Question

Question 1 Suppose we are given the data 2 -1 22: 2 -2 0 2 0 -1 -2 -3 30 2 2 2 1 0 -1 0 -1 We aim at fitting the linear model Y; = Bo + Bizil + B22i2 + Ei, i = 1, 2, ..., 7. (1) Find the least square estimate B; (2) Find the R2 statistic; (3) Find ô2 and Cov(); (4) Find a 95% confidence interval for B1; (5) Test Ho: B1 = B2 = 0 at the level of a = 0.05; (6) Find a 95% confidence interval for the mean response E(Y) = Bo+Bızı + B222, where z; is the sample mean of zj, j = 1,2. (7) Find a 95% prediction interval for a new response Yo corresponding to (71, 72). .

Answer 1

Answer:

Given Data

> y=C(1,0,1,0,-1,0,-1) > zl=c(2,2,2,0,-1,-2,-3) > z2=c(-1,-2,-3,0,2,2,2) > ## Linear model > model=lm (yz1+z2) > summary (model) Call: Im (formula = y - zl + z2) Residuals: 1 2 5.200e-01 -6.400e-01 3 4 5 2.000e-01 -2.776e-17 -5.200e-01 7 6.400e-01 -2.000e-01 Coefficients: Estimate Std. Error t value Pr(>01) (Intercept) -7.553e-17 2.268e-01 0.000 1.000 zl 1.600e-01 3.059e-01 0.523 0.629 22 -1.600e-01 3.059e-01 -0.523 0.629 Residual standard error: 0.6 on 4 degrees of freedom Multiple R-squared: 0.64, Adjusted R-squared: F-statistic: 3.556 on 2 and 4 DF, p-value: 0.1296 0.46

1) From the above, B^=( -7.553e-17 1.600e-01 -1.600e-01 )'

2) R^2 statistic=0.64.

3) sigma^2=0.6^2=0.36

> X=cbind (rep (1, length (y)), zl, z2) > # Cobariance of B^ is > 0.36*solve (t (X)***X) zl 22 0.05142857 0.0000 0.0000 zl 0.00000000 0.0936 0.0864 22 0.00000000 0.0864 0.0936

4)

> confint (model) 2.5 % 97.5 $ (Intercept) -0.6296386 0.6296386 zl -0.6894289 1.0094289 22 -1.0094289 0.6894289 > |

5)

Analysis of Variance Source Regression Residual Error Total DF SS MS 2 2.5600 1.2800 4 1.4400 0.3600 6 4.0000 F P 3.56 0.130

P-value is 0.130 and greater than alpha=0.05. Hence, fail to reject the null hypothesis.

6)

> ## 6) Confidence interval > predict (model, data.frame (zl=mean (zl), z2=mean (22)), interval = "confidence") fit lwr upr 1 -7.553247e-17 -0.6296386 0.6296386 > ## 7) Prediction interval > predict (model, data.frame (zl=mean (zl), z2=mean (22)), interval = "prediction") fit lwr upr 1 -7.553247e-17 -1.780887 1.780887

## R command

y=c(1,0,1,0,-1,0,-1)
z1=c(2,2,2,0,-1,-2,-3)
z2=c(-1,-2,-3,0,2,2,2)

## Linear model
model=lm(y~z1+z2)
summary(model)
## 3

X=cbind(rep(1, length(y)), z1, z2)
# Cobariance of B^ is
0.36*solve(t(X)%*%X)

##4
confint(model)

## 6) Confidence interval
predict(model, data.frame(z1=mean(z1), z2=mean(z2)), interval = "confidence")

## 7) Prediction interval
predict(model, data.frame(z1=mean(z1), z2=mean(z2)), interval = "prediction")

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