** Time Series Approach **
Project Objective
Construct a time series plot, find an accurate forecasts based on MSEQuestion 1: Using the naive method (most recent value) as the forecast for the next week, compute the following measures of forecast accuracy.
Step 1: Import data
week <- 1:6
values <- c(17, 13, 15, 11, 17, 14)###Step 2: Most Recent Value as Forecast
forecast_a <- values[-length(values)] #Excludes the last value
actual_a <- values[-1] #Excludes the first value###Step 3: Calculate MSE, MAE, MAPE, and forecast value for week 7
#Calculate the Mean Squre Error (MSE)
mse_a <- mean((actual_a - forecast_a)^2)
mse_a #MSE = 16.2## [1] 16.2#Calculate the Mean Absolute Error (MAE)
mae_a <- mean(abs(actual_a - forecast_a))
mae_a #MEA = 3.8## [1] 3.8#Calculate the Mean Absolute Percentage Error (MAPE)
mape_a <- mean(abs((actual_a - forecast_a) / actual_a)) * 100
mape_a #MAPE = 27.44%## [1] 27.43778#Forecast value for week 7
forecast_week7_a <- tail(values, 1)
forecast_week7_a## [1] 14Interpretation: The measures of forecast value accuracy of week 7 is 14###Step 4: Ave of All Data as Forescast
#Part B. Ave of All Data as Forescast
cumulative_avg <- cumsum(values[-length(values)])/(1:(length(values) - 1))
cumulative_avg## [1] 17.0 15.0 15.0 14.0 14.6forecast_b <- cumulative_avg
actual_b <- values[-1] #Excludes the last value
mse_b <- mean((actual_b - forecast_b)^2)
mse_b## [1] 8.272Interpretation: The mean squared error is 8.27###Step 5: Avg of forecast value week 7
#Avg of forecast value week 7
forecast_week7_b <- mean(values) #Avg of all week forecast week 7
forecast_week7_b ## [1] 14.5Interpretation: The measures of forecast value accuracy of week 7 is 14.5###Step 6: Comparison and Result
#Comparison
better_method <- ifelse(mse_a < mse_b, "Most Recent Value", "Avg of All Data")
#Result
list(
MSE_Most_Recent_Value = mse_a,
Forecast_Week7_Most_Recent = forecast_week7_a,
MSE_Avg = mse_b,
Forecast_Week7_Avg = forecast_week7_b,
Better_Method = better_method
)## $MSE_Most_Recent_Value
## [1] 16.2
##
## $Forecast_Week7_Most_Recent
## [1] 14
##
## $MSE_Avg
## [1] 8.272
##
## $Forecast_Week7_Avg
## [1] 14.5
##
## $Better_Method
## [1] "Avg of All Data"Interpretation: The three-month moving average approach provides more accurate forecasts based on MSEQuestion 2: Use this data to construct a time series plot and compare the three-month moving average approach with the exponential smoothing forecast using alpha = 0.2.
###Step 1: Import data and descriptive statistics:
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(zoo)##
## Attaching package: 'zoo'## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numeric#Time series data
df <- data.frame(month = c(1,2,3,4,5,6,7,8,9,10,11,12),
values = c(240, 352, 230, 260, 280, 322, 220, 310, 240, 310, 240, 230))
#Descriptive statistics
summary(df) ## month values
## Min. : 1.00 Min. :220.0
## 1st Qu.: 3.75 1st Qu.:237.5
## Median : 6.50 Median :250.0
## Mean : 6.50 Mean :269.5
## 3rd Qu.: 9.25 3rd Qu.:310.0
## Max. :12.00 Max. :352.0Interpretation: The values of Alabama building contracts is 269.5 millions.###Step 2: Create time series plot
plot(df$month, df$values, type = "o", col = 'black', xlab = "Month", ylab = "Values",
main = "Values of Alabama building contracts (in $ millions)")
###Step 3: Mannualy caculate the Three-month Moving Avg
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 4: Calculate the square error and MSE
#Caculate the square error
df <- df %>%
mutate(
square_error = ifelse(is.na(avg_values3), NA, (values - avg_values3)^2)
)
#Compute MSE
mse <- mean(df$square_error, na.rm = TRUE)
mse ## [1] 2040.444MSE = 2040.44###Step 5: Exponential Smoothing, Comparison, and Result
#Exponential Smoothing
alpha <- 0.2
exp_smooth <- rep(NA, length(df$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 #Output = 84.04## [1] 84.04221#Comparison
better_method <- ifelse(mse < mse_exp_smooth, "Three-month Moving Avg", "Exponential Smoothing")
#Result
list(
MSE_Three_month_Moving_Avg = mse,
MSE_Avg = mse_exp_smooth,
Better_Method = better_method
)## $MSE_Three_month_Moving_Avg
## [1] 2040.444
##
## $MSE_Avg
## [1] 84.04221
##
## $Better_Method
## [1] "Exponential Smoothing"Interpretation : Exponential Smoothin is better model##Question 3: Construct a time series plot, develop the linear trend equation for this time series, and using the linear trend equation from question 3B, forecast the average interest rate for period 25
###Step 1: Import data
library(readxl)
library(ggplot2)
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 rowssummary(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: The average interest rate (%) for a 30-year fixed-rate mortgage over a 20-year period is 5.08###Step 2: Constrcut a time series
ggplot(df, aes(x= Period, y = Interest_Rate)) +
geom_line() +
geom_point() +
xlab("Period") +
ylab("Interest_Rate") +
ggtitle("Time series lot of interest rate")
Interpretation: We observe a decreasing trend from period 1 to 22, followed by an increasing trend from period 22 to 25.###Step 3: Develop a linear trend, find MSE and MAPE value, calculate the residuals
#Develop a linear trend
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#Resul: Interest_Rate = 6.70 - 0.13*Period
#To find MSE and MAPE value
df$predicted_rates <- predict(model)
#Calculate the residuals
df$residuals <- df$Interest_Rate - df$predicted_rates
#Calculate the MSE
mse <- mean(df$residuals^2)
cat("Mean Square Error (MSE):", mse,"\n")## Mean Square Error (MSE): 0.989475#Calculate the MAPE
df$percentage_error <- abs(df$residuals / df$Period) * 100
mape <- mean(df$percentage_error)
cat("Mean Absolute Percentage Error (MAPE):", mape , "%\n")## Mean Absolute Percentage Error (MAPE): 12.43593 %###Step 4: Find the orecast average interest rate for period 25
forecast_period25 <- predict(model, newdata = data.frame(Period =25))
forecast_period25## 1
## 3.472942Forecast rate in period 25 is 3.47