chapter 11: Inference for Distributions of Categorical Data
Inference for Relationships
Location and Party affiliation
Lesson 11 - R Chapter 11 Review:
Presentation transcript:

WARM UP Pick the 4 papers from back of room and then check your answer to 12 (d-i)

WARM-UP: Examine whether the grade you got on the AP Statistics Test is independent of what class period you are in. ABCF 1 st Period rd Period th Period16843 X 2 Test of Independence P-Value = X 2 cdf (8.75, E99, 6) = H 0 : Test Grades and Class Period are independent. H a : The Grade you earn on the Test is associated to what class period you are in. X 2 = Since the P-Value is NOT less than α = 0.05 you will Fail to reject H 0. There is no evidence to conclude that Test Grades and Class Period are related. 1.SRS - Stated X 2.All Expected Counts are 1 or greater. √ 3.No more than 20% of the Expected Counts are less than 5. X

The Chi-Square Test for Homogeneity A test comparing the distribution of counts for TWO or MORE Populations on the ONE categorical variable. -GOF tests only one Population on only ONE categorical variable. -Homogeneity represents Multiple GOF tests. df = (#Rows – 1) x (#Cols. – 1) H 0 : The distribution of the one variable is equivalent among the populations H a : The distribution of the one variable is NOT equivalent among the populations

P-Value = X 2 cdf (X 2, E99, df) NOTE: The Chi-Square Tests for Homogeneity and for Independence are performed exactly the same way!

WARM-UP: Examine whether the distribution of grades is equivalent for each period of AP Statistics. X 2 Test of HOMOGENEITY P-Value = X 2 cdf (8.75, E99, 6) =.1881 H 0 : The distribution of A,B,C, and F’s in all three periods is spread equivalently. H a : The distribution of A,B,C, and F’s in all three periods is NOT spread equivalently. X 2 = 8.75 Since the P-Value is NOT less than α = 0.05 we fail to reject H 0. There is no evidence to conclude that Test Grades are NOT distributed equally among the 3 classes. 1.SRS - Stated X 2.All Expected Counts are 1 or greater. √ 3.No more than 20% of the Expected Counts are less than 5. X ABCF 1 st Period rd Period th Period

An SRS of 120 voters from AR and an SRS of 115 voters from TX was taken to determine whether there was a significant difference in how people, as of that moment, would vote with regards to Obama. Definitely Would Mostly Likely Probably would Not Definitely Would Not Arkansas Texas X 2 = X 2 Test ofHomogeneity P-Value = X 2 cdf (11.277, E99, 3) = H 0 : The Distribution of how people would vote today in the State of Arkansas is equal to that of Texas. H a : The Distribution of how people would vote today in the State of Arkansas is NOT equal to that of Texas Since the P-Value is less than α = 0.05 the data IS significant. REJECT H 0. Support is different between AR and TX. 1.SRS – stated 2.All Expected Counts are 1 or greater. 3.No more than 20% of the Expected Counts are less than 5.

#18 Medical researchers followed an SRS of 6272 Swedish men for 30 years to see if there was an association between the amount of fish in their diet and Prostate Cancer. Is there any evidence of such an association? Fish Consumption Total Subjects Prostate Cancer Never12414 Small part of diet Moderate part Large part54942 NO Prostate Cancer X 2 Test ofIndependence P-Value = X 2 cdf (3.677, E99, 3) = H 0 : There is NO relationship between fish consumption and the development of Prostate Cancer. H a : There is relationship between fish consumption and the development of Prostate Cancer. X 2 = 3.677

Fish Consumption Total Subjects Prostate Cancer Never12414 Small part of diet Moderate part Large part54942 NO Prostate Cancer X 2 Test ofIndependence P-Value = X 2 cdf (3.677, E99, 3) = H 0 : There is NO relationship between fish consumption and the development of Prostate Cancer. H a : There is relationship between fish consumption and the development of Prostate Cancer. X 2 = Since the P-Value is NOT less than α = 0.05 there is NO evidence to reject H 0. There is NO relationship between fish consumption and Prostate Cancer. CONDITIONS 1.SRS - Stated √ 2.All Expected Counts are 1 or greater. √ 3.No more than 20% of the Expected Counts are less than 5. √ WARM – UP Medical researchers followed 6272 Swedish men for 30 years to see if there was an association between the amount of fish in their diet and Prostate Cancer. Is there any evidence of such an association?

WARM – UP Does ones regional location have an affect on their Political affiliation? To begin to investigate this situation data from 177 voters was analyzed. DemocratRepublican West3927 Northeast3515 Southeast1744 Political Affiliation Location a.) Find the Proportion of Democrats in each region. Democrats in each region. b.) Make a Bar Chart for the Prop. c.) Find the Expected Values for each cell. each cell % of Dem N NW SE Regional Location

Homework: Page 630: #15 omit h, 16, 17, 20, 21

EXAMPLE: Is the Distribution of colors in a package of PLAIN M&M’s statistically equivalent to the Distribution of colors in a package of PEANUT M&M’s? A random package of plain and peanut M&M’s are selected and analyzed. BrownBlueOrangeGreenRedYellow PLAIN PEANUT X 2 Test of Homogeneity P-Value = X 2 cdf (4.967, E99, 5) = H 0 : The Distribution of colors in the Plain Packet of M&M’s is equivalent to that of the Peanut M&M’s. H a : The Distribution of colors in the Plain Packet of M&M’s is NOT equivalent to that of the Peanut M&M’s. X 2 =

EXAMPLE: Is the Distribution of colors in a package of PLAIN M&M’s statistically equivalent to the Distribution of colors in a package of PEANUT M&M’s? A random package of plain and peanut M&M’s are selected and analyzed. BrownBlueOrangeGreenRedYellow PLAIN PEANUT X 2 Test of Homogeneity P-Value = X 2 cdf (4.967, E99, 3) = H 0 : The Distribution of colors in the Plain Packet of M&M’s is equivalent to that of the Peanut M&M’s. H a : The Distribution of colors in the Plain Packet of M&M’s is NOT equivalent to that of the Peanut M&M’s. X 2 = Since the P-Value is NOT less than α = 0.05 there is NO evidence to reject H 0. No evidence that the Distributions are NOT equivalent. Although the results are uncertain. CONDITIONS 1.SRS - Stated √ 2.All Expected Counts are 1 or greater. X 3.No more than 20% of the Expected Counts are less than 5. X