# 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.

 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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