R Markdown

#Times series data
months <- 1:6
values <- c(17, 13, 15, 11, 17, 14)


#Part A. Most recent values as forecast
forecasts_a <- values[-length(values)] #exclude the last value
actual_a <- values[-1] #exclude the first value
mse_a <- mean((actual_a - forecasts_a)^2)
mse_a #16.2
## [1] 16.2
forecast_month_7_a <- tail(values, 1)
forecast_month_7_a #14
## [1] 14
#Part B. Average all of the data as forecast
cumulative_averages <- cumsum(values[-length(values)])/(1:(length(values)-1))
cumulative_averages
## [1] 17.0 15.0 15.0 14.0 14.6
forecasts_b <- cumulative_averages
actual_b <- values[-1] #exclude the first value
mse_b <- mean((actual_b - forecasts_b)^2)
mse_b #8.272
## [1] 8.272
forecast_month_7_b <- mean(values, 1)
forecast_month_7_b #14.5
## [1] 14.5
#Part C. Which method is better
better_method <- ifelse(mse_a < mse_b, "Most recent values", "Average of all data")
list(
  MSE_most_recent_value = mse_a, 
  forecast_month_7_most_recent = forecast_month_7_a, 
  MSE_average_of_all_data = mse_b, 
  forecast_month_7_average = forecast_month_7_b, 
  Better_method = better_method
)
## $MSE_most_recent_value
## [1] 16.2
## 
## $forecast_month_7_most_recent
## [1] 14
## 
## $MSE_average_of_all_data
## [1] 8.272
## 
## $forecast_month_7_average
## [1] 14.5
## 
## $Better_method
## [1] "Average of all data"
### Example 2 ###

library(dplyr)
## 
## 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
library(zoo)
## 
## Attaching package: 'zoo'
## The following objects are masked from 'package:base':
## 
##     as.Date, as.Date.numeric
df <- data.frame(month = c(1,2,3,4,5,6,7,8,9,10,11,12), 
                 data = c( 240, 352, 230, 260, 280, 322, 220, 310, 240, 310, 240, 230))
#Times series plot
plot(df$month, df$data, type = "o", col = "blue", xlab = "Month", ylab = "Building Contracts in millions", 
        main = "Alabama Building Contracts")

#Interpretation: The time-series plot exhibits a horizontal pattern. 

#Part B. Three-month moving average
df$avg_contracts <- c(NA, NA, NA, 
                      (df$data[1] + df$data[2] + df$data[3]) / 3,
                       (df$data[2] + df$data[3] + df$data[4]) / 3,
                       (df$data[3] + df$data[4] + df$data[5]) / 3,
                       (df$data[4] + df$data[5] + df$data[6]) / 3,
                       (df$data[5] + df$data[6] + df$data[7]) / 3,
                       (df$data[6] + df$data[7] + df$data[8]) / 3,
                       (df$data[7] + df$data[8] + df$data[9]) / 3,
                       (df$data[8] + df$data[9] + df$data[10]) / 3,
                       (df$data[9] + df$data[10] + df$data[11]) / 3)

#calculate the squared errors (only for the months where average is available)
df <- df %>%
mutate(
  squared_error = ifelse(is.na(avg_contracts), NA, (data - avg_contracts)^2)
) 

#compute MSE (excluding the initual months with NA)
mse <- mean(df$squared_error, na.rm = TRUE)
mse
## [1] 2040.444
#Exponential smoothing
alpha = 0.2
exp_smooth <- rep(NA, length(df$data))
exp_smooth[1] <- df$data[1] 
for(i in 2:length(df$data)) {
  exp_smooth[i] <- alpha * df$data[i-1] + (1 - alpha) * exp_smooth[i-1]
}
mse_exp_smooth <- mean((df$data[2:12] - exp_smooth[2:12])^2)
mse_exp_smooth
## [1] 2593.762
#Compare
better_method <- ifelse(mse_a < mse_b, " Three-month moving average", "Exponential smoothing")
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] "Exponential smoothing"
### Example 3 ###
library(readxl)
library(ggplot2)

#Load the data
df1 <- read_excel("Mortgage.xlsx")

#Part A. Times series plot
ggplot(df1, aes(x = Period, y = Interest_Rate)) +
  geom_line() +
  geom_point() +
  xlab ("year") +
ylab("Interest Rate") +
  ggtitle("Times Series Plot of Interest Rate on Mortage") 

#Interpretation: We observe a decrease at first then increasing pattern in the time series plot. 

#Part B: Develope a linear trend equation
model <- lm(Interest_Rate ~ Period, data = df1)
summary(model)
## 
## Call:
## lm(formula = Interest_Rate ~ Period, data = df1)
## 
## 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
#Results - linear trend equation: Revenue = 6.70 - 0.13 * Period

#Part C. Forecast for period 25 (i.e, 2024)
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25 #3.47
##        1 
## 3.472942