Question

1. The rowan (Sorbus aucuparia) is a tree that grows in a wide range of altitudes. To study how the tree adapts to its varying habitats, researchers collected twigs with attached buds from 12 trees growing at various altitudes in North Angus, Scotland. The buds were brought back to the laboratory and measurements were made of the dark respiration rate. The accompanying table shows the altitude of origin (in meters) of each batch of buds and the dark respiration rate (expressed as µl of oxygen per hour per mg dry weight of tissue)

Altitude of origin X (m)	Respiration rate Y (µl/hr x mg)	XY	X2	Y2
90	0.11	9.9	8100	0.0121
230	0.20	46.0	52900	0.0400
240	0.13	31.2	57600	0.0169
260	0.15	39.0	67600	0.0225
330	0.18	59.4	108900	0.0324
400	0.16	64.0	160000	0.0256
410	0.23	94.3	168100	0.0529
550	0.18	99.0	302500	0.0324
590	0.23	135.7	348100	0.0529
610	0.26	158.6	372100	0.0676
700	0.32	224.0	490000	0.1024
790	0.37	292.3	624100	0.1369
Sum:5200	2.52	1253.4	2760000	0.5946

a. make a scatter plot of the data

b. Compute the correlation coefficient

c. Calculate the least-squares regression line of Y on X ( Yhat = a + bx )

d. Predict the respiration rate at altitude of origin, 720.

Answer 1

a.) The scatterplot is plotted below:

b.) To determine correlation coefficient.

We know that, correlation coefficient is given by :

Substituting given values, we get,

  

  

  

  

  

Hence, correlation coefficient is 0.13744

c) Least square regression line of Y on X is given as,

where a and b are given by,

Altitude of origin X	Respiration rate Y	XY	X2	Y2	Xi-X̅	(Xi-X̅)^2	(Yi-Y̅)	(Xi-X̅)(Yi-Y̅)
90	0.11	9.9	8100	0.0121	-343.333	117877.8	-0.1	34.33333333
230	0.2	46	52900	0.04	-203.333	41344.44	-0.01	2.033333333
240	0.13	31.2	57600	0.0169	-193.333	37377.78	-0.08	15.46666667
260	0.15	39	67600	0.0225	-173.333	30044.44	-0.06	10.4
330	0.18	59.4	108900	0.0324	-103.333	10677.78	-0.03	3.1
400	0.16	64	160000	0.0256	-33.3333	1111.111	-0.05	1.666666667
410	0.23	94.3	168100	0.0529	-23.3333	544.4444	0.02	-0.466666667
550	0.18	99	302500	0.0324	116.6667	13611.11	-0.03	-3.5
590	0.23	135.7	348100	0.0529	156.6667	24544.44	0.02	3.133333333
610	0.26	158.6	372100	0.0676	176.6667	31211.11	0.05	8.833333333
700	0.32	224	490000	0.1024	266.6667	71111.11	0.11	29.33333333
790	0.37	292.3	624100	0.1369	356.6667	127211.1	0.16	57.06666667
5200	2.52	1253.4	2760000	0.5946		506666.7		161.4

X̅=	433.3333
Y̅=	0.21

    

Therefore, substituting values from above table, we get,

Now substituting value of b in a,

Hence, Least square regression line of Y on X is  

d)

Now, to predict the respiration rate at altitude of origin, 720. i.e., To predict Y given X=720

Using the least square regression line euation computed in (c), we get,

Therefore, the respiration rate at altitude of origin 720 µl/hrxmg is 0.30128 m.