**Predicting the mortgage interest rate on FreddieMac website

Project Objective

Question 3: Construct a time series plot. What type of pattern exists in the data?

Step 1: Install and load required libraries

###install.packages(“ggplot2”)

#Load the libraries
library(ggplot2)

## Warning: package 'ggplot2' was built under R version 4.4.2

library(readxl)

## Warning: package 'readxl' was built under R version 4.4.2

Step 2: Import the data

# Load the data
df <- read_excel(file.choose())

###Step 3: Summarize the data

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.   :8.053

Interpretation:  On average the interest rate for a 30-year-fixed-rate mortgage over a 20-year period is 5.08

Step 4: Construct 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")

Interpretation: Interpretation:We observe a general downward trend in interest rates over the observed periods, with noticeable fluctuations and a sharp increase toward the end.

Step 5: Develop a linear 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

Interpretation: Result - estimated linear trend equation: Interest_rate = 6.69 - 0.13*Period 
The R-square is 0.45
The overall model is significant as p-value < 0.05

Step 6: To find the MSE and MAPE values, Calculate the residuals, Calculate the Mean Squared Error (MSE)

df$predicted_interest_rate <- predict(model)

df$residuals <- df$Interest_Rate - df$predicted_interest_rate

mse <- mean(df$residuals^2)
cat("Mean Squared Error (MSE):", mse, "\n")

## Mean Squared Error (MSE): 0.989475

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 %

Interpretation:
Mean Absolute Percentage Error (MAPE): 15.79088 %
Mean Squared Error (MSE): 0.989475

Step 6: Forecast the number of interest rate in period 25

forecast_period_25 <- predict (model, newdata = data.frame(Period = 25))
forecast_period_25

##        1 
## 3.472942

Interpretation: The forecasted amount of interest rate in period 25 is 3.47