Class Exercise 16: Chapter 17
Erica Ashley Carandang
2024-11-22
Question 1: Naive Approach
# Time Series Data
week <- 1:6 #independent variable (Time)
Values <- c(17, 13, 15, 11, 17, 14) #dependent variable (sales) # Most Recent Value as Forecast
forecast_a <- Values[-length(Values)] #exclude last value
actual_a <- Values[-1] #exclude first sale # Part A: Mean Absolute Error
mae_a <- mean(abs(actual_a - forecast_a))
mae_a #3.8## [1] 3.8 # Part B: Mean Squared Error
mse_a <- mean((actual_a - forecast_a)^2)
mse_a #16.2## [1] 16.2 # Part C: Mean Absolute Percentage Error
MAPE <- mean(abs(actual_a - forecast_a) / actual_a) * 100 # actual_a as denominator
MAPE #27.44## [1] 27.43778 # Part D: Forecast sales for week 7
forecast_week7_a <- tail(Values, 1)
forecast_week7_a #14## [1] 14 #interpretation: The number of product projected to be sold on week 7.#Question 2: Smoothing Approach (Moving Average and Exponential Smoothing Approach)
#Time Series Data
library(dplyr)## Warning: package 'dplyr' was built under R version 4.4.2##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union df <- data.frame(month=c(1,2,3,4,5,6,7,8,9,10,11,12),
Contracts=c(240, 352, 230, 260, 280, 322, 220, 310, 240, 310, 240, 230)) #in millions
summary(df)## month Contracts
## Min. : 1.00 Min. :220.0
## 1st Qu.: 3.75 1st Qu.:237.5
## Median : 6.50 Median :250.0
## Mean : 6.50 Mean :269.5
## 3rd Qu.: 9.25 3rd Qu.:310.0
## Max. :12.00 Max. :352.0 #sales over 12 month period is 269.5 #Part A: Graph Time Series Plot
plot(df$month, df$Contracts, type = "o", col = "blue",
xlab = "Month", ylab = "Contracts (in millions)",
main = "Alabama Building Contracts")
#Interpretation: Time series plot displays a horizontal pattern and is steady on its mean. #Part B: Three Month Moving approach
df$avg_contracts3 <- c( NA, NA, NA,
(df$Contracts[1] + df$Contracts[2] + df$Contracts[3]) / 3,
(df$Contracts[2] + df$Contracts[3] + df$Contracts[4]) / 3,
(df$Contracts[3] + df$Contracts[4] + df$Contracts[5]) / 3,
(df$Contracts[4] + df$Contracts[5] + df$Contracts[6]) / 3,
(df$Contracts[5] + df$Contracts[6] + df$Contracts[7]) / 3,
(df$Contracts[6] + df$Contracts[7] + df$Contracts[8]) / 3,
(df$Contracts[7] + df$Contracts[8] + df$Contracts[9]) / 3,
(df$Contracts[8] + df$Contracts[9] + df$Contracts[10]) / 3,
(df$Contracts[9] + df$Contracts[10] + df$Contracts[11]) / 3
)
summary(df)## month Contracts avg_contracts3
## Min. : 1.00 Min. :220.0 Min. :256.7
## 1st Qu.: 3.75 1st Qu.:237.5 1st Qu.:263.3
## Median : 6.50 Median :250.0 Median :274.0
## Mean : 6.50 Mean :269.5 Mean :273.7
## 3rd Qu.: 9.25 3rd Qu.:310.0 3rd Qu.:284.0
## Max. :12.00 Max. :352.0 Max. :287.3
## NA's :3 #Calculate Squared Errors
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_contracts3), NA, (Contracts - avg_contracts3)^2))
mse_Contracts <- mean(df$squared_error, na.rm = TRUE)
mse_Contracts## [1] 2040.444 #Exponential Smoothing
alpha <- 0.2
exp_smooth <- rep (NA, length(df$Contracts))
exp_smooth[1] <- df$Contracts[1] #starting point
for(i in 2: length (df$Contracts)) {
exp_smooth[i] <- alpha * df$Contracts[i-1] + (1 - alpha) * exp_smooth[i-1]
}
mse_exp_smooth <- mean((df$Contracts[2:12] - exp_smooth[2:12])^2)
mse_exp_smooth #2593.76## [1] 2593.762 #Better Method
better_method <- ifelse(mse_Contracts < mse_exp_smooth, "Three-Month Moving Average", "Exponential Smoothing")
better_method #"Three-Month Moving average" is more accurate with a lower error.## [1] "Three-Month Moving Average" list(
MSE_Moving_Average = mse_Contracts,
MSE_Exponential_Smoothing = mse_exp_smooth,
Better_Method = better_method
)## $MSE_Moving_Average
## [1] 2040.444
##
## $MSE_Exponential_Smoothing
## [1] 2593.762
##
## $Better_Method
## [1] "Three-Month Moving Average" #Results
#MSE Moving AVG: 2040.44
#MSE Exp. AVG: 2593.76
#Better Method: "Three-Month Moving Average"
#Interpretation: The "Three-Month Moving Average provides a more accurate
#forecast since it has a smaller MSE value.#Question 3: Linear Trend Approach
#File Upload
library(readxl)
library(ggplot2)## Warning: package 'ggplot2' was built under R version 4.4.2 Mortgage <- read_excel("C:/Users/caran/Downloads/Mortgage.xlsx")
summary(Mortgage)## Year Period Interest_Rate
## Min. :2000-01-01 00:00:00 Min. : 1.00 Min. :2.958
## 1st Qu.:2005-10-01 18:00:00 1st Qu.: 6.75 1st Qu.:3.966
## Median :2011-07-02 12:00:00 Median :12.50 Median :4.863
## Mean :2011-07-02 18:00:00 Mean :12.50 Mean :5.084
## 3rd Qu.:2017-04-02 06:00:00 3rd Qu.:18.25 3rd Qu.:6.105
## Max. :2023-01-01 00:00:00 Max. :24.00 Max. :8.053#Graph Time Series Plot
ggplot(Mortgage, aes(x = Year, y = Interest_Rate)) +
geom_line() +
geom_point() +
xlab("Year") +
ylab("Interest Rates") +
ggtitle("US Mortgage Interest Rate Average")
#Question 4: Linear Trend Equation for this Time Series
model <- lm(Interest_Rate ~ Period, data = Mortgage)
summary(model)##
## Call:
## lm(formula = Interest_Rate ~ Period, data = Mortgage)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.3622 -0.7212 -0.2823 0.5015 3.1847
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.69541 0.43776 15.295 3.32e-13 ***
## Period -0.12890 0.03064 -4.207 0.000364 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.039 on 22 degrees of freedom
## Multiple R-squared: 0.4459, Adjusted R-squared: 0.4207
## F-statistic: 17.7 on 1 and 22 DF, p-value: 0.0003637 #6.70 + -0.13*period#Question 5: Linear Trend Equation for Question 3B Period 25
Mortgage_prediction <- predict(model, newdata = data.frame(Period = 25))
Mortgage_prediction## 1
## 3.472942 #3.47