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Data Session 7: Notes
 
Session 7 Part A Part B Part C Part D Homework
 
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A B C D 
Homework

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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.

<< back to Problem D1


 

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.

<< back to Problem D2


 

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.

<< back to Problem D3


 

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

<< back to Problem D4


 

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.

<< back to Problem D5


 

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

<< back to Problem D6


 

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.

<< back to Problem D7


 

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.

<< back to Problem D8


 

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