Question 1
Time series data
week <- 1:6 # This is the independent variable - time
values <- c(17,13,15,11,17,14) # dependent variablePart A. most values as forecast
forecast_a <- values[-length(values)] #Excludes the last values
actual_a <- values[-1] # Excludes the first values
mse_a <- mean((actual_a - forecast_a)^2)
mse_a #Mean square error is 16.2## [1] 16.2#forecast the values for week 7
forecast_week7_a <- tail(values, 1)
forecast_week7_a ## [1] 14#Inrepretation: The number of week 7 is 14
Part B. Average of all data as forecast
Note: We’re still working with the same dataseet in line 4&5
cumulative_averages <- cumsum(values[-length(values)]) / (1:(length(values) - 1))
cumulative_averages## [1] 17.0 15.0 15.0 14.0 14.6forecast_b <- cumulative_averages
actual_b <- values[-1] #Exclude the first value
mse_b <- mean((actual_b - forecast_b)^2)
mse_b #Mean square error is 8.272## [1] 8.272forecast the values for week 7
forecast_week7_b <- mean(values) #Average of week 7
forecast_week7_b## [1] 14.5#Inrepretation: The number of week 7 is 14.5
#Part c. comparison
better_method <- ifelse(mse_a < mse_b, "Most Recent Value", "Average of All Data")Results
list(
MSE_Most_Recent_Value = mse_a,
Forecast_Week7_Most_Recant = forecast_week7_a,
MSE_Average = mse_b,
Forecast_Week7_Average = forecast_week7_b,
Better_Method = better_method
) ## $MSE_Most_Recent_Value
## [1] 16.2
##
## $Forecast_Week7_Most_Recant
## [1] 14
##
## $MSE_Average
## [1] 8.272
##
## $Forecast_Week7_Average
## [1] 14.5
##
## $Better_Method
## [1] "Average of All Data"Qustion 2
Method 2 : Moving Average and Exponential Smoothing Approach
Part A. Moving Average:
Note: Install the “dplyr” package if you don’t have it. Just uncomment it
install.packages(“dplyr”)
install.packages(“zoo”)
Load the libraries
library(dplyr)## Warning: 套件 'dplyr' 是用 R 版本 4.4.2 來建造的##
## 載入套件：'dplyr'## 下列物件被遮斷自 'package:stats':
##
## filter, lag## 下列物件被遮斷自 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(zoo)## Warning: 套件 'zoo' 是用 R 版本 4.4.2 來建造的##
## 載入套件：'zoo'## 下列物件被遮斷自 'package:base':
##
## as.Date, as.Date.numericTime Series Data
df <- data.frame(month=c(1,2,3,4,5,6,7,8,9,10,11,12),
contract=c(240, 352, 230, 260, 280, 322, 220, 310, 240, 310, 240, 230))Descriptive statistics
summary(df)## month contract
## Min. : 1.00 Min. :220.0
## 1st Qu.: 3.75 1st Qu.:237.5
## Median : 6.50 Median :250.0
## Mean : 6.50 Mean :269.5
## 3rd Qu.: 9.25 3rd Qu.:310.0
## Max. :12.00 Max. :352.0Interpretation: The average contract over the 12 month period is 250.0
Time series plot:
plot(df$month, df$contract, type = "o", col = "blue", xlab = "month", ylab="contract(million)",
main = "Alabama building contracts")
#Inteprtation: The time series plot exhibits as norizontal pattern as it is # steady on the moon
Manually calculate the three-month moving average
df$avg_contract3 <- c(NA, NA, NA,
(df$contract[1] + df$contract[2] + df$contract[3]) / 3,
(df$contract[2] + df$contract[3] + df$contract[4]) / 3,
(df$contract[3] + df$contract[4] + df$contract[5]) / 3,
(df$contract[4] + df$contract[5] + df$contract[6]) / 3,
(df$contract[5] + df$contract[6] + df$contract[7]) / 3,
(df$contract[6] + df$contract[7] + df$contract[8]) / 3,
(df$contract[7] + df$contract[8] + df$contract[9]) / 3,
(df$contract[8] + df$contract[9] + df$contract[10]) / 3,
(df$contract[9] + df$contract[10] + df$contract[11]) / 3
)Calculate the square errors (only for month were moving average is available)
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_contract3), NA, (contract - avg_contract3)^2)
)Compute MSE (excluding the initialcmoths with Na)
mse <- mean(df$squared_error, na.rm = TRUE)
mse #Output the Mse - 2040.44## [1] 2040.444Paart B. Exponential Smoothing
Note; We’re using the same data set fror line 13
alpha <- 0.2
exp_smooth <- rep(NA, length(df$contract))
exp_smooth[1] <- df$contract[1] # starting point
for(i in 2 : length(df$contract)){
exp_smooth[i] <- alpha * df$contract[i-1] + (1 - alpha) * exp_smooth[i-1]
}
mse_exp_smooth <- mean((df$contract[2:12] - exp_smooth[2:12])^2)
mse_exp_smooth #output the MSE - 2536.443## [1] 2593.762Qustion 3~5
Method 3: Linear Trend Regression Approach
Note: Install the “ggplot2” package if you don’t have it. Just uncomment it
#install.packages(“ggplot2”)
Load the libararies
library(readxl) ## Warning: 套件 'readxl' 是用 R 版本 4.4.2 來建造的library(ggplot2)## Warning: 套件 'ggplot2' 是用 R 版本 4.4.2 來建造的#Load the data # If using excel:
df <- read_excel(file.choose())
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 rowsDescriptive 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. :8.053On average the number of interest rate over 20-year period
Construct a time series plot
ggplot(df, aes(x = Period, y = Interest_Rate)) +
geom_line() +
geom_point() +
xlab("Period") +
ylab("Interest") +
ggtitle(" average interest rate for a 30-year fixed-rate mortgage over a 20-year period")
Interpretation: we observe an decreasing pattern or trend in the time series polt
Develope a lonear 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: 0.0003637Result - estimated linear trend equation: Interest_Rat = 6.69541 + -0.12890*Period or
T_hat = 6.69541 + -0.12890*t
The R-square is 0.4459(Moderately fit the data)
The overall model is significant as p-value < 0.05
To find the MSE and Mape values
Caculate the fitted values from the miodel
df$predicted_Interest_Rate <- predict(model)Calculate the risiduals
df$residuals <- df$Interest_Rate - df$predicted_Interest_RateCalculate the Mean squared error (MSE)
mse <- mean(df$residuals^2)
cat("Mean squared error (Mse)", mse, "\n")## Mean squared error (Mse) 0.989475Bonus 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 %Focecast the number of Interest_Rate in 2024 (i.e., Period 25).
forecast_Period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_Period_25## 1
## 3.472942