? Alt marijuana. The result s are given in the following table. 1500 people to A

Question

? Alt marijuana. The result s are given in the following table. 1500 people to AGES LEGALIZATION OF MARIJUANA 17-27 28 38 39- 49 FOR a) 170 d) 155 GAINST UNDECIDED c) 130 D125 b) 100 Row Totals 130 50 & Over Column Totals b) 100 k) 130 400 500 1500 20. The alternative hypothesis is: a) A person's age and their view on the legalization of marijuana are independent. b) A person's age and their view on the legalization of marijuana have a negative relationship A person's age and their view on the legalization of marijuana have a weak relationship. d) A person's age and their view on the legalization of marijuana are dependent 21· The expected value for cell k is: a) 130 b) 133.33 c) 140 d 116.67 e 106.67 22. Degrees of Freedom is: b)8 a)12 c)6 d) 4 e)10 23. At ?-190, the critical value for this test is: a) 12.59 b) 16.81 c) 21.03 d) 15.51 24. The value of the chi-square test statistic is: a) 18.21 b) 17.14 c) 17.48 d)21.89 25. Can the social researcher conclude that age and legalization of marijuana are depender a) yes b) no c) undecided 26. At a large public state college in Virginia, the mean Verbal SAT score of all ater ? 600 witha population standard deviation: ? 96. A random sample of 64 students! population of students. What is the probability that the sample mean Verbal SAT score is at most 620? e) d) 41.69% c) 58.31% b) 95.25% a) 4.75% utinaluney of High School students of driving

Explanation / Answer

20)

option D is correct

21)

expected value for cell k=(row total*column total)/grand total=106.67 ; option E

22)

degree of freedom=(row-1)*(column-1)=(4-1)*(3-1)=6

option C

23)

critical value =16.81

option B

24)

value of chi square test statistic =17.48

option C

25)

Yes

26)

P(Xbar<620)=P(Z<(620-600)*sqrt(64)/96)=P(Z<1.67)=95.25%

option B

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