Question

R programming: it is the whole question. no more data

Problem Five: Below are some statistical summaries of the two variables “Age as the predictor and MaxHR as the response. The sample size for this data is 303. summary (Age) Min. ist Qu Median 29.00 48.00 56.00 54.44 61.00 77.00 var(Age) Mean Max [1] 81.69742 0 23182.6 20.a >summary (MaxHR) Min. lst Qu. Median Mean 3rd Qu 71.0 133.5 153.0 149.6 166.0 202.0 var(MaxHR) [1] 523.2658 [1] 2443182 sum (MaxHR *Age) a) Use the statistical summaries to calculate Sx, Sy, Syy SST b) Calculate the covariance between Age and MaxHR c) Calculate the linear correlation coefficient between Age and MaxHR d) What are the values of slope and the y-intercept values of the SLR using Age as the MaxHR as the response? predictor and e) Use the equation of the SLR to predict the MaxHR of someone who is 60 years old.

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Answer #1

Considering x=Age and y=MaxHR

Given

,x_bar=54.44,

sum x =x_bar*303 =16495.32

Var(x) =81.697

81.697= rac{sum x^{2}}{n}-mu ^{2} => 81.697 = 22-54.442 303 => 2922759.4

y_bar=149.6

,sum y= y_bar*303 =149.6*303= 45328.8

Similarly var(Y)=523.2658 = rac{sum y^{2}}{n}-mu ^{2} => 523.2658 = y-= (523.2658 + 149.62) x 303 = 6939738.02

a)Sxx = (ΣΤ)-= 922759.4 (54.4 × 303)2 303 = 26073.32

Syy =(149.6 x 303)2 303 158549.5

Sxy= _ Σ.Μ 5444 × 1496 × 3032 303 ry 2443 184 Σ.ry 24517.9

SST =5328 4 ΣΥ2 ·2 45328.82 149.62 = 96 6758808.32 -μ- 303

b)Cov(x,y) =(sum xy-rac{(sum xsum y)/n}{n-1} = rac{2443182-rac{16495.32 imes 45328.8}{303}}{303-1}=-81.185

c)Correlation coefficient,r=))-81.185 -0.3927 SrSy 9.039 x 22.875

d)Equation is y=b0+b1*x

b1 = Sxy/Sxx = -24517.9/26073.32 = -0.94 and b0 =y_bar-b1*x_bar =149.6-(-0.94)*54.44 =200.77

Equation is y= 200.77-0.94*x

When age =60,MaxHR =200.77-0.94*60 = 144.35

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