1 Example 1

Find a straight line that best fits the following data:

$$\begin{array}{c|cccccccccccc} x & 1 & 6 & 11 & 16 & 20 & 26 \\ \hline y & 13 & 16 & 17 & 23 & 24 & 31 \end{array}$$ Solution

Let the straight line be given by the equation $Y = a + bX$. To determine the values of $a$ and $b$ for this line, we use the normal equations:

$$\begin{array}{cccccc} \displaystyle \sum_{i=1}^{n} y_i &=& na & + &\displaystyle b \sum_{i=1}^{n} x_i \\ \displaystyle \sum_{i=1}^{n} y_i x_i &= & a\displaystyle\sum_{i=1}^{n}x_i & + & \displaystyle b\sum_{i=1}^{n}x_i^{2} \end{array}$$

can be written as

$$\begin{array}{cccccc} \displaystyle \sum y & = &na &+& \displaystyle b \sum x \\ \displaystyle \sum yx & = & \displaystyle a \sum x &+& \displaystyle b \sum x^{2} \end{array}$$

We need to calculate $\displaystyle \sum y$, $\displaystyle \sum x$, $\displaystyle \sum xy$, and $\displaystyle \sum x^2$, which can be obtained from the following table:

$$\begin{array}{ccccccccccccc} i & x & y & x^2 & xy \\ \hline 1 & 1 & 13 & 1 & 13 \\ 2 & 6 & 16 & 36 & 96 \\ 3 & 11 & 17 & 121 & 187 \\ 4 & 16 & 23 & 256 & 368 \\ 5 & 20 & 24 & 400 & 480 \\ 6 & 26 & 31 & 676 & 806 \\ \hline n = 6 & \displaystyle \sum x = 80 & \displaystyle \sum y = 124 & \displaystyle \sum x^2 = 1490 & \displaystyle \sum xy = 1950 \\ \hline \end{array}$$

For example, $$\begin{align*} \displaystyle \sum x & = \sum_{i=1}^6 x_i \\ & = x_1 + x_2 + x_3 + x_4 + x_5 + x_6 \\ & = 1 + 6 + 11 + 16 + 20 + 26 \\ & = 80, \end{align*}$$ $$\begin{align*} \displaystyle \sum y & = \sum_{i=1}^6 y_i \\ & = y_1 + y_2 + y_3 + y_4 + y_5 + y_6 \\ & = 13 + 16 + 17 + 23 + 24 + 31 \\ & = 124, \end{align*}$$ $$\begin{align*} \displaystyle \sum y & = \sum_{i=1}^6 x_i^2 \\ & = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 \\ & = 1^2 + 6^2 + 11^2 + 16^2 + 20^2 + 26^2 \\ & = 1 + 36 + 121 + 256 + 400 + 676 \\ & = 1490, \end{align*}$$ and $$\begin{align*} \displaystyle \sum xy & = \sum_{i=1}^6 x_i y_i \\ & = x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4 + x_5 y_5 + x_6 y_6 \\ & = 13 + 96 + 187 + 368 + 480 + 806 \\ & = 1950 \end{align*}$$

By substituting the values of $\displaystyle \sum y$, $\displaystyle \sum x$, $\displaystyle \sum xy$, and $\displaystyle \sum x^2$ into the normal equations, we obtain:

$$\begin{array}{cccccc} \displaystyle 124 & =& 6 a &+& 80 b && (1)\\ \displaystyle 1950 & = &80 a &+& 1490 b && (2) \end{array}$$ Now we solve equations (1) and (2).

Multiplying equation (1) by 80 and equation (2) by 6, i.e.

$$\begin{array}{cccccc} \displaystyle 124 & =& 6 a &+& 80 b && \times \quad 80 \\ \displaystyle 1950 & =& 80 a &+& 1490 b && \times \quad 6 \end{array}$$ we have $$\begin{array}{cccccc} \displaystyle 9920 & = & 480 a &+& 6400 b && (3) \\ \displaystyle 11700 & = & 480 a &+&8940 b && (4) \end{array}$$

Subtracting (3) from (4), we obtain

$$\begin{array}{ccllll} \displaystyle & 1780 & = & 2540 b & & \\ \displaystyle \Rightarrow & b & = & 1780/2540 & =& 0.7008 \end{array}$$

By substituting the value of $b$ in equation (2), we have

$$\begin{array}{ccclll} & 24 & = & 6a+80\times 0.7008 \\ & 124 & = & 6a+56.064 \\ & 67.936 & = & 6a \\ \Rightarrow & a & = & 11.3227 \end{array}$$

with these values for $a$ and $b$, the equation for the line of best fit is: $Y=11.3227+0.7008 X$.

