nasal length nasal width 609 2

nasal length nasal width
609 241
629 222
620 233
564 207
645 247
493 189
606 226
660 240
630 215
672 231
778 263
616 220
727 271
810 284
778 279
823 272
755 268
710 278
701 238
803 255
855 308
838 281
830 288
864 306
635 236
565 204
562 216
580 225
596 220
597 219
636 201
559 213
615 228
740 234
677 237
675 217
629 211
692 238
710 221
730 281
763 292
686 251
717 231
737 275
816 275

In the kangaroo noses tab in the data file, We have data on thenasal length and width of grey kangaroos in mm. Use this data tocomplete the following exercises.

a) Run a regression and report your findings with nasal lengthas the dependent variable and nasal width as the independentvariable, i.e. length = b0 + b1 × width

b) Interpret the estimate for b1.

c) Test the hypothesis that the ?1 coefficient is equal to1.

d) Construct a 95% confidence interval for the b0coefficient.

Answer:

nasal length nasal width (X-Mx)^2 (Y-My)^2 (X-Mx)(Y-My) Y’ (Y-Y’)^2
609 241 6146.56 9.957531 247.3956 221.6067 376.0983
629 222 3410.56 490.8686 1293.884 227.359 28.71881
620 233 4542.76 124.4464 751.8844 224.7705 67.72497
564 207 15227.56 1380.535 4584.996 208.6642 2.769526
645 247 1797.76 8.090864 -120.604 231.9608 226.1778
493 189 37791.36 3042.135 10722.24 188.2437 0.571973
606 226 6625.96 329.6242 1477.862 220.7439 27.62649
660 240 750.76 17.26864 113.8622 236.275 13.8758
630 215 3294.76 850.0464 1673.529 227.6466 159.9366
672 231 237.16 173.0686 202.5956 239.7263 76.14875
778 263 8208.36 355.1131 1707.307 270.2132 52.03078
616 220 5097.96 583.4909 1724.707 223.62 13.10464
727 271 1568.16 720.6242 1063.04 255.545 238.8569
810 284 15030.76 1587.58 4884.929 279.4168 21.00543
778 279 8208.36 1214.135 3156.907 270.2132 77.20722
823 272 18387.36 775.3131 3775.707 283.1558 124.4517
755 268 4569.76 568.5575 1611.884 263.5982 19.37627
710 278 510.76 1145.446 764.8844 250.6556 747.7165
701 238 184.96 37.89086 -83.7156 248.0671 101.3462
803 255 13363.36 117.602 1253.618 277.4035 501.9188
855 308 28089.76 4076.113 10700.33 292.3594 244.6287
838 281 22680.36 1357.513 5548.773 287.47 41.86062
830 288 20334.76 1922.335 6252.218 285.1691 8.014112
864 306 31187.56 3824.735 10921.73 294.9479 122.1489
635 236 2745.76 66.51309 427.3511 229.0847 47.82182
565 204 14981.76 1612.469 4915.04 208.9518 24.52034
562 216 15725.16 792.7353 3530.707 208.089 62.58448
580 225 11534.76 366.9353 2057.307 213.266 137.6871
596 220 8353.96 583.4909 2207.818 217.8678 4.54634
597 219 8172.16 632.802 2274.062 218.1554 0.713353
636 201 2641.96 1862.402 2218.196 229.3723 804.9863
559 213 16486.56 970.6686 4000.373 207.2261 33.3376
615 228 5241.76 261.002 1169.662 223.3324 21.7863
740 234 2766.76 103.1353 -534.182 259.284 639.2789
677 237 108.16 51.20198 74.41778 241.1644 17.34212
675 217 153.76 737.4242 336.7289 240.5892 556.4486
629 211 3410.56 1099.291 1936.284 227.359 267.6167
692 238 21.16 37.89086 -28.3156 245.4786 55.92905
710 221 510.76 536.1798 -523.316 250.6556 879.4543
730 281 1814.76 1357.513 1569.573 256.4078 604.7742
763 292 5715.36 2289.091 3617.04 265.8991 681.2596
686 251 1.96 46.84642 -9.58222 243.7529 52.52048
717 231 876.16 173.0686 -389.404 252.6689 469.5404
737 275 2460.16 951.3798 1529.884 258.4211 274.859
816 275 16537.96 951.3798 3966.596 281.1425 37.73038
Mean 687.4 244.1556
Sum 30933 10987 377508.8 40195.91 108576.2 10987 8968.053

a) Y’=b0+b1*Width

b) To interpret, for every additonal unit of input output valueincreases by 0.287612 units.

c)

To test hypothesis:

and

Df= n-2= 43

The test statistic:

P-value= 0.000

The test statistic is significant and rejects H0 at coefficientvalue equal to 1. THere is insufficient evidence to support theclaim that the coeffciient value is equal to 1.

d) 95% Confidence interval for b0:

t critical:


 
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