Question

Suppose doctor measuree the he ht, and head ircumfere o 11 children anc ob ains the data below The corelat on c em ent ie 0.896 and the less18 a es reareeeis lire is 0 2 2 + 11.872 Сото 1 te parts a throu eo 27.75 25.5 26.75 25.25 29 26.75 26 28.75 27 27 27 Height, x Head Circumference, y 17,5 16,9 17,2 17.1 17 172 17.1 17.4 17,3 17.4 17.4 (a) Compute the ocecient or detenminaion, R R2-D%(Round to one oornal piece ฮ..dec.) b Gona eres dua piot o enty he equirements o lhe less squares regression model. Aresiduel piol is a sce 1er cieg am ₩ith the escaals the vert al axia 6nd the e len o y var able on the honzontel a Ree duals e e m uted to each observa5on using the t lowing ormue. resicusl observed y precicted y y-y whie the residua s can be ondロy uang the ranrssıon cquation 1 ar the prcdcted va ues o v and then p onec by hanc, t s gcncral easier to usc chnology to roducc a residual plat dro y TOT thn data set. hoase te oorrect graph bclow. A. C. D. a, (с 1 er ret e cce ent of deler na on an oorn ent on he a e uscy o e ineer moce r a poto residuals 8gainst he explanato ver able 8 0 S 9 Gecen be patern auch es a curve, che he explanatory and response ver able may no be neer y elaled. Approximately | | % of the variation n (Round Lo one decirlal plece ษ8 needed.} ▼ | is exalained byth, least-squares mgresson madal. Accornirg to the renual plot. tha linaar modal appears to be

Answer 1

(a) for simple linear regression y=a+bx,

the coefficient of determination R²=r*r=0.896*0.896=0.803 ,

where r is correlation between x and y

(b) right choice is B.

the residual plot is given image

(c) Approximately 80.3% of the variation in dependent variable(y) is explained by the least squrare regression model. According to residual plot , the linear model apprears to be appropriate or good fit.

A residual plot is a graph that shows the residuals on the vertical axis and the independent variable on the horizontal axis. If the points in a residual plot arerandomly dispersed around the horizontal axis, a linear regression model is appropriate for the data; otherwise, a non-linear model is more appropriate.

following information has been generted using ms-excel

the fitted regression y^=11.872+0.202*x

b=cov(x,y)/var(x)

a=mean(y)-b*mean(x)

Observation	x	y	Predicted Y	Residuals
1	27.75	17.5	17.4775	0.0225
2	25.5	16.9	17.023	-0.123
3	26.75	17.2	17.2755	-0.0755
4	25.25	17.1	16.9725	0.1275
5	28	17.5	17.528	-0.028
6	26.75	17.2	17.2755	-0.0755
7	26	17.1	17.124	-0.024
8	26.75	17.4	17.2755	0.1245
9	27	17.3	17.326	-0.026
10	27	17.4	17.326	0.074
11	27	17.4	17.326	0.074
mean=	16.35227	21.98864
var=	0.645661	0.032893
cov(x,y)=	0.130579
corr(x,y)=r=	0.896026
R2=r*r=	0.802816
a=mean(y)-b*mean(x)=	11.872
b=cov(x,y)/var(x)	0.202