# Create a data frame to store the data
X <- c(1,  6, 11, 16, 20, 26)
Y <- c(13, 16,  17, 23, 24, 31)
data <- data.frame(X, Y)
data

# Create a scatterplot of Y versus X
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
  )

# Fit a linear regression model to the data
linear_model <- lm(Y ~ X, data = data)
 
# Summarize the linear regression model
summary(linear_model)

## 
## Call:
## lm(formula = Y ~ X, data = data)
## 
## Residuals:
##       1       2       3       4       5       6 
##  0.9764  0.4724 -2.0315  0.4646 -1.3386  1.4567 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 11.32283    1.17618   9.627 0.000651 ***
## X            0.70079    0.07464   9.389 0.000717 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.536 on 4 degrees of freedom
## Multiple R-squared:  0.9566, Adjusted R-squared:  0.9457 
## F-statistic: 88.16 on 1 and 4 DF,  p-value: 0.0007169

# Calculate the predicted Y for a range of X
predicted_values <- predict(
  linear_model, 
  newdata = data.frame(X = seq(min(X), 
                      max(X), 
                      length.out = 5)
                      )
 )

# Plot the data points and the predicted Y values
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
  )
lines(seq(min(X), 
          max(X), 
          length.out = 5), 
      predicted_values, col = "red"
      )

2 Example 2

Find a straight line that best fits the following data:

$$\begin{array}{c|cccccccccccc} x & 6 & 7 & 8 & 9 & 11 \\ \hline y & 5 & 4 & 3 & 2 & 1 \end{array}$$

Solution

Let the straight line be given by the equation $Y = a + bX$. To determine the values of $a$ and $b$ for this line, we use the normal equations:

$$\begin{array}{cccccc} \displaystyle \sum_{i=1}^{n} y_i &=& na & + & \displaystyle b \sum_{i=1}^{n} x_i \\ \displaystyle \sum_{i=1}^{n} y_i x_i &=& \displaystyle a \sum_{i=1}^{n}x_i & + & \displaystyle b\sum_{i=1}^{n}x_i^{2} \end{array}$$ can be written as

$$\begin{array}{cccccc} \displaystyle \sum y & = & na & + & \displaystyle b \sum x \\ \displaystyle \sum yx & = & \displaystyle a \sum x & + & \displaystyle b \sum x^{2} \end{array}$$

We need to calculate $\displaystyle \sum y$, $\displaystyle \sum x$, $\displaystyle \sum xy$, and $\displaystyle \sum x^2$, which can be obtained from the following table:

$$\begin{array}{ccccccccccccc} i & x & y & x^2 & xy \\ \hline 1 & 6 & 5 & 36 & 30 \\ 2 & 7 & 4 & 49 & 28 \\ 3 & 8 & 3 & 64 & 24 \\ 4 & 9 & 2 & 81 & 18 \\ 5 & 11 & 1 & 121 & 11 \\ \hline n =5& \displaystyle \sum x = 41 & \displaystyle \sum y = 15 & \displaystyle \sum x^2 = 351 & \displaystyle \sum xy = 111 \\ \hline \end{array}$$

By substituting the values of $\displaystyle \sum y$, $\displaystyle \sum x$, $\displaystyle \sum xy$, and $\displaystyle \sum x^2$ into the normal equations, we obtain:

$$\begin{array}{cccccc} \displaystyle 15 & = & 5 a & + & 41 b & & (1)\\ \displaystyle 111 & = & 41 a & + & 351 b & & (2) \end{array}$$

Now we solve equations (1) and (2).

