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AP Statistics Thursday , 24 April 2014

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)

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AP Statistics Thursday , 24 April 2014

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  1. AP StatisticsThursday, 24 April 2014 • OBJECTIVETSW 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 • χ2, 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 • χ2 test • Wear your Falcon Statisticst-shirt. Questions?

  5. 1) E 2) E 3) A 4) C 5) D 6) C 7) D 8) C 9) B 10) D 11) A 12) E χ2 AP Review

  6. χ2Worksheet 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. H0: The observed counts of random digits equal the expected counts of random digits Ha: The observed counts of random digits are not equal to the expected counts of random digits χ2= 18, df = 9, p = 0.03517,  = 0.05 p <  ∴Reject H0. There is sufficient evidence to suggest that the observed counts of random digits are not equal to the expected counts of random digits.

  7. χ2Worksheet 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. H0: Anxiety level and the need to succeed are independent Ha: Anxiety level and the need to succeed are not independent (or are dependent) χ2= 29.6, df = 4, p-value = 0.000005903,  = 0.05 p <  ∴Reject H0. There is sufficient evidence to suggest that anxiety level and the need to succeed are dependent.

  8. χ2Worksheet 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. H0: The counts of observed scores on the AP test equal the counts of expected scores on the AP test. Ha: The counts of observed scores on the AP test do not equal the counts of expected scores on the AP test. χ2= 162.91, df = 4, p = 0,  = 0.05 p <  ∴Reject H0. 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.

  9. χ2Worksheet 4)χ2= 16.7, df = 4, p = 0.0022,  = 0.05 p <  ∴Reject H0. There is sufficient evidence to suggest that there is an association between the number of hours volunteered and the type of volunteer. 5) χ2 = 628.08, df = 8, p = 0,  = 0.05 p <  ∴Reject H0. There is sufficient evidence to suggest that the proportion of adults who would give each answer is different for each country. 6) χ2 = 16.43, df = 5, p = 0.0057,  = 0.05 p <  ∴Reject H0. There is sufficient evidence to suggest that the distribution of peanut M&M’s is different from the distribution of plain M&M’s.

  10. Mixed Hypothesis Review #2 1) t = 3.06 p-value = .0019 df = 39 2) t = 1.82 p-value = .0368 df = 65.71 3)χ2= 5.96 p-value = .1137 df = 3 4) t = -3.06 p-value = .0030 df = 21 5) z = .95 p-value = .3445 6)χ2= 8.6 p-value = .1261 df = 5 7) t = 2.40 p-value = .0370 df = 4 8) z = 1.40 p-value = .1623

  11. Mixed Hypothesis Review #3 1) t = 1.89 p-value = .0702 df = 25.32 2)(0.298, 1.102) df = 49.07 3)χ2= 9.23 p-value = .1002 df = 5 4) t = 2.437 p-value = .0224 df = 7 5) t = 2.358 p-value = .0192 df = 234 6) z = .758 p-value = .2242

  12. D D A B C A 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 H0: The direction that a lost hiker tends to walk is independent of the hiker’s experience level. Ha: 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 H0. There is not sufficient evidence to suggest that the direction a lost hiker would tend to walk is dependent on his/her experience level. QUIZ: Chi-Square

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