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## AP Statistics Thursday , 24 April 2014

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**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**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.**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.**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?**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**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.**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.**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.**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.**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**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**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