Multiplying equation (1) by 41 and equation (2) by 5, i.e.

$$\begin{array}{cccccc} \displaystyle 15 & = & 5 a & + & 41 b & & \times \quad 41 \\ \displaystyle 111 & = & 41 a & + & 351 b & & \times \quad 5 \end{array}$$

we have $$\begin{array}{cccccc} \displaystyle 615 & = & 205 a & + & 168 b && (3) \\ \displaystyle 555 & = & 205 a & + & 1755 b && (4) \end{array}$$

Subtracting (3) from (4), we obtain

$$\begin{array}{ccllllc} \displaystyle & -60 & = & 74 b & & \\ \displaystyle \Rightarrow & b & = & -60/74 & =& -0.8108 \end{array}$$

By substituting the value of $b$ in equation (2), we have $$\begin{array}{cclllllllll} & 15 & = & 5a+41\times (-0.8108) \\ & 15 & = & 5a - 33.2428 \\ \Rightarrow & a & = & 9.6486 \end{array}$$

with these values for $a$ and $b$, the equation for the line of best fit is: $Y=9.6486-0.8108 X$.

# Create a data frame to store the data
X <- c(6, 7, 8, 9, 11)
Y <- c(5, 4, 3, 2, 1)
data <- data.frame(X, Y)
data

# Create a scatterplot of Y versus X
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
  )

# Fit a linear regression model to the data
linear_model <- lm(Y ~ X, data = data)
 
# Summarize the linear regression model
summary(linear_model)

## 
## Call:
## lm(formula = Y ~ X, data = data)
## 
## Residuals:
##        1        2        3        4        5 
##  0.21622  0.02703 -0.16216 -0.35135  0.27027 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  9.64865    0.65370   14.76 0.000675 ***
## X           -0.81081    0.07802  -10.39 0.001901 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.3002 on 3 degrees of freedom
## Multiple R-squared:  0.973,  Adjusted R-squared:  0.964 
## F-statistic:   108 on 1 and 3 DF,  p-value: 0.001901

# Calculate the predicted Y for a range of X
predicted_values <- predict(
  linear_model, 
  newdata = data.frame(X = seq(min(X), 
                      max(X), 
                      length.out = 5)
                      )
 )

# Plot the data points and the predicted Y values
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
  )
lines(seq(min(X), 
          max(X), 
          length.out = 5), 
      predicted_values, col = "red"
      )

3 Example 3

Find a straight line that best fits the following data:

$$\begin{array}{c|c} \hline \textrm{Year} & \textrm{Twitter users} \\ \hline 11.00 & 68 \\ 11.25 & 85 \\ 11.50 & 101 \\ 11.75 & 117 \\ 12.75 & 185 \\ \hline \end{array}$$

Solution

Let the straight line be given by the equation $Y = a + bX$. To determine the values of $a$ and $b$ for this line, we use the normal equations:

$$\begin{array}{cccccc} \displaystyle \sum_{i=1}^{n} y_i &=& na & + & \displaystyle b \sum_{i=1}^{n} x_i \\ \displaystyle \sum_{i=1}^{n} y_i x_i &=& \displaystyle a \sum_{i=1}^{n}x_i & + & \displaystyle b\sum_{i=1}^{n}x_i^{2} \end{array}$$

can be written as

$$\begin{array}{cccccc} \displaystyle \sum y & = & na &+& \displaystyle b \sum x \\ \displaystyle \sum yx & = & \displaystyle a \sum x &+& \displaystyle b \sum x^{2} \end{array}$$ We need to calculate $\displaystyle \sum y$, $\displaystyle \sum x$, $\displaystyle \sum xy$, and $\displaystyle \sum x^2$, which can be obtained from the following table:

$$\begin{array}{ccccccccccccc} i & x & y & x^2 & xy \\ \hline 1 & 11 & 68 & 121 & 748 \\ 2 & 11.25 & 85 & 126.5625 & 956.25 \\ 3 & 11.5 & 101 & 132.25 & 1161.5 \\ 4 & 11.75 & 117 & 138.0625 & 1374.75 \\ 5 & 12.75 & 185 & 162.5625 & 2358.75 \\ \hline n = 5& \displaystyle \sum x = 58.25 & \displaystyle \sum y = 556 & \displaystyle \sum x^2 = 680.4375 & \displaystyle \sum xy = 6599.25 \\ \hline \end{array}$$

By substituting the values of $\displaystyle \sum y$, $\displaystyle \sum x$, $\displaystyle \sum xy$, and $\displaystyle \sum x^2$ into the normal equations, we obtain:

$$\begin{array}{cccccc} \displaystyle 556 & = & 5 a &+& 58.25 b && (1)\\ \displaystyle 6599.25 & = & 58.25 a &+& 680.4375 b & & (2) \end{array}$$

