Question 1:

The Naive Method

Enter Data

month <- 1:6
value <- c(17,13,15,11,17,14)

Mean Absolute Eror

forecast_a <- value[-length(value)]
actual_a <- value[-1]
mae <- mean(abs(actual_a - forecast_a ))
mae
## [1] 3.8
# The Mean Absolute error is 3.8

Mean Squared Error

mse <- mean((actual_a - forecast_a)^2)
mse 
## [1] 16.2
#The Mean Squared Error is 16.2

Mean absolute percentage error

MAPE <- mean(abs ((actual_a - forecast_a) /actual_a) *100)
MAPE
## [1] 27.43778

Forecast for week 7

forecast_week7_a <- tail(value, 1)
forecast_week7_a
## [1] 14
# The value for week 7 is 14

Question 2:

Step 1: Load in packages

#Question 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)
## Warning: package 'zoo' was built under R version 4.4.2
## 
## Attaching package: 'zoo'
## The following objects are masked from 'package:base':
## 
##     as.Date, as.Date.numeric

Enter the data

df <- data.frame(month=c(1:12),
                 data=c(240, 352, 230, 260, 280, 322, 220, 310, 240, 310, 240,
                        230))

Descriptive Stats

summary(df)
##      month            data      
##  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
#Interpretation: the average value for the 12 months is 250

Time Series Plot

plot(df$month,df$data, type = "o", col = "purple", xlab = "Month", ylab = "data",
     main = "Values of Alabama building contracts plot")

#Interpreation: The pattern shown in the plot is a seasonal pattern

Three month average

df$avg_data3 <- c(NA, NA, NA,
                  df$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)

Squared Error

df <- df %>%
  mutate(
    squared_error = ifelse(is.na(avg_data3), NA, (data - avg_data3)^2))

Calculate MSE

mse <- mean(df$squared_error, na.rm = TRUE)
mse 
## [1] 143065.2
#The mse result is 143065.2

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 < mse_exp_smooth, "Exponential Smoothing",
                        "Three-Month Moving Average" )

Result

list( 
  MSE_Moving_Average = mse,
  MSE_Exponential_Smoothing = mse_exp_smooth,
  Better_Method = better_method
)
## $MSE_Moving_Average
## [1] 143065.2
## 
## $MSE_Exponential_Smoothing
## [1] 2593.762
## 
## $Better_Method
## [1] "Three-Month Moving Average"

Question 3: Linear trend approach

Step 1: Load in libraries

#install package 
#install.packages("ggplot2")
#load the libraries
library(readxl) 
library(ggplot2)

Load/import data

df_m <- read_excel("C:/Users/reyre/Downloads/class16/Mortgage.xlsx"
)
summary(df_m)
##       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

Make time series plot

ggplot(df_m, aes(x = Period, y = Interest_Rate)) + 
  geom_line() +
  geom_point() +
  xlab("period") +
  ylab("Interest_Rate") +
  ggtitle("Time Series Plot")

Question 4: Make linear trend equation

model <- lm(Interest_Rate ~ Period, data = df_m)
summary(model)
## 
## Call:
## lm(formula = Interest_Rate ~ Period, data = df_m)
## 
## 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

Question 5: Forecast period 25

forecast_period25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period25
##        1 
## 3.472942