Question 1: Compute the forecast accuracy and determine which method
is better
Step 1: Identify the values
week <- 1:6
values <- c(17, 13, 15, 11, 17, 14)
Step 2: Compute the most recent value as forecast
forecasts_a <- values[-length(values)] #exclude the last value
actual_a <- values[-1] #exclude the first value
mse_a <- mean((actual_a - forecasts_a)^2)
mse_a
## [1] 16.2
forecast_week_7_a <- tail(values, 1)
forecast_week_7_a
## [1] 14
# Interpretations:
# The mean square error for the most recent value is 16.2
#The most recent forecast for week 7 is the value 14
Step 3. Average all of the data as forecast
cumulative_averages <- cumsum(values[-length(values)]) / (1:(length(values) - 1))
cumulative_averages
## [1] 17.0 15.0 15.0 14.0 14.6
forecasts_b <- cumulative_averages
actual_b <- values[-1] #exclude the first value
mse_b <- mean((actual_b - forecasts_b)^2)
mse_b
## [1] 8.272
forecast_week_7_b <- tail(values) #average of all months as forecast for month 7
forecast_week_7_b
## [1] 17 13 15 11 17 14
Interpretations:
The mean square error for the average of all data is 8.28 and the average forecast for week 7 is 15
Step 3. Which method is better
better_method <- ifelse(mse_a < mse_b, "Most Recent Value", "Average of All Data")
list(
MSE_most_recent_value = mse_a,
forecast_week_7_most_recent = forecast_week_7_a,
MSE_average_of_all_data = mse_b,
forecast_week_7_average = forecast_week_7_b,
Better_Method = better_method
)
## $MSE_most_recent_value
## [1] 16.2
##
## $forecast_week_7_most_recent
## [1] 14
##
## $MSE_average_of_all_data
## [1] 8.272
##
## $forecast_week_7_average
## [1] 17 13 15 11 17 14
##
## $Better_Method
## [1] "Average of All Data"
Interpretation: After running these lines of code, we can conclude that the better method is the average for all data method
Question 2. Construct a time series plot & Compare the
three-month moving average
Step 1. Install the following packages
#install.packages("dplyr")
#install.packages("zoo")
library(dplyr)
##
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
##
## filter, lag
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union
library(zoo)
##
## Attaching package: 'zoo'
## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numeric
Step 2. Construct a time series plot
df <- data.frame(month=c(1,2,3,4,5,6,7,8,9,10,11,12),
data=c(240,352,230,260,280,322,220,310,240,310,240,230))
# Time series plot
plot(df$month, df$data, type = "o", col = "blue", xlab = "Month", ylab = "Contracts Received",
main = "Monthly Percentage of All Contracts Received")
Interpretation: The time-series plot exhibits a horizontal pattern
Step 3. Compare the three-month moving average approach
df$avg_contract <- 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$data[11]) / 3)
mse <- mean(df$squared_error, na.rm = TRUE)
## Warning in mean.default(df$squared_error, na.rm = TRUE): argument is not
## numeric or logical: returning NA
mse
## [1] NA
Interpretation: From cleaning up the MSE our result came down to -5.70
Step 5. Exponential Smoothing
alpha = 0.2
exp_smooth <- rep(NA, length(df$data))
exp_smooth[1] <- df$data[1] #starting point
for(i in 2:length(df$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] -68993.72
Interpretation: By smoothing the forecast using 0.2 alpha, our result came down to -68993.72 which suggests that our time series plot is in a negative trend
Step 6. Compare
better_method <- ifelse(mse < mse_exp_smooth, "Three-month moving average", "Exponential smoothing")
list(
MSE_moving_average = mse,
mse_exponential_smoothing = mse,
Better_Method = better_method
)
## $MSE_moving_average
## [1] NA
##
## $mse_exponential_smoothing
## [1] NA
##
## $Better_Method
## [1] NA
Interpretation: After running the line of code, we can conclude that the better method is Exponential smoothing
Question 3a. Find the average interest rate for a 30-year-fixed-rate
mortgage
Step 1. Install and load the required packages
library(readxl)
library(ggplot2)
# Load the data
df1 <- read_excel("Mortgage.xlsx")
Step 2. Construct a time series plot
ggplot(df1, aes(x = Period, y = Interest_Rate)) +
geom_line() +
geom_point() +
xlab("Year") +
ylab("Interest Rate ($ millions") +
ggtitle("Time Series Plot of Mortgage Interest Rate")
Interpretation:
We observe an decreasing pattern or trend in the time series plot
Question 3b. Develop a linear trend equation
model <- lm(Interest_Rate ~ Period, data = df1)
summary(model)
##
## Call:
## lm(formula = Interest_Rate ~ Period, data = df1)
##
## 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: Interest Rate = 6.70 - 0.13*Period or y = 6.70 + -0.13t
Question 3c. Forecast the average interest rate for Period 25 (ie;
2024)
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25
## 1
## 3.472942
Interpretation:
The result for the forecast regarding period 25 is $34,729.42