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.

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= $\frac{\sum x^{2}}{n}-\mu ^{2}$ => 81.697 = $\frac{\sum x^{2}}{303}-54.44 ^{2}$ => $\sum x^{2} = 922759.4$

y_bar=149.6

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

Similarly var(Y)=523.2658 = $\frac{\sum y^{2}}{n}-\mu ^{2}$ => 523.2658 = $\sum y^{2}=(523.2658+149.6^{2})\times 303$ = 6939738.02

a)Sxx = $\sum x^{2}-\frac{(\sum x)^{2}}{n} =922759.4-\frac{(54.4\times 303)^{2}}{303}=26073.32$

Syy =$\sum y^{2}-\frac{(\sum y)^{2}}{n} =6939738.02-\frac{(149.6\times 303)^{2}}{303}=158549.5$

Sxy= $\sum xy-\frac{\sum x\sum y}{n}= 2443184-\frac{54.44\times 149.6\times 303^{2}}{303} = -24517.9$

SST =$\frac{\sum y^{2}}{n}-\mu ^{2}=\frac{45328.8^{2}}{303}-149.6^{2}= 6758808.32$

b)Cov(x,y) =$(\sum xy-\frac{(\sum x\sum y)/n}{n-1} = \frac{2443182-\frac{16495.32\times 45328.8}{303}}{303-1}=-81.185$

c)Correlation coefficient,r=$\frac{Cov(x,y))}{SxSy} = \frac{-81.185}{9.039\times 22.875}= -0.3927$

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