Question

3. This problem uses the wblake data set in the alr4 package. This data set includes samples of small mouth bass collected in West Bearskin Lake, Minnesota, in 1991. Interest is in predicting length with age. Complete this problem without using Im) in R. (a) Do the regression of length on age, and report the estimates, their standard errors and the estimate of variance. Interpret Bo and B. (b) Obtain a 90% confidence interval for A from the data. Interpret this interval. (c) Obtain a prediction and a 90% prediction interval for a small mouth bass at age 2, Interpret this interval.

Answer 1

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Output:

BRStudio File Edit Code View Plots Session Build Debug Profile Tools Help * Addins Source Console Terminal # install.packages("alr4") library (alr4) data (wblake) names (wblake) [1] "Age" "Length" "Scale" # Part (a) solution : XwblakeŞAge; Y-wblake$Length xbar mean (X)ybar-mean (Y)inlength (Y) sSxy sum((x-xbar) (Y-ybar)) SSxxsum ( (x-xbar)2) SSyy sum((Y-ybar) 2) (bl-SSxy/SSxx) (bo ybar-b1*xbar) # Reg Estimates [1] 30.32389 [1] 65.52716 # Reg of length on age sum ((Y-Y.hat)"2 ) ; SSR = sum ( (Y.hat-ybar)"2 ) ; SST = SSR+SSE # Coefficient of Deter > SSE (R.SqSSR/SST) = [1] 0.816477 (MSESSE/ (n-2)) [1] 820.5847 # Estimate of variance (SE. b1 = sart (MSE/ssxx) ) [1] 0.6877291 > ( SE .b0 sqrt (MSE * ( (1/n)+(xbar"2/ssxx)))) [1] 3.197388 # SE of slope bl # SE of intercept b0 > #summary: > #1. Regression estimates are, b0 = 65.5272 and b1 = 30.3239 > #2. For eve ry unit increase in small mouth base age, we expect > # small bass length increases by 30.32mm > #3. Standard error of reg coeff are, se (b0)= 3.1974 and se (b1)= 0.6877 > #4. R^2-0.8165; 81.65% variation explained by the explanatory variable > #5. Estimated variance is 821.585.

> # Part (b) solution: > # Calculate a 90% confidence interval >a=1-0.90 > t . Star = round (at ( 1-a/2, n-2 ) , 3) for beta1. # Confidence level 0.90 # t critical value MEt.star*SE.bl # 90% CI for beta1 b1-c (ME,-ME) [1] 29.19051 31.45727 > #summary: > # There is a 90% confidence that, for every unit increase in our predictor > #(age) a small mouth bass grows between 29.1905 mm and 31.4573 mm > # Part (c) solution : > # Cal a prection and a 90% preidction interval for a small mouth at age 2 >x0 = 2 # Predicted Y when X 2 [1] 126.1749 ME t.star sqrt (MSE) *sqrt (1+ (1/n)+((x0-xbar) 2)/ssxx) Y.Pred -c (ME,-ME) [1] 78.84695 173.50293 > #summary: > # At age 2, the predicted length of a samll mouth bass is 126.17 mm > # There is a 90% confidence that, for every unit increase in our predictor > #(age) a small mouth bass grows between 78.847 mm and 173.503 mm .

Code in text format:

install.packages("alr4")
library(alr4)
data(wblake)
names(wblake)

# Part (a) solution:
X = wblake$Age; Y = wblake$Length
xbar = mean(X); ybar = mean(Y); n = length(Y)
SSxy = sum((X-xbar)*(Y-ybar))
SSxx = sum((X-xbar)^2); SSyy = sum((Y-ybar)^2)

(b1 = SSxy/SSxx); (b0 = ybar-b1*xbar) # Reg Estimates

Y.hat = b0+b1*X # Reg of length on age

SSE = sum((Y-Y.hat)^2); SSR = sum((Y.hat-ybar)^2); SST = SSR+SSE
(R.Sq = SSR/SST) # Coefficient of Deter

(MSE = SSE/(n-2)) # Estimate of Variance
(SE.b1 = sqrt(MSE/SSxx)) # SE of slope b1
(SE.b0 = sqrt(MSE*((1/n)+(xbar^2/SSxx)))) # SE of intercept b0

#------------------------------------------------------------------------
#
# Part (b) solution:
# Calculate a 90% confidence interval for beta1.
a = 1-0.90 # Confidence level = 0.90
t.star = round(qt(1-a/2,n-2),3) # t critical value
ME = t.star*SE.b1
b1-c(ME,-ME) # 90% CI for beta1

#-------------------------------------------------------------------------
#
# Part(c) solution:
# Cal a prection and a 90% preidction interval for a small mouth at age 2
x0 = 2
(Y.Pred = b0+b1*x0) # Predicted Y when X=2
ME = t.star*sqrt(MSE)*sqrt(1+(1/n)+((x0-xbar)^2)/SSxx)
Y.Pred - c(ME,-ME)