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AP Stat Chapter 26-27 Review Chi Square & Linear-regression t-test

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1. You wonder if a bag of M & M’s is made up of 1/6 orange, brown, yellow, green, blue & red. Your friend weights each color in the bag and finds 10 grams of orange, 11 grams of brown, 9 grams of yellow, 7 grams of green, 13 grams of blue & 10 grams of red. What could you conclude from running a Chi Squared test of the expected colors of M & M’s? Answer: Can’t do a Chi Squared because M & M’s were not counted. 2.How many df are there for a chi-square test of homogeneity based on a table that has 5 rows & 6 columns? Answer:df = 20 3.Find the number of times you expect each face of a dice to come up if you roll it 90 times. Answer:15

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A new blood pressure medication is tested on 30 volunteers from 5 different age groups. The Results are shown on the table above. 4.If a Chi-square test were done, what kind of test would be conducted? Answer: Chi-square test of homogeneity 5.Find the expected number of successful cases of 40-49 year olds. Answer: (30/150) x 70 = 14 6.Suppose we are testing an SAT prep program. At the end we want to see if the amount of points gained depends on the number of hours a student spent preparing for the SAT. What kind of test would we do? Answer: Linear Regression t-test 20-2930-3940-4950-5960 & over Succeeded in lowering blood pressure 1012172021 Did not succeed in lowering blood pressure 201813109

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FlavorGrapeLemonLimeOrangeStrawberry Frequency530470420610585 Trix cereal comes in five fruit flavors, and each flavor has a different shape. A curious Student methodically sorted an entire box of the cereal & found the distribution of flavors For the pieces of cereal in the box as shown above: Is there evidence that the flavors are evenly distributed? P: The distribution of the different flavors in a box of Trix H: Ho: The flavors are evenly distributed Ha: The flavors in the box are not evenly distributed. A:Expected Counts: 2615/5 = 523 for each flavor ECF’s >5, data is counted data, sample is representative of a typical box of Trix N:Chi-square GOF test T:ECF’s shown above, df = 4

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O: M:Reject Ho at the.05 level S:There is strong evidence that the distribution of Trix is not the same by flavor. Orange & strawberry have more than expected & lemon & lime have less than expected.

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Observed/Expected10 th Graders11 th Graders12 th Graders No Extracurriculars8/17.9615/14.8525/15.19 Sports Team15/11.6010/9.596/9.81 Club12/10.109/8.356/8.55 Band8/5.614/4.643/4.75 Choir5/3.743/3.092/3.17 Other4/2.992/2.472/2.53 A sample of the non-academic interests of students at a high school is shown below. The observed counts are shown versus the expected counts a.Show how this distribution would be changed to meet the ECF condition Observed/Expected10 th Graders11 th Graders12 th Graders No Extracurriculars8/17.9615/14.8525/15.19 Sports15/11.6010/9.596/9.81 Club12/10.109/8.356/8.55 Band/Choir/Other17/12.349/10.207/10.45

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A hypothesis test is run where the null hypothesis is the distribution of non-academic activities is the same across grade levels against the alternative hypothesis that the distribution of non-academic activities is not the same across grade levels. b.What kind of test is this? Answer: χ 2 Homogeneity Test c.The results of the test are shown below. State the conclusion. Answer: The distribution of non-academic activities is not the same across grade levels. It appears that students do less non-academic activities as they progress through high school.

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20-2930-3940-4950-5960 & over Succeeded in lowering blood pressure 1012172022 Did not succeed in lowering blood pressure 201813108 Back to our blood pressure medication example. Test whether the blood pressure medicine Is equally effective in treating all age groups: P:The distribution of effectiveness of blood pressure medication across different age groups H:Ho:The medicine is equally effective for all age groups Ha:The medicine is not equally effective for all age groups

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20-2930-3940-4950-5960 & over Succeeded in lowering blood pressure 16 Did not succeed in lowering blood pressure 14 A:All ECF’s are greater than 5, data is counted, sample is representative N:Chi-square test for homogeneity T:ECF’s shown above, df = 4 O:

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M:Reject Ho at the.05 level S:There is strong evidence that the medicine is more effective for different age groups. It appears that the medicine is more effective in treating older patients.

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A group of students volunteered for a students where they drank a randomly assigned number of cans of beer. 30 minutes later, a police officer measured their blood alcohol content (BAC). A linear regression is run & the output is shown below: 1.Write the equation of the LSRL 2.Test the hypothesis for a linear association between beers consumed & BAC 3.Perform a 95% CI to predict the rate of change between beers consumed & BAC.

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1.Predicted BAC = -.0127006 +.0179638(beers consumed) 2.P: β = linear association between beers consumed & BAC H:H o : β = 0 H a : β ≠ 0 A:The scatterplot is straight. The residual plot is random. The residuals are normally distributed. Each student’s BAC is independent. N:Lin Reg t-test T:n = 16, df = 14, b 1 =.0179628, SE =.002402 O:Draw distribution with t = -7.48 & 7.48, P <.0001 M:Reject Ho at the.05 level S:There is strong evidence of a linear association between beers consumed & BAC

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3.N: Lin Reg CI I: CI =.0179638 +- 2.145(.002402) CI =.0179638 +-.0052 (.0128,.0232) We are 95% confident that as BAC will increase between.01238 &.0232 for each additional beer consumed.

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