1) For your sample, use R to find the mean of x, the mean of y, the standard dev

Question

1) For your sample, use R to find the mean of x, the mean of y, the standard deviation of x, the standard deviation of y, and the correlation between x and y. (You must give R code for credit.)

> set.seed(0003653838)

> x=rnorm(500)

> y=2*x+rnorm(500)

> Xbar=mean(x)

> Xbar

[1] 0.0712552

> Ybar=mean(y)

> Ybar

[1] 0.2370446

> sdx=sd(x)

> sdx

[1] 0.9841084

> sdy=sd(y)

> sdy

[1] 2.277837

> corxy=cor(x,y)

> corxy

[1] 0.9040743

2) Find the equation of the regression line to predict y from x.

> lm(y~x)

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x

    0.08794      2.09259

Y = -0.08794 + 2.09259*x

3) We select a point (x1,y1) from the parent population of your data. Suppose x = 1. What is the probability that y is greater than 3? (You may find this either by using theory or based on your data.)

4) Find a 95% confidence interval for the slope coefficient of the regression line predicting y from x.

> confint(lm(y~x), level = 0.95)

                  2.5 %    97.5 %

(Intercept) 0.002089795 0.1737838

x           2.005495780 2.1796802

The 95% confidence interval is (2.005, 2.180)

** I believe my answers are correct for Questions 1, 2 and 4. I am stuck on #3, so please help with that and provide any suggestions if I have missed anything on the other three questions.

Explanation / Answer

Here I write R- for above problem as:

x=rnorm(500)

y=2*x+rnorm(500)

xbar=mean(x)

xbar

ybar=mean(y)

ybar

sdx=sd(x)

sdx

sdy=sd(y)

sdy

corxy=cor(x,y)

corxy

l=lm(y~x)

summary(l)

confint(l,level=0.95)

And the output is:

Also for third question The R-code is:

x=rnorm(500)

y=rnorm(500)

z={}

for(i in 1:500)

{

if(x[i]==1&&y[i]<=3){z[i]=1}

else{z[i]=0}

}

p=sum(z)/500

p

Thus we get probability is 0.

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