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i. Using the 5-step process, test the null hypothesis that the two variables are independent. Use an alpha level of 0.01, and state each step
i. Using the 5-step process, test the null hypothesis that the two variables are independent. Use an alpha level of 0.01, and state each step of your hypothesis test. (25 points) i. Step 1 ii. Step 2 iii. Step 3 iv. Step 4 v. Step 53. Hirschi [1969) examined the association between the intimacy of a boy's communication with his father and the number of the boy's iends who had been picked up by police. Hirschi hypothesized that intimate communications with fathers would reduce the number of delinquent friends. The table below.r summarizes Hirschi's analysis. Intimacy of Number of Friends Picked U n - Police Communication High 191:! 35 63 Medium 1 34 3 3 1'? Lour 3? 53 33 a. \"mat are the independent and dependent variables? (3 points) Independent Intimacy of Communication Dependent No. of Friends Picked Up by Police b. How many observations or cases are there? [3 points] Number of observations = Sum of all cells There are Ql observations or cases. c. 'What are the column marginals? (3 points) None =19t] +134 + 31' =4l. neTwo = 35 + 33 + 53 = 225. Three ormore= 53 +22 +33 =223. The colnmn marginals are: None = 461, DueTwo = 22d, and three or more = 223. d. What are the row marginals'?r {3 points) High = 1913 + 35 + 53 = 333. Medium =134 + 83 + T? = 349. Low: 3? + 53 + 33 = 223. The row marginals are: High = 33.3:| hiedinm = 349', and LOW = 223. e. What is the size of this contingency table? (3 points) The size of this contingency table is 333. f. How mart}r boys had high intimacy.r of communication and no friends picked up by police? [3 points} The number of hogs had high intimacy of communication and no friends picked up by police is 190. g. How mam,r boys had loin.r intimacy of communication and no friends picked up by police? (3 points) The number of boys had 10\"! intimacy of communication and no friends picked up by police is 37. h. How mart}r degrees of 'eedom are there in this table? (3 points) df= (total row mitotal column - 1} Substitute: df: (3 - mo -1)- df= 2*2. df= 4. There are 4 degrees of freedom in the table
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