A using the 05 level of significance test the null

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(a) Using the .05 level of significance, test the null hypothesis that survival rates are independent of the passengers’ accommodations (cabin or steerage). Statistical hypothesis: H 0 : type of accommodations and survival rates are independent H 1 : H 0 is false Decision Rule: We will reject the null hypothesis at 0.05 level of significance if x 2 > 3.84, given the degrees of freedom:
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TOPIC 7 EXERCISES = (c-1) (r-1) = (2-1) (2-1) = 1 f e ( columntotal )( rowtotal ) grand total f e (cabin, survived) = ( 579 )( 485 ) 1291 = 280815 1291 = 217.52 f e (steerage, survived) = ( 712 )( 485 ) 1291 = 345320 1291 = 276.48 f e (cabin, not survived) = ( 579 )( 806 ) 1291 = 466674 1291 = 361.48 f e (steerage, not survived) = ( 712 )( 806 ) 1291 = 573872 1291 = 444.52 Survived/Accommodations Cabin Steerage Total Yes f 0 f e 299 217.52 186 276.48 485 No f 0 f e 280 361.48 526 444.52 806 Total 579 712 1291 X 2 = ( 299 217.52 ) 2 217.52 + ( 186 267.48 ) 2 267.48 + ( 280 231.48 ) 2 361.48 + ( 516 444.52 ) 2 444.52 = 6638.99 217.52 + 6638.99 267.48 + 6638.99 361.48 + 6638.99 444.52 = 30.52 + 24.82 + 18.37+ 14.94 = 88.65 The null hypothesis will be rejected because x 2 of 88.65 > critical x 2 of 3.84. There seems to be a correlation between the type of accommodation and the survival rate .
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TOPIC 7 EXERCISES (b) Assuming a significant c2, estimate the strength of the relationship. Ø 2 c = 88.65 1291 ( 2 1 ) = 88.65 1291 = 0.07 Strength of relationship = medium (c) To more fully appreciate the importance of this relationship, calculate an odds ratio to determine how much more likely a cabin passenger is to have survived than a steerage passenger. OR = 299 / 280 186 / 526 = 1.07 0.35 = 3.06 The findings note, a cabin passenger is 3.06 times more likely to survive than a steerage passenger. Chapter 19 question 19.14 In a classic study, Milgram et al. “lost” stamped envelopes with fictitious addresses (Medical Research Association, Personal Address, Friends of Communist Party, and Friends of Nazi Party).* One hundred letters with each address were distributed among four locations (shops, cars, streets, and phone booths) in New Haven Connecticut, with the following results: (a) Using the .05 level of significance, test the null hypothesis that address does not matter in the underlying population. f e = ( row total )( columntotal ) Grandtotal f e (medical, returned) = ( 100 )( 193 ) 400 = 19300 400 = ¿ 48.25 f e (personal, returned) = ( 100 )( 193 ) 400 = 19300 400 = ¿ 48.25 f e (communist, returned) = ( 100 )( 193 ) 400 = 19300 400 = ¿ 48.25
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TOPIC 7 EXERCISES f e (Nazi, returned) = ( 100 )( 193 ) 400 = 19300 400 = ¿ 48.25 f e (medical, not returned) = ( 100 )( 207 ) 400 = 20700 400 = ¿ 51.75 f e (personal, not returned) = ( 100 )( 207 ) 400 = 20700 400 = ¿ 51.75 f e (communist, not returned) = ( 100 )( 207 ) 400 = 20700 400 = ¿ 51.75 f e (Nazi, not returned) = ( 100 )( 207 ) 400 = 20700 400 = ¿ 51.75 x 2 = ( 72 48.5 ) 2 48.5 + ( 71 48.5 ) 2 48.5 + ( 25 48.5 ) 2 48.5 + ( 25 48.5 ) 2 48.5 + ( 28 51.75 ) 2 51.75 + ( 29 51.75 ) 2 51.75 + ( 75 51.75 ) 2 51.75 + ( 75 51. 51.75 = 11.69 + 10.73 + 11.2 + 11.2 + 10.9 + 10 + 10.44 + 10.44 = 86.6 df = (c-1)(r-1)=3 Reject the null hypothesis (b) p <0.05. (c) Assuming c2 is significant; estimate the strength of this relationship. Ø 2 = x 2 n ( k 1 ) = 86.6 400 ( 2 1 ) = 0.22 (d) How might these results be reported in the literature? There appears to be a relationship between addresses and the number of letters returned. (e) Collapse the original 4 x 2 table to a 2 x 2 table by combining the results for the two neutral addresses and for the two inflammatory addresses. Calculate the odds ratio for returned letters.
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TOPIC 7 EXERCISES OR = 143 / 57 50 / 150 = 2.51 0.33 = 7.6 Chapter 19 question 19.16 A social scientist cross-classifies the responses of 100 randomly
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