#Question 1. Naive Approach
#Project Objective To forecast for the next week, using Naive method install.packages(“ggplot2”)
#Time Series Data ### Step 1. Input the data
week <- 1:6
values <- c(17,13,15,11,17,14)
forecast_a <- values[-length(values)]
actual_a <- values[-1]
mse_a <- mean((actual_a - forecast_a)^2)
mse_a
## [1] 16.2
### Step 4. Forecast the sales for the week 7
forecast_week7_a <- tail(values, 1)
forecast_week7_a
## [1] 14
mae_a <- mean(abs(actual_a - forecast_a))
mae_a
## [1] 3.8
mape_a <- mean(abs((actual_a - forecast_a) / actual_a)) * 100
mape_a
## [1] 27.43778
#Question 2. Moving Average and Smoothing Exponential
#Project Objective To Determine which Forecast is better to use between Exponential Smoothing and Moving Average
#Part A. Moving Average
#install.packages("dplyr")
#install.packages("zoo")
library(dplyr)
##
## 載入套件:'dplyr'
## 下列物件被遮斷自 'package:stats':
##
## filter, lag
## 下列物件被遮斷自 'package:base':
##
## intersect, setdiff, setequal, union
library(zoo)
## Warning: 套件 'zoo' 是用 R 版本 4.4.2 來建造的
##
## 載入套件:'zoo'
## 下列物件被遮斷自 'package:base':
##
## as.Date, as.Date.numeric
df <- data.frame(month=c(1,2,3,4,5,6,7,8,9,10,11,12),
values=c(240,352,230,260,280,322,220,310,240,310,240,230))
summary(df)
## month values
## 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
plot(df$month, df$values, type = "o", col = "red", xlab = "Month", ylab = "$ Millions",
main = "Values of Alabama Building contracts for 12 months")
df$avg_values3 <- c(NA, NA, NA,
(df$values[1] + df$values[2] + df$values[3]) / 3,
(df$values[2] + df$values[3] + df$values[4]) / 3,
(df$values[3] + df$values[4] + df$values[5]) / 3,
(df$values[4] + df$values[5] + df$values[6]) / 3,
(df$values[5] + df$values[6] + df$values[7]) / 3,
(df$values[6] + df$values[7] + df$values[8]) / 3,
(df$values[7] + df$values[8] + df$values[9]) / 3,
(df$values[8] + df$values[9] + df$values[10]) / 3,
(df$values[9] + df$values[10] + df$values[11]) / 3)
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_values3), NA, (values - avg_values3)^2)
)
mse <- mean(df$squared_error, na.rm = TRUE)
mse
## [1] 2040.444
alpha <- 0.2
exp_smooth <- rep(NA, length(df$values))
exp_smooth[1] <- df$values[1]
for(i in 2: length(df$values)) {
exp_smooth[i] <- alpha * df$values[i-1] + (1 - alpha) * exp_smooth[i-1]
}
mse_exp_smooth <- mean((df$values[2:12] - exp_smooth[2:12])^2)
mse_exp_smooth
## [1] 2593.762
better_method <- ifelse(mse < mse_exp_smooth, "Three-Month Moving Average", "Exponential Smoothing")
#List the Result
list(
MSE_Moving_Average = mse,
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"
#Interpretation: The Three-Month Moving Average provides more accurate forecasts than Exponential Smoothing because
#it has a lower MSE (2040.44 < 2593.76), making this a better method to minimize errors and forecast more accurately
#Question 3. Construct Time Series Plot and define its pattern
#install.packages("ggplot2")
#install.packages("readxl")
library(ggplot2)
library(readxl)
## Warning: 套件 'readxl' 是用 R 版本 4.4.2 來建造的
df <- read_excel("Mortgage.xlsx")
df
## # A tibble: 24 × 3
## Year Period Interest_Rate
## <dttm> <dbl> <dbl>
## 1 2000-01-01 00:00:00 1 8.05
## 2 2001-01-01 00:00:00 2 6.97
## 3 2002-01-01 00:00:00 3 6.54
## 4 2003-01-01 00:00:00 4 5.83
## 5 2004-01-01 00:00:00 5 5.84
## 6 2005-01-01 00:00:00 6 5.87
## 7 2006-01-01 00:00:00 7 6.41
## 8 2007-01-01 00:00:00 8 6.34
## 9 2008-01-01 00:00:00 9 6.03
## 10 2009-01-01 00:00:00 10 5.04
## # ℹ 14 more rows
colnames(df)
## [1] "Year" "Period" "Interest_Rate"
summary(df)
## 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
ggplot(df, aes(x = Period, y = `Interest_Rate`)) +
geom_line() +
geom_point() +
xlab("Period") +
ylab("Interest Rate") +
ggtitle("Interest Rate of Mortgage")
#Question 4. Develop the linear trend equation for this time series
model <- lm(`Interest_Rate` ~ Period, data = df)
summary(model)
##
## Call:
## lm(formula = Interest_Rate ~ Period, data = df)
##
## 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
# Calculate the fitted values from the model
df$predicted_interest_rat <- predict(model)
# Calculate the residuals
df$residuals <- df$Interest_Rate - df$predicted_interest_rat
# Calculate the Mean Squared Error (MSE)
mse <- mean(df$residuals^2)
cat("Mean Squared Error (MSE):", mse, "\n")
## Mean Squared Error (MSE): 0.989475
# BONUS SECTION: Calculate Mean Absolute Percentage Error (MAPE)
df$percentage_error <- abs(df$residuals / df$Interest_Rate) * 100
mape <- mean(df$percentage_error)
cat("Mean Absolute Percentage Error (MAPE)", mape, "%\n")
## Mean Absolute Percentage Error (MAPE) 15.79088 %
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25
## 1
## 3.472942