Question

4, The accompanying data on x = head circumference z score (a comparison score with peers of the same age - a positive score suggests a larger size than for peers) at age 6 to 14 months and y = volume of cerebral grey matter (in ml) at age 2 to 5 years were read from a graph in an article.

Cerebral Grey Matter (ml) 2-5 yr	Head Circumfer- ence z Scores at 6-14 Months
680	-0.71
690	1.3
700	-0.26
720	0.29
740	0.34
740	1.6
750	1.2
750	2.1
760	1.2
780	1.2
790	2.1
810	2.2
815	2.9
820	2.3
825	0.94
835	2.45
840	2.4
845	2.3

(a) What is the value of the correlation coefficient? (Give the answer to two decimal places.)
r =   


(b) Find the equation of the least-squares line. (Give the answer to two decimal places.)
=  

(c) Predict the volume of cerebral grey matter for a child whose head circumference z score at age 12 months was 1.7. (Give the answer to two decimal places.)

  

5 .Representative data read from a plot on runoff sediment concentration for plots with varying amounts of grazing damage, measured by the percentage of bare ground in the plot, are given for steeply sloped plots.

Steeply Sloped Plots
Bare ground (%)	7	7	14	21	28	35	28	42	49	56	49
Concentration	90	225	270	540	450	450	810	720	990	1080	900

(a) Using the data for steeply sloped plots, find the equation of the least-squares line for predicting y runoff sediment concentration using x = percentage of bare ground. (Give the answer to two decimal places.)
=  

(b) What would you predict runoff sediment concentration to be for a steeply sloped plot with 17% bare ground? (Round your answer to the nearest whole number.)

Answer 1

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	25.85	13890	17.51402778	48950.0	723.27
mean	1.44	771.67	SSxx	SSyy	SSxy

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.78

..............

b)

sample size ,   n =   18          
here, x̅ = Σx / n=   1.44   ,     ȳ = Σy/n =   771.67  
                  
SSxx =    Σ(x-x̅)² =    17.5140          
SSxy=   Σ(x-x̅)(y-ȳ) =   723.3          
                  
estimated slope , ß1 = SSxy/SSxx =   723.3   /   17.514   =   41.2964
                  
intercept,   ß0 = y̅-ß1* x̄ =   712.3604          
                  
so, regression line is   Ŷ =   712.36 +   41.30 *x

/......................

c)

Predicted Y at X=   1.7   is                      
Ŷ =   712.36042   +   41.296421   *   1.7   = 782.56

.................

5)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	336	6525	2966.727273	1088713.6	52375.91
mean	30.55	593.18	SSxx	SSyy	SSxy

sample size ,   n =   11          
here, x̅ = Σx / n=   30.55   ,     ȳ = Σy/n =   593.18  
                  
SSxx =    Σ(x-x̅)² =    2966.7273          
SSxy=   Σ(x-x̅)(y-ȳ) =   52375.9          
                  
estimated slope , ß1 = SSxy/SSxx =   52375.9   /   2966.727   =   17.6544
                  
intercept,   ß0 = y̅-ß1* x̄ =   53.9189          
                  
so, regression line is   Ŷ =   53.92   +   17.65   *x

...............

B)

Predicted Y at X=   17   is                  
Ŷ =   53.91892   +   17.654440   *   17   =   354

................

THANKS

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