We will now solve equations (1) and (2) using Cramer’s rule. Let us proceed with the calculations:

$$a = \dfrac{\det \left(\begin{array}{cc} 556 & 58.25 \\ 6599.25 & 680.4375 \end{array} \right)}{\det \left(\begin{array}{cc}5 & 58.25 \\ 58.25 & 680.4375 \end{array} \right)} = -666.6370$$

and

$$b = \dfrac{\det \left(\begin{array}{cc} 5 & 556 \\ 58.25 & 6599.25 \end{array} \right)}{\det \left(\begin{array}{cc}5 & 58.25 \\ 58.25 & 680.4375 \end{array} \right)} = 66.7671$$

with these values for $a$ and $b$, the equation for the line of best fit is: $Y=-666.6370+66.7671 X$.

# Create a data frame to store the data
Year <- c(11.00, 11.25, 11.50, 11.75, 12.75 )
Users <- c(68, 85, 101, 117, 185)
data <- data.frame(Year, Users)
data

# Create a scatterplot of Y versus X
plot(
  data$Year, 
  data$Users, 
  main = "Users vs. Year", 
  xlab = "Year", 
  ylab = "Users"
  )

# Fit a linear regression model to the data
linear_model <- lm(Users~Year, data = data)
 
# Summarize the linear regression model
summary(linear_model)

## 
## Call:
## lm(formula = Users ~ Year, data = data)
## 
## Residuals:
##       1       2       3       4       5 
##  0.1986  0.5068 -0.1849 -0.8767  0.3562 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -666.6370     5.5204  -120.8 1.25e-06 ***
## Year          66.7671     0.4732   141.1 7.85e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.6393 on 3 degrees of freedom
## Multiple R-squared:  0.9998, Adjusted R-squared:  0.9998 
## F-statistic: 1.991e+04 on 1 and 3 DF,  p-value: 7.85e-07

# Calculate the predicted Y for a range of X
predicted_values <- predict(
  linear_model, 
  newdata = data.frame(Year = seq(min(Year), 
                      max(Year), 
                      length.out = 5)
                      )
 )

# Plot the data points and the predicted Y values
plot(
  data$Year, 
  data$Users, 
  main = "Users vs. Year", 
  xlab = "Year", 
  ylab = "Users"
  )
lines(seq(min(Year), 
          max(Year), 
          length.out = 5), 
      predicted_values, col = "red"
      )

4 Example 4

Fit a second-degree parabola to the following data:

$$\begin{array}{c|cccccccccccc} x & y \\ \hline 0 & 1 \\ 1 & 3 \\ 2 & 4 \\ 3 & 5 \\ 4 & 6 \\ \hline \end{array}$$

Solution

Let $Y = a+ bX + cX^2$ be the second-degree parabola, and we have to determine $a$, $b$ and $c$. Normal equations for the second-degree parabola are

$$\begin{array}{ccccccccccccc} \displaystyle \sum y & = & \displaystyle na & + & \displaystyle b \sum x & + & \displaystyle c \sum x^2 \\ \displaystyle \sum xy & = & \displaystyle a \sum x & + & \displaystyle b \sum x^2 & + & \displaystyle c \sum x^3 \\ \displaystyle \sum x^2y & = & \displaystyle a \sum x^2 & + & \displaystyle b \sum x^3 & + & \displaystyle c \sum x^4 \\ \end{array}$$

To solve the above normal equations, we need $\displaystyle \sum y$, $\displaystyle \sum x$, $\displaystyle \sum xy$, $\displaystyle \sum x^2 y$, $\displaystyle \sum x^2$, $\displaystyle \sum x^3$, and $\displaystyle \sum x^4$, which are obtained from the following table:

$$\begin{array}{ccccccccc} i & x & y & xy & x^2 & x^2y & x^3 & x^4 \\ \hline 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 3 & 3 & 1 & 3 & 1 & 1 \\ 3 & 2 & 4 & 8 & 4 & 16 & 8 & 16 \\ 4 & 3 & 5 & 15 & 9 & 45 & 27 & 81 \\ 5 & 4 & 6 & 24 & 16 & 96 & 64 & 256 \\ \hline n = 5& \displaystyle \sum x = 10 & \displaystyle \sum y = 19 & \displaystyle \sum xy = 50 & \displaystyle \sum x^2 = 30 & \displaystyle \sum x^2y = 160 & \displaystyle \sum x^3 = 100 & \displaystyle \sum x^4 = 354\\ \hline \end{array}$$

