Assignment5

Question problems

• Regress Days on Index using simple linear regression, what are the estimates of your fitted regression line?

Data X as Index Y as Days

y <- c(91, 105, 106, 108, 88, 91, 58, 82, 81, 65, 61, 48, 61, 43, 33, 36)
x <- c(16.7, 17.1, 18.2, 18.1, 17.2, 18.2, 16.0, 17.2, 18.0, 17.2, 16.9, 17.1, 18.2, 17.3, 17.5, 16.6)

cbind(x,y)

##          x   y
##  [1,] 16.7  91
##  [2,] 17.1 105
##  [3,] 18.2 106
##  [4,] 18.1 108
##  [5,] 17.2  88
##  [6,] 18.2  91
##  [7,] 16.0  58
##  [8,] 17.2  82
##  [9,] 18.0  81
## [10,] 17.2  65
## [11,] 16.9  61
## [12,] 17.1  48
## [13,] 18.2  61
## [14,] 17.3  43
## [15,] 17.5  33
## [16,] 16.6  36

• Make a scatterplot of the data, Title the plot and label the axes appropriately

  #Answer

Plot Days vs Index

plot(x,y, main= " Days vs Index", xlab="Index", ylab= "Days")

Model_1 object that represents the regression of Days on Index

model_1 <- lm(y~x)

Parameters of simple linear regression

Intercept : -193.0

Slop : 15.3

model_1

## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##      -193.0         15.3

• Add the least squares line to the scatterplot

#plotting the graph
plot(x,y, main="Simple Linear Regression")
abline(model_1)

• # min and max of predictor variable Days

  min(x)

  max(x)

  # Create a sequence of values that span min to max Days with spacing of 5.0

  new_x <- seq(16, 18.2, 0.02)

  # Create the confidence and prediction interval at each of new_x values

  conf <- predict(model_1, data.frame(x=new_x), interval=“confidence”)

  pred <- predict(model_1, data.frame(x=new_x), interval=“prediction”)

  lines(new_x,conf[,2], col=“blue”)

  lines(new_x,conf[,3], col=“blue”)

  lines(new_x,pred[,2], col=“orange”)

  lines(new_x,pred[,3], col=“orange”)

  legend(“topleft”, inset=c(0, 0), legend=c(“Linear regression line”,“Confidence interval”, “Prediction interval”), lty=1, col=c(“black”,“blue”,“orange”), cex=.8

• What is the value of R²?

  #Answer

summary(model_1)$r.square

## [1] 0.1584636

• Test for the signifiance of the regression at a 0.05 level of signficance assuming the response is Normally distributed, what is your conclusion?

summary(model_1)$coefficients

##               Estimate Std. Error   t value  Pr(>|t|)
## (Intercept) -192.98383 163.503283 -1.180306 0.2575450
## x             15.29637   9.420975  1.623650 0.1267446

#Answer

Looking at the summary table above, we can notice that for this linear regression model, t₀=1.62365 and for t with 95% confidence and degree of freedom equal to 14,we have that _(0.025,14)=2.145.

Since t₀ value is between -t_(0.025,14) and t_(0.025,14), therefore we fail to reject H₀, which means that our variable “Days” does not have linear relationship with “Index”.

• Regardless of whether you conclude that the regression is signficant above, make a scatterplot of the data showing the fitted regression line, confidence interval, and prediction interval

  #Answer

• Calculate a 95% confidence interval on the mean number of days the ozone level exceeds 20ppm when the meteorological index is 17.0. Comment on the meaning of this interval?

  ss_res <- sum((fitted(model_1)-y)^2) #(sum of sequare Error)
  msr <- summary(model_1)$sigma^2
  sxx <- sum(x^2)- sum(x)^2/length(x) #sxx
  t_0.025 <- 2.145 # t-test value from tables
  y_17 <- model_1$coefficients[1] + model_1$coefficients[2] *(17) # E(Y|x=17)
  
  # simplifying the equation (the parts under the square root)
  s <- (17-mean(x))^2 
  v <- s/sxx

  # get the confidince interval
  lower_cI <- y_17 - t_0.025 * sqrt(msr * ((1/length(x))+ v))
  upper_cI <- y_17 + t_0.025 * sqrt(msr * ((1/length(x))+ v))

  #Answer

  lower_cI

  ## (Intercept) 
  ##    52.52604

  upper_cI

  ## (Intercept) 
  ##    81.58271

• Calculate a 95% prediction interval on the mean number of days the ozone level exceeds 20ppm when the meterological index is 17.0. Comment on the meaning of this interval? Compare the width of the prediction interval to that of the confidence interval and comment

lower_pI <- y_17 - t_0.025 * sqrt(msr * (1+(1/length(x))+ v))
upper_pI <- y_17 + t_0.025 * sqrt(msr * (1+(1/length(x))+ v))

#Answer

lower_pI

## (Intercept) 
##    13.98675

upper_pI

## (Intercept) 
##     120.122