Question Number 1

Time Series Forecasting

week <- 1:6
values <- c(17, 13, 15, 11, 17, 14)

Part A. Most recent value as forecast

forecasts_a <- values[-length(values)] 
actual_a <- values[-1]
mse_a <- mean((actual_a - forecasts_a)^2)
mse_a # Result: 16.2
## [1] 16.2
forecast_week_7_a <- tail(values,1)
forecast_week_7_a
## [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
forecast_b <- cumulative_averages
actual_b <- values[-1]
mse_b <- mean((actual_b - forecast_b)^2)
mse_b
## [1] 8.272
forecast_week_7_b <- tail(values)
forecast_week_7_b
## [1] 17 13 15 11 17 14

Part C: Which method is better.

better_method <-ifelse(mse_a < mse_b, "Most Recent Value", "Average of All Data")
list(
  MSE_most_recent_value = mse_a,
  forecast_week_7_most_recent = forecast_week_7_a,
  MSE_average_of_all_data = mse_b,
  forecast_week_7_average = forecast_week_7_b,
  Bettter_Method = better_method
)
## $MSE_most_recent_value
## [1] 16.2
## 
## $forecast_week_7_most_recent
## [1] 14
## 
## $MSE_average_of_all_data
## [1] 8.272
## 
## $forecast_week_7_average
## [1] 17 13 15 11 17 14
## 
## $Bettter_Method
## [1] "Average of All Data"

Question Number 2- Class Exercise 16

Load the libraries

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))

Time Series Plot

plot(df$month, df$data, type = "o", col= "blue", xlab = "Month", ylab = "Contratcs",
     main = "Monthly Contracts")

The Time-Series Plot shows 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 square errors

df<- df %>%
  mutate(
    sqaured_error= ifelse(is.na(avg_contracts), NA, (data - avg_contracts)^2)
  )

Compute MSE

mse <- mean(df$sqaured_error, na.rm = TRUE)
mse # Result: 2040.44
## [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 # Result: 2593.76
## [1] 2593.762

Compare

better_method <- ifelse(mse < mse_exp_smooth, "Three Month 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] "Three Month average"
Result: Better Method ~ Three Month Average

Question Number 3

library(readxl)
library(ggplot2)

Load the data

df1 <- read_excel("Mortgage.xlsx")

Part A: Time 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 decreasing pattern at first then it increases.

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
Linear Trend Equation: Interest Rate = 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 # Answer: 3.47
##        1 
## 3.472942