Class Exercise 16: Chapter 17
Question 1 : Using the naive method (most recent value) as the forecast for the next week, compute the following measures of forecast accuracy.
week <- 1:6
values <- c(17, 13, 15, 11, 17, 14)
forecast <- values[-length(values)]
actual <- values[-1]a. Mean absolute error
mae <- mean(abs(actual - forecast))
mae## [1] 3.8b. Mean squared error.
mse <- mean((actual - forecast)^2)
mse ## [1] 16.2c. Mean absolute percentage error.
mape <- mean(abs((actual - forecast)/actual))* 100
mape ## [1] 27.43778d. What is the forecast for week 7?
forecast_week7 <- tail(values, 1)
forecast_week7## [1] 14Question 2
A. Construct a time series plot.
#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, unionlibrary(zoo)##
## Attaching package: 'zoo'## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numericdf <- 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))
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. :352.0plot(df$month, df$values, type = "o", col = "blue", xlab ="Month", ylab = "Values",
main = "The values of Alabama building contracts Plot")
The time series plot exhibits a horizontal pattern as it is steady on the mean B.
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)
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_values3), NA, (values - avg_values3)^2)
)
mse <- mean(df$squared_error, na.rm = TRUE)
mse## [1] 2040.444alpha <- 0.2
exp_smooth <- rep(NA, length(df$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## [1] 2593.762better_method <- ifelse(mse < mse_exp_smooth, "three-month moving average", "exponential smoothing")
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"2040.444(MSE_Moving_Average) < 2593.762 (MSE_Exponential_Smoothing)
The moving average(with k = 3) provided more accurate forecasts than the exponential smoothing(with alpha = 0.2) Question 3 Construct a time series plot. What type of pattern exists in the data?
#install.packages("ggplot2")
#install.packages("readxl")
library(ggplot2)
library(readxl)
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. :8.053ggplot(df, aes(x= Period, y = Interest_Rate)) +
geom_line() +
geom_point()+
xlab("Period") +
ylab("Interest_Rate") +
ggtitle("time series plot of mortgage")
Question 4 Develop the linear trend equation for this time series.
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: 0.0003637Question 5 Using the linear trend equation from question 3B, forecast the average interest rate for period 25 (i.e., 2024).
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25## 1
## 3.472942