Substituting the values of $\displaystyle \sum y$, $\sum x$, $\displaystyle \sum xy$, $\displaystyle \sum x^2 y$, $x^2 $, $\sum x^3$, and $\displaystyle \sum x^4$ into the above normal equations, we have:

$$\begin{array}{ccccccccccccc} \displaystyle \sum y & = & \displaystyle na & + & \displaystyle b \sum x & + & \displaystyle c \sum x^2 \\ \displaystyle \sum xy & = & \displaystyle a \sum x & + & \displaystyle b \sum x^2 & + & \displaystyle c \sum x^3 \\ \displaystyle \sum x^2y & = & \displaystyle a \sum x^2 & + & \displaystyle b \sum x^3 & + & \displaystyle c \sum x^4 \\ \end{array}$$ $$\begin{array}{ccccccccccccc} 19 & = & 5 a & + & 10 b & + & 30 c & & (1) \\ 50 & = & 10 a & + & 30 b & + & 100 c & & (2) \\ 160 & = & 30 a & + & 100 b & + & 354 c & & (3) \\ \end{array}$$

Now, we solve equations (1), (2) and (3).

Multiplying equation (1) by 2, we get

$$\begin{array}{ccccccccccccc} 38 & = & 10 a &+& 20 b &+& 60 c && (4) \end{array}$$

Subtracting equation (4) from equation (2), we get

$$\begin{array}{ccccccccccccc} 50 & = & 10 a & + & 30 b & + & 100 c & & (2) \\ 38 & = & 10 a & + & 20 b & + & 60 c & & (4) \\ \hline 12 & = & 0 a & + & 10 b & + & 40 c & & (5) \\ \end{array}$$

Multiplying equation (2) by 3, we get

$$\begin{array}{ccccccccccccc} 150 & = & 30 a &+& 90 b &+& 300 c && (6) \end{array}$$

Subtracting equation (6) from equation (3), we get

$$\begin{array}{ccccccccccccc} 160 & = & 30 a & + & 100 b & + & 354 c & & (3) \\ 150 & = & 30 a & + & 90 b & + & 300 c & & (6)\\ \hline 10 & = & 0 a & + & 10 b & + & 54 c & & (7) \\ \end{array}$$

Now we solve equation (5) and (7).

Subtracting equation (5) from equation (7), we get

$$\begin{array}{ccccccccccccc} 12 & = & 0 a & + & 10 b & + & 40 c & & (5) \\ 10 & = & 0 a & + & 10 b & + & 54 c & & (7) \\ \hline 2 & = & & & 0 b & - & 14 c & & \\ \end{array}$$

$$\begin{array}{cclllllllllllll} & 2 & = & - 14c \\ & c& = & -2/14 \\ \Rightarrow & c& = & -0.1429 \\ \end{array}$$

Substituting the value of $\mathrm{c}$ in equation (7), we get

$$\begin{array}{ccclllllllllll} & 10 & = & 10 b &+ & 54 (-0.1429 ) \\ & 10 & = & 10 b &- & 7.7166 \\ & 10 b & = & 10 &+ & 7.7166 \\ \Rightarrow & b & = & 1.77166 \\ \end{array}$$

Substituting the value of $b$ and $a$ in equation (1), we get

$$\begin{array}{ccccccccccccc} & 19 & = & 5 a & + & 10 (1.77166) & + & 30 (-0.1429) \\ & 19 & = & 5 a & + & 17.717 & - & 4.287 \\ \Rightarrow & a & = & 1.114 \end{array}$$

Thus, the second degree of parabola of best fit is

$$Y=1.114+1.7717 X-0.1429 X^{2}$$

# Create a data frame to store the data
X <- c(0, 1, 2, 3, 4)
Y <- c(1, 3 ,4, 5, 6)
data <- data.frame(X, Y)
data

# Create a scatterplot of Y versus X
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
  )

# Fit a quadratic regression model to the data
quadratic_model <- lm(Y ~ X + I(X^2), data = data)
 
# Summarize the quadratic regression model
summary(quadratic_model)

## 
## Call:
## lm(formula = Y ~ X + I(X^2), data = data)
## 
## Residuals:
##        1        2        3        4        5 
## -0.11429  0.25714 -0.08571 -0.14286  0.08571 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)  
## (Intercept)  1.11429    0.22497   4.953   0.0384 *
## X            1.77143    0.26650   6.647   0.0219 *
## I(X^2)      -0.14286    0.06389  -2.236   0.1548  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.239 on 2 degrees of freedom
## Multiple R-squared:  0.9923, Adjusted R-squared:  0.9846 
## F-statistic: 128.5 on 2 and 2 DF,  p-value: 0.007722

# Calculate the predicted Y for a range of X
predicted_values <- predict(
  quadratic_model, 
  newdata = data.frame(X = seq(min(X), 
                      max(X), 
                      length.out = 5)
                      )
  )

# Plot the data points and the predicted Y values
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
  )
lines(seq(min(X), 
          max(X), 
          length.out = 5), 
      predicted_values, col = "red"
      )

5 Example 5

Fit the power curve $Y = a X^b$ to the following data:

$$\begin{array}{c|cccccccccccc} x & y \\ \hline 6 & 9 \\ 2 & 11 \\ 10 & 12 \\ 5 & 8 \\ 8 & 7 \\ \hline \end{array}$$

Solution

Let the power curve be $Y=aX^b$, and the normal equations for estimating $a$ and $b$ are:

$$\begin{array}{ccccccccccccc} \displaystyle \sum u & = & n A & + & \displaystyle b\sum v \\ \displaystyle \sum uv & = & \displaystyle A \sum v & + & \displaystyle b \sum v^{2} \end{array}$$

where $u=\log y, v= \log x$, and $A=\log a$.

Note: Here we are using $\log$ with base $10$.

To find the values of $a$ and $b$ from the above normal equations, we require $\displaystyle \sum u$, $\displaystyle \sum v$, $\displaystyle \sum uv$, and $\displaystyle \sum v^2$, which are calculated in the following table:

$$\begin{array}{ccccccccc} i & x & y & u=\log y & v=\log x & uv & v^2 \\ \hline 1 & 6 & 9 & 0.954242509 & 0.77815125 & 0.742545002 & 0.605519368 \\ 2 & 2 & 11 & 1.041392685 & 0.301029996 & 0.313490435 & 0.090619058 \\ 3 & 10 & 12 & 1.079181246 & 1 & 1.079181246 & 1 \\ 4 & 5 & 8 & 0.903089987 & 0.698970004 & 0.631232812 & 0.488559067 \\ 5 & 8 & 7 & 0.84509804 & 0.903089987 & 0.763199578 & 0.815571525 \\ \hline n = 5& \displaystyle \sum x = 31 & \displaystyle \sum y = 47 & \displaystyle \sum u = 4.823004468 & \displaystyle \sum v= 3.681241237 & \displaystyle \sum uv = 3.529649074 & \displaystyle \sum v^2 = 3.000269018 \\ \hline \end{array}$$

Substituting the values $\displaystyle \sum u=4.8230, \displaystyle \sum v =3.6812, \displaystyle \sum uv =3.5296$ and $\displaystyle \sum v^{2}=3.0003$ into the above normal equations, we obtain:

$$\begin{array}{ccccccccccccc} 4.8230 & = & 5 A & + & 3.6812 b & & (1) \\ 3.5296 & = & 3.6812 A & + & 3.0003 b & & (2) \end{array}$$

Now, we solve the equation (1) and equation (2).

Multiplying equation (1) by $3.6812$ and equation (3) by $5$, i.e.,

$$\begin{array}{cccllllllllllc} 4.8230 & = 5 A & + & 3.6812 b && (1) \times 3.6812 \\ 3.5296 & = 3.6812 A & + & 3.0003 b && (2) \times 5 \end{array}$$

we have

$$\begin{array}{ccccccccccccc} 17.7544 & = & 18.4060 A & + & 13.5512 b & & (3) \\ 17.6480 & = & 18.4060 A & + & 15.0015 b & & (4) \end{array}$$

Subtracting equation (4) from equation (3), we have

$$