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Statistics

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Apr 3, 2024

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Statistics Assignment Page 1 of 2 ChiSquare Hypothesis Tests for Association or Independence 1. Calculate the expected values for the cells marked A, B, and C in the following table: (2 points) A=(120/386) (104)=32.33 1 0 4 B=(144/386) (170)=63.42 1 7 0 C=(122/386) (112)=35.4 1 1 2 120 144 122 3 8 6 2. You decide to research a certain topic. You collect data and organize them into a 5 X 7 table. You find that a chisquare test would be valid for your data, and find that this table has a X 2 of 34.5. df=4*6=24 A. What conclusion would you draw about the relationship between these two variables if α = .05? (2 points) Since .10>p>.05, we conclude that we cannot reject the null hypothesis that the data are independent. B. What conclusion would you draw about the relationship between these two variables if α = .10? (2 points) Since .1>p, we have enough evidence to reject the null that the data are independent. Therefore, we have evidence to support the alternate that the data are associated. 3. Calculate degrees of freedom, X 2 , and an exact P value for the following table: (2 points) 79 93 103 44 68 99 df=(3-1)(2-1)=2 X2 = 30.2, df=2, p=2.77E-7, or .00000028 4. You want to know if there's an association between college students' spring break destinations and what year they're in. You take a random sample of 405 college students and record the following data: Amusement Parks Mexico Home Other Freshman 23 21 43 21 Sophomore 34 23 14 26 Junior 25 30 23 26 Senior 27 33 17 19 A. Set up your null and alternative hypotheses. (2 points) H0: the variables are independent (There is no association between college students' spring break destinations and what year they're in.) Ha: the variables are not independent (There is an association between college students’ spring break destinations and what year they’re in.)

B. What are your degrees of freedom? (2 points) df=(4-1)(4-1)=3*3= 9, df=9 C. Using your TI83, compute the expected values for each cell. (2 points) EXPECTED VALUES TABLE Amusement Parks Mexico Home Other Freshman 29.07 28.53 25.87 24.53 Sophomore 26.12 25.63 23.23 22.03 Junior 27.99 27.63 24.91 23.62 Senior 25.84 25.36 23 21.81 D. Will a chisquare hypothesis test of independence be valid in this situation? Explain. (2 points) Yes; the expected values are greater than at least 5. E. What are your test statistic ( X 2 ) and P value? (2 points) X2=27.36 , p-value=.00122 F. What can you conclude if α = .01? (2 points) We have evidence, since p< α , that allows us to reject the null hypothesis which stated that the variables are independent, or that there is no association between college students' spring break destinations and what year they're in. Thus, we have evidence to support the alternative that there is an association between college students’ spring break destinations and what year they’re in. 5. A hotel wants to know if there's a relationship between gender and the way customers make room reservations. A manager takes a random sample of 160 reservations. She records whether they were made by a man or a woman, and also records how the reservation was made. She gets the following data: Phone Fax Email Men 28 9 37 Women 45 12 29 Perform a chisquare significance test of association with α = .05. Be sure to include your null and alternative hypotheses, a justification for the use of this test, your test statistic calculations, your P value, and your conclusion. (5 points) H0: the variables are independent (There's no relationship between gender and the way customers make room reservations.) Ha: the variables are not independent (There's a relationship between gender and the way customers make room reservations.) We can use chi-square test because expected values are greater than 5. Table of expected values: Phone Fax Email Men 33.76 9.71 30.53 Women 39.24 11.29 35.48 df=(2-1)(3-1)=2, X2=4.48, p-value=.106

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