#** Basic Times Series Analysis **
Excersice Objectives
Use Different methods for Basic Times Series Analysis : Naive Approach, Moving Averages/Exponential Smoothing, and Linear Trend Regression approach.Q1 - Using naive method (most recent value) as the forecast for the next week, compute the following measures of forecast accuracy (MAE, MSE, MAPE) & forecast the value for week 7?
Step (1) Time Series Data
week <- 1:6 # this is the independent variable -time
Value <- c(17,13,15,11,17,14) # dependent variabl Step (2) Most recent value as forecast
forecast <- Value[-length(Value)] #excludes the last sale (14)
actual <- Value[-1] #excludes the first sale (17)Step (3) Part a. Calculating Mean Absolute Error:
mae <- mean(abs(actual - forecast))
mae # Mean Absolute Error is 3.80## [1] 3.8Step (4) Part b. Calculating Mean Square Error:
mse <- mean((actual - forecast)^2)
mse # mean square error is 16.20## [1] 16.2Step (5) Part c. Calculating Mean Perccentage Absolute Error:
errors <- (abs(actual - forecast) / actual) * 100
mape <- mean(errors)
mape## [1] 27.43778Step (6) Part d. Forecast the Value for Week 7
forecast_week7 <- tail(Value, 1)
forecast_week7 #Interpretation : In week 7 value would be 14## [1] 14Q2 - (a.) Construct a time series plot. What type of pattern exists in the data? Upload a screenshot of your time series plot.
Step (1) 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, unionlibrary(zoo)##
## Attaching package: 'zoo'## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numeric###Step (2) Import the 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))Step (3) Descriptive Stats
summary(df)
Step (4) Create Time Series Plot
plot(df$month, df$values, type = "o", col = "pink", xlab = "Month", ylab = "Contract Values (in $ millions)",
main = "12 Month Alabama building contracts (in $ millions)")
#Interpretation: Their is fluctuation that goes up and down but not consistent gradual shift upward or downward making it a horizontal pattern.Q2 - (b). Compare the three-month moving average approach with the exponential smoothing forecast using alpha = 0.2. Which provides more accurate forecasts based on MSE? Upload a screenshot of your code outputs with proper explanation of your answers.
Step (1) Manually Calculate the Three-Week - Moving Average Method
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 (2) Calculate the square errors (only for months avaliable)
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_values3), NA, (values - avg_values3)^2)
)Step (3) Compute MSE (exclude the initial weeks w/NA)
mse <- mean(df$squared_error, na.rm = TRUE)
mse #Output the MSE = 2040.444## [1] 2040.444###Step (4) Exponential Smoothing Method (Note we are still using the same data & alpha <- 0.2)
alpha <- 0.2
exp_smooth <- rep(NA, length(df$values))
exp_smooth[1] <- df$values[1] #start point
for(i in 2: length(df$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 the MSE = 2593.76## [1] 2593.762Step (5) Comparison Between Moving Average & Exponential Smoothing
better_method <-ifelse(mse < mse_exp_smooth, "Three-Month Average", "Exponential Smoothing")
#results
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"#Interpretation: Moving average gives a more accurate forecastQ3 - (a.) Construct a time series plot. What type of pattern exists in the data?
Step (1) Load Libraries
library(ggplot2)
library(readxl)Step (2)Load the Data
df <- read_excel("Mortgage.xlsx")
df## # A tibble: 24 × 2
## Period Interest_Rate
## <dbl> <dbl>
## 1 1 8.05
## 2 2 6.97
## 3 3 6.54
## 4 4 5.83
## 5 5 5.84
## 6 6 5.87
## 7 7 6.41
## 8 8 6.34
## 9 9 6.03
## 10 10 5.04
## # ℹ 14 more rowsStep (3) Descriptive Stats:
summary(df)## Period Interest_Rate
## Min. : 1.00 Min. :2.958
## 1st Qu.: 6.75 1st Qu.:3.966
## Median :12.50 Median :4.863
## Mean :12.50 Mean :5.084
## 3rd Qu.:18.25 3rd Qu.:6.105
## Max. :24.00 Max. :8.053#Interpretation: on average the interest rate over a 20-year period is 5.08%Step (4) Constructing a Time Series Plot
ggplot(df, aes(x= Period, y=Interest_Rate))+
geom_line() +
geom_point() +
xlab("Period") +
ylab("Interest Rate") +
ggtitle("Time Series Plot of Interest Rate (%) over a 20-year period")
#Interpretation : We observed a decreasing pattern or trend in the time series plot that fluctuates upward after a period of time.Q4 - Develop the linear trend equation for this time series.
Step (1) Develop a Linar 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: 0.0003637#Result: T_hat = 6.70 - 0.13*tQ5 - Using the linear trend equation from question 3B, forecast the average interest rate for period 25 (i.e., 2024).
Step (1) Find MSE and MAPE values: Calculate the fitted values from the model
df$predicted_Interest_Rate <- predict(model)Step (2) Calculate the rResiduals
df$residuals <- df$Interest_Rate - df$predicted_Interest_RateStep (3)Calculate the MSE (Mean Squared Error)
mse <- mean(df$residuals^2)
cat("Mean Squared Error (MSE):", mse, "\n") # MSE = 0.99## Mean Squared Error (MSE): 0.989475Step (4) Calculate the MAPE (MEAN ABSOLUTE PERCENTAGE ERROR)
df$percentage_error <- abs(df$residuals / df$Interest_Rate) *100
mape <- mean(df$percentage_error)
cat("Mean Absolute Percentage Error (MAPE):",mape, "%\n") #MAPE = 15.79%## Mean Absolute Percentage Error (MAPE): 15.79088 %Step (5)Forcast the number of the average interest rate for period 25 (i.e., 2024).
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25 # 3.47## 1
## 3.472942