options(repos = c(CRAN = "https://cloud.r-project.org"))
options(repos = c(CRAN = "https://cloud.r-project.org"))Question 1. Naive Approach
Project Objective
To forecast for the next week, using Naive method (Most Recent Value)Step 1. Input the data
week <- 1:6 # This is the independent variable - time
values <- c(17,13,15,11,17,14) # Dependent variableData Description: A Description of the features are presented in the table below
Variable | Definition
1.Week | Time period by weeks
2.Value | Coordinated value for each week from week 1 to week 6Step 2. Exclude the Last and First Value
forecast_a <- values[-length(values)] #Excludes the last value
actual_a <- values[-1] #Excludes the first valueI ### Step 3. Find the MSE (Mean Squared Error)
mse_a <- mean((actual_a - forecast_a)^2)
mse_a ## [1] 16.2Interpretation:
The Mean Squared error is 16.2Step 4. Forecast the sales for the week 7
forecast_week7_a <- tail(values, 1)
forecast_week7_a## [1] 14Interpretation: The value for week 7 is 14Step 5. Find the MAE (Mean Absolute Error)
mae_a <- mean(abs(actual_a - forecast_a))
mae_a## [1] 3.8Interpretation: the Mean Absolute Error is 3.8Step 6. Find the Mean Absolute Percentage Error (MAPE)
mape_a <- mean(abs((actual_a - forecast_a) / actual_a)) * 100
mape_a## [1] 27.43778Interpretation: The MAPE is 27.44Question 2. Moving Average and Smoothing Exponential
Project Objective
To Determine which Forecast is better to use between Moving Average and Exponential SmoothingPart A. Moving Average
Step 1. Install and load the packages
install.packages("dplyr")## Installing package into 'C:/Users/DELL/AppData/Local/R/win-library/4.4'
## (as 'lib' is unspecified)## package 'dplyr' successfully unpacked and MD5 sums checked## Warning: cannot remove prior installation of package 'dplyr'## Warning in file.copy(savedcopy, lib, recursive = TRUE): problem copying
## C:\Users\DELL\AppData\Local\R\win-library\4.4\00LOCK\dplyr\libs\x64\dplyr.dll
## to C:\Users\DELL\AppData\Local\R\win-library\4.4\dplyr\libs\x64\dplyr.dll:
## Permission denied## Warning: restored 'dplyr'##
## The downloaded binary packages are in
## C:\Users\DELL\AppData\Local\Temp\RtmpwvM4vv\downloaded_packagesinstall.packages("zoo")## Installing package into 'C:/Users/DELL/AppData/Local/R/win-library/4.4'
## (as 'lib' is unspecified)## package 'zoo' successfully unpacked and MD5 sums checked## Warning: cannot remove prior installation of package 'zoo'## Warning in file.copy(savedcopy, lib, recursive = TRUE): problem copying
## C:\Users\DELL\AppData\Local\R\win-library\4.4\00LOCK\zoo\libs\x64\zoo.dll to
## C:\Users\DELL\AppData\Local\R\win-library\4.4\zoo\libs\x64\zoo.dll: Permission
## denied## Warning: restored 'zoo'##
## The downloaded binary packages are in
## C:\Users\DELL\AppData\Local\Temp\RtmpwvM4vv\downloaded_packageslibrary(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, unionlibrary(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.numericStep 2. Input the Time Series Data
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))Data Description
Variable | Defintion
Month | The time period over 12 months
Values | The monetary value of Alabama building contracts (in $ millions)Step 3. Descriptive Statistics
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. :352.0Interpretation: The mean is 269.5, Median is 250.0Step 4. Time Series Plot
plot(df$month, df$values, type = "o", col = "blue", xlab = "Month", ylab = "$ Millions",
main = "Values of Alabama Building contracts in 12 months")
Interpretation: The time series plot exhibits a horizontal pattern as it is steady on the meanStep 5. Manually calculate the Three-Month Moving Average
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)Step 6. Calculate the Squared Errors (only for months were moving average is available)
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_values3), NA, (values - avg_values3)^2)
)Step 7. Compute Mean Squared Error (MSE) (excluding the intial moths with NA)
mse <- mean(df$squared_error, na.rm = TRUE)
mse## [1] 2040.444Interpretation: the MSE - 2040.44#Part B. Exponential Smoothing
Step 8. Input alpha value and
alpha <- 0.2Step 9. Find the Mean Squared Error (MSE)
exp_smooth <- rep(NA, length(df$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] 2593.762Interpretation: Output the MSE - 2593.76Step 10. Comparison
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 accuratelyLinear Trend Regression Approach
Project Object.
Forecast the average interest mortgage rate for the next year period 2024Question 3. Construct Time Series Plot and define its pattern
Step 1. Install and Load the packages
install.packages("ggplot2")## Installing package into 'C:/Users/DELL/AppData/Local/R/win-library/4.4'
## (as 'lib' is unspecified)## package 'ggplot2' successfully unpacked and MD5 sums checked
##
## The downloaded binary packages are in
## C:\Users\DELL\AppData\Local\Temp\RtmpwvM4vv\downloaded_packagesinstall.packages("readxl")## Installing package into 'C:/Users/DELL/AppData/Local/R/win-library/4.4'
## (as 'lib' is unspecified)## package 'readxl' successfully unpacked and MD5 sums checked
##
## The downloaded binary packages are in
## C:\Users\DELL\AppData\Local\Temp\RtmpwvM4vv\downloaded_packageslibrary(ggplot2)## Warning: package 'ggplot2' was built under R version 4.4.2library(readxl)## Warning: package 'readxl' was built under R version 4.4.2Step 2. Import the data
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 rowscolnames(df)## [1] "Year" "Period" "Interest_Rate"Data Description
Variable |Definition
Period | Time period in yearsStep 3. Descriptive Statistics
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. :8.053Interpretation: On average the number of interest rate of mortgage over a 20-year period is 5.08Step 4. Construct a time series plot
ggplot(df, aes(x = Period, y = `Interest_Rate`)) +
geom_line() +
geom_point() +
xlab("Period") +
ylab("Interest Rate") +
ggtitle("Interest Rate of Mortgage")
Interpretation: The Time Series Plot exhibits a Trend Pattern as there are gradual shifts to lower values over a long period of time. We observe a decreasing pattern or trend in this time series plot.Question 4. Develop the linear trend equation for this time series
Step 5. Devlop a linear trend equation
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: 0.0003637Interpretation: The Estimated linear trend equation: interest rate = 6.70 - 0.13*Period OR T_hat = 6.70 - 0.13*t
# The R-Square is 0.45 (Moderately fits the data)
# The overall model is significant as p-value < 0.05Question 5. Forecast the Average interest rate for the period 25.
Step 6. To find the MSE and MAPE values
# 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 %Interpretation: The MSE is 0.99. The MAPE is 15.79%Step 7. Forecast the number of interest rate in Period 25 (i.e., period 25)
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25## 1
## 3.472942Interpretation: The Forecasted number of interest rate in 2024 is 3.47