step by step answer and Interpretation using R software.

dataset can be found is R

library(sleuth3)

ex2016

16. Bumpus Natural Selection Data. Hermon Bumpus analyzed various characteristics of some house sparrows that were found on the ground after a severe winter storm in 1898. Some of the sparrows survived and some perished. The data on male sparrows in Display 20.17 are survival status (1 = survived, 2 = perished), age (1 = adult, 2 = juvenile), the length from tip of beak to tip of tail (in mm), the alar extent (length from tip to tip of the extended wings, in mm), the weight in grams, the length of the head in mm, the length of the humerus (arm bone, in inches), the length of the femur (thigh bones, in inches), the length of the tibio-tarsus (leg bone, in inches), the breadth of A subset of data on 51 male sparrows that survived (SV = 1) and 36 that perished (SV = 2) during a severe winter storm: Age (AG) is 1 for adults, 2 for juveniles; TZ is total length; AE DISPLAY 20.17 is alar extent; WT is weight; BH is length of beak and head; HL is length of humerus; FL is length of femur; 17 is length of tibio-tarsus; SK is width of skull; and KL is length of keel of sternum SV AG TL AE WT BH HL FL TT SK KL 1 1 154 241 24.5 31.2 0.687 0.668 1.022 0.587 0.830 1 160 252 26.9 30.8 0.736 0.709 1.180 0.602 0.841 155 243 26.9 30.6 0.733 0.704 1.151 0.602 0.846 154 245 24.3 31.7 0.741 0.688 1.146 0.584 0.839 156 247 24.1 31.5 0.715 0.706 1.129 0.575 0.821 162 247 27.6 31.8 0.731 0.719 1.113 0.597 0.869 NNN NN - - - 163 246 25.8 31.4 0.689 0.662 1.073 0.604 0.836 161 246 24.9 30.5 0.739 0.726 1.138 0.580 0.803 160 242 26.0 31.0 0.745 0.713 1.105 0.600 0.803 162 246 26.5 31.5 0.720 0.696 1.092 0.606 0.809 the skull in inches, and the length of the sternum in inches. (A subset of this data was discussed in Exercise 2.21.) Analyze the data to see whether the probability of survival is associated with physical charac- teristics of the birds. This would be consistent, according to Bumpus, with the theory of natural selection: those that survived did so because of some superior physical traits. Realize that (i) the sampling is from a population of grounded sparrows, and (ii) physical measurements and survival are both random. (Thus, either could be treated as the response variable.)Reference: Bumpus Natural Selection Data. In 1899, biologist Hermon Bumpus presented as evidence of natural selection a comparison of numerical characteristics of moribund house sparrows that were collected after an uncommonly severe winter storm and which had either perished or survived as a result of their injuries. Display 2.15 shows the length of the humerus (arm bone) in inches for 59 of these sparrows, grouped according to whether they survived or perished. Analyze these data to summarize the evidence that the distribution of humerus lengths differs in the two populations. Write a brief paragraph of statistical conclusion, using the ones in Section 2.1 as a guide, including a DISPLAY 2.15 Humerus lengths of moribund male house sparrows measured by Hermon Bumpus, groupe according to survival status Humerus Lengths (inches) of 35 Males That Survived 0.687, 0.703, 0.709, 0.715, 0.721, 0.723, 0.723, 0.726, 0.728, 0.728, 0.728, 0.729, 0.730, 0.730, 0.733, 0.733, 0.735, 0.736, 0.739, 0.741, 0.741, 0.741, 0.741, 0.743, 0.749, 0.751, 0.752, 0.752, 0.755, 0.756, 0.766, 0.767, 0.769, 0.770, 0.780 Humerus Lengths (inches) of 24 Males That Perished 0.659, 0.689, 0.702, 0.703, 0.709, 0.713, 0.720, 0.720, 0.726, 0.726, 0.729, 0.731, 0.736, 0.737, 0.738, 0.738, 0.739, 0.743, 0.744, 0.745, 0.752, 0.752, 0.754, 0.765