1. 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 – counts as a test grade. –Covers the entire school year. –Thursday, 01 May 2014: Part I (25 questions) –Friday, 02 May 2014: Part II (25 questions) AP TEST –Friday Afternoon, 09 May 2014. COOKOUT: Monday, 19 May 2014 –$3.00, due by Wednesday, 14 May 2014

2. TEST TOPICS: Chi-Square Inference Test Identify the characteristics of the three types of chi-square inference tests. –Goodness of Fit –Independence –Homogeneity Determine degrees of freedom. Given a set of data, determine the chi-square test statistic using the formula.

3. TEST TOPICS: Chi-Square Inference Test Perform a full write-up chi-square test. –Assumptions –Hypotheses –Necessary values χ 2, p-value, df, α –Compare p and α –Conclusion Given a statistic situation, identify the Type I and Type II errors.

4. TEST TOPICS: Chi-Square Inference Test Identify when to use – –1-sample t test on means (Matched Pairs Test) –2-sample t test on means –1-sample z test on proportions –2-sample z test on proportions – χ 2 test Wear your Falcon Statistics t-shirt.Questions?

5. χ 2 AP Review 1)E 2)E 3)A 4)C 5)D 6)C 7)D 8)C 9)B 10)D 11)A 12)E

6. χ 2 Worksheet 1)Assumptions: * Reasonably random sample * We have counts of last digits & we expect each digit to occur at least once * 50/10 = 5 ≥ 5  approximately normal distr. H 0 : The observed counts of random digits equal the expected counts of random digits H a : The observed counts of random digits are not equal to the expected counts of random digits χ 2 = 18, df = 9, p = 0.03517,  = 0.05 p <  ∴ Reject H 0. There is sufficient evidence to suggest that the observed counts of random digits are not equal to the expected counts of random digits.

7. 2)Assumptions: * Reasonable SRS * We have counts of anxiety levels and the levels of the need to succeed & we expect each to occur at least once * Expected values: 18.75, 37.5, 18.75, 25, 50, 25, 6.25, 12.5, 6.25 all are ≥ 5  approx. normal distr. H 0 : Anxiety level and the need to succeed are independent H a : Anxiety level and the need to succeed are not independent (or are dependent) χ 2 = 29.6, df = 4, p-value = 0.000005903,  = 0.05 p <  ∴ Reject H 0. There is sufficient evidence to suggest that anxiety level and the need to succeed are dependent. χ 2 Worksheet

8. 3)Assumptions: * Reasonable SRS * We have counts of each of the scores on the AP test and we expect each to occur at least once * Expected values: 81.855, 117.7, 132.68, 105.93, 96.835 all are ≥ 5  approx. normal distr. H 0 :T he counts of observed scores on the AP test equal the counts of expected scores on the AP test. H a :T he counts of observed scores on the AP test do not equal the counts of expected scores on the AP test. χ 2 = 162.91, df = 4, p = 0,  = 0.05 p <  ∴ Reject H 0. There is sufficient evidence to suggest that the counts of observed scores on the AP test do not equal the counts of expected scores on the AP test. χ 2 Worksheet

9. 4) χ 2 = 16.7, df = 4, p = 0.0022,  = 0.05 p <  ∴ Reject H 0. There is sufficient evidence to suggest that there is an association between the number of hours volunteered and the type of volunteer. 5) χ 2 = 628.08, df = 8, p = 0,  = 0.05 p <  ∴ Reject H 0. There is sufficient evidence to suggest that the proportion of adults who would give each answer is different for each country. 6) χ 2 = 16.43, df = 5, p = 0.0057,  = 0.05 p <  ∴ Reject H 0. There is sufficient evidence to suggest that the distribution of peanut M&M’s is different from the distribution of plain M&M’s. χ 2 Worksheet

10. Mixed Hypothesis Review #2 1) t = 3.06p-value =.0019df = 39 2) t = 1.82p-value =.0368df = 65.71 3) χ 2 = 5.96p-value =.1137df = 3 4) t = -3.06p-value =.0030df = 21 5) z =.95p-value =.3445 6) χ 2 = 8.6p-value =.1261df = 5 7) t = 2.40p-value =.0370df = 4 8) z = 1.40p-value =.1623

11. Mixed Hypothesis Review #3 1) t = 1.89 p-value =.0702df = 25.32 2) (0.298, 1.102)df = 49.07 3) χ 2 = 9.23p-value =.1002df = 5 4) t = 2.437p-value =.0224df = 7 5) t = 2.358p-value =.0192df = 234 6) z =.758p-value =.2242

12. QUIZ: Chi-Square 1)D 2)D 3)A 4)B 5)C 6)A 7)Assumptions: *Reasonable SRS *We have counts of lost hikers and where they tend to walk and we expect each to occur at least once. *Expected counts: 18, 48, 54, 12, 32, 36 ≥ 5 H 0 : The direction that a lost hiker tends to walk is independent of the hiker’s experience level. H a : The direction that a lost hiker tends to walk is dependent on his/her experience level. df = 2, p = 0.47127,  = 0.05 p >  ∴ Fail to reject H 0. There is not sufficient evidence to suggest that the direction a lost hiker would tend to walk is dependent on his/her experience level.