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KEY: In-Class Questions from Old Examinations
MSC 287 Dr. Stafford Spring 2004
Question 1
Item Answer Explanation 1 R Weight is a ratio- scaled variable because it is continuous, zero has meaning, and the ratio has meaning. 2 I (^) Interval scale is appropriate for shoe size 3 N Jersey number is just a label; it implies no order or ranking. Thus it is a nominal scaled variable 4 O Rank ordering is implied by the number of stars. Thus ordinal is the appropriate scale 5 D shirt size and scale of satisfaction are ordinal-scaled variables; thus answer is D = two 6 E sports shoe size, score on SAT exam, and ambient temperature are all interval-scaled variables; thus E = some other value is the correct answer 7 C (^) Only number of people in line is a ratio-scaled variable; thus the answer is C = one 8 R Since mean and standard deviation require division , ratio scale is the appropriate one 9 O Shoe size is a discrete interval-scaled variable; number of students waiting in line is a discrete ratio-scaled variable 10 T We have only a finite, discrete number of classes into which we put such variables
Question 2
The cough-syrup-stained “frequency table” is shown below. Let us first compute each of the letter values.
Class A B C D E F G H J 4 - 6 K^ L^ 0.30 M N 20 56 0.25 0. 10 - 12 16 R S 0. T U V W X Total Y Z
Question 2 {continued}
We know that A, B, C, and D are column headings, and that each will be one of fi , cf (^) i , rf (^) i , or crfi. Since the A and B columns contain counts rather than frequencies, these columns are the fi and cfi columns; but which is which? The cfi values are ever increasing in value as we move down the column. In column A, we have a value of 20 before a value of 16; thus A cannot be the cumulative frequency, and therefore must be the frequency column, or A = f (^) i. Then B = cf (^) i. The crfi values are ever increasing in value as we move down the column. In column C, we have a value of 0.30 before a value of 0.25; thus C cannot be the cumulative relative frequency, and therefore must be the relative frequency column, or C = rf (^) i. Then D = crfi. Examining the “Class” column, it should be clear that: E = “1-3"; N = “7- 9"; and T = “13-15". At this point in the puzzle solution, the table looks like the next table below.
Class M (^) i f (^) i cfi rf (^) i crf (^) i 1 - 3 2 F G H J 4 - 6 5 K L 0.30 M 7 - 9 (^8) 20 56 0.25 0. 10 - 12 11 16 R S 0. 13 - 15 14 U V W X Total Y Z
We know that the sum of the relative frequencies is 1.0, so Z = 1.00 and X = 1.00. Let us next use the HINT: crfi = crf (^) i-1 + rf (^) i. From this, X = 0.9 + W, or W = 1.0 - 0.9 = 0.1. Likewise, S = 0.90 - 0.70 = 0.20. Then M = 0.70 - 0.25 = 0.45 ; and J = M - 0.30 = 0.45 - 0.30 = 0.15. Since crf 1 = rf 1 , H = J = 0.15. At this point in the puzzle solution, the table looks like this:
Class Mi f (^) i cf (^) i rf (^) i crf (^) i 1 - 3 2 F G 0.15 0. 4 - 6 5 K L 0.30 0. 7 - 9 8 20 56 0.25 0. 10 - 12^11 16 R^ 0.20^ 0. 13 - 15 14 U V 0.10 1. Total Y 1.
Using the HINT: rf (^) i = f (^) i /n, and using the givens in the “7-9" row, 0.25 = 20/Y; solving; Y = 20/0.25 = 80. Further, V = Y = 80. Using the HINT: cfi = cf (^) i-1 + f (^) i , R = 56 + 16 = 72. Likewise, V = U + R, or 80 = U + 72; thus U = 8. Further, 56 = L + 20, or L = 36. K = (0.30)(80), or K = 24. G + K = L, or G = L - K = 36 - 24 = 12. F = G = 12. The final solution to this puzzle is as follows:
