Solutions for Session 7, Part D

See solutions for Problems: D1 | D2 | D3 | D4 | D5 | D6| D7 | D8

 Problem D1 Overall, there is an upward trend; that is, the points generally go up and to the right. This corresponds to the positive association between height and arm span.

Problem D2

 a. The line does a reasonably good job. Some points are above the line, some are below it, and some are on the line, but all are generally pretty close. b. It looks like it may be possible for another line to be, overall, "closer" to these points.

 Problem D3 Answers will vary. The lines Height = Arm Span and Height = Arm Span - 1 each seem to do a good job of dividing the points fairly evenly above and below the line, and matching the overall trend of data. It is difficult to distinguish between them without a more mathematical test. Each is clearly better than Height = Arm Span + 1, which lies above a majority of the points.

Problem D4

Here is the completed table:

Person #

Arm Span (X)

Height (Y)

YL = X

Error = Y - YL

Distance = |Y - YL|

 7 162 170 162 8 8 8 165 166 165 1 1 9 170 170 170 0 0 10 170 167 170 -3 3 11 173 185 173 12 12 12 173 176 173 3 3 13 177 173 177 -4 4 14 177 176 177 -1 1 15 178 178 178 0 0 16 184 180 184 -4 4 17 188 188 188 0 0 18 188 187 188 -1 1 19 188 182 188 -6 6 20 188 181 188 -7 7 21 188 192 188 4 4 22 194 193 194 -1 1 23 196 184 196 -12 12 24 200 186 200 -14 14

Problem D5

Here is the completed table:

Person #

Arm Span (X)

Height (Y)

YL = X - 1

Error = Y - YL

Distance = |Y - YL|

 7 162 170 161 9 9 8 165 166 164 2 2 9 170 170 169 1 1 10 170 167 169 -2 2 11 173 185 172 13 13 12 173 176 172 4 4 13 177 173 176 -3 3 14 177 176 176 0 0 15 178 178 177 1 1 16 184 180 183 -3 3 17 188 188 187 1 1 18 188 187 187 0 0 19 188 182 187 -5 5 20 188 181 187 -6 6 21 188 192 187 5 5 22 194 193 193 0 0 23 196 184 195 -11 11 24 200 186 199 -13 13

For the model YL = X - 1, the total vertical distance is 7 + 4 + ... + 13 = 100. Surprisingly, according to this measure of fit, the two lines are equally good. This suggests that another measure of best fit may be useful.

Problem D6

Here is the completed table:

Person #

Arm Span (X)

Height (Y)

YL = X

Error = Y - YL

(Error)2
=
(Y - YL)2

 13 177 173 177 -4 16 14 177 176 177 -1 1 15 178 178 178 0 0 16 184 180 184 -4 16 17 188 188 188 0 0 18 188 187 188 -1 1 19 188 182 188 -6 36 20 188 181 188 -7 49 21 188 192 188 4 16 22 194 193 194 -1 1 23 196 184 196 -12 144 24 200 186 200 -14 196

Problem D7

Here is the completed table:

Person #

Arm Span (X)

Height (Y)

YL =
X - 1

Error = Y - YL

(Error)2
=
(Y - YL)2

 7 162 170 161 9 81 8 165 166 164 2 4 9 170 170 169 1 1 10 170 167 169 -2 4 11 173 185 172 13 169 12 173 176 172 4 16 13 177 173 176 -3 9 14 177 176 176 0 0 15 178 178 177 1 1 16 184 180 183 -3 9 17 188 188 187 1 1 18 188 187 187 0 0 19 188 182 187 -5 25 20 188 181 187 -6 36 21 188 192 187 5 25 22 194 193 193 0 0 23 196 184 195 -11 121 24 200 186 199 -13 169

The sum of squared errors (SSE) is 49 + 16 + ... + 169 = 772. Since this is less than the sum of squared errors for the line Height = Arm Span (which was 784), the line Height = Arm Span - 1 is a slightly better fit.

Problem D8

 a. The best model is YL = X - .7, because it has the smallest SSE. The worst model is YL = X + 1, because it has the largest SSE. b. As all of these lines have the same slope, if we changed the slope, we might find ways to reduce the SSE. c. No, we cannot reduce the SSE to zero unless all the data points lie on a straight line, which these 24 points clearly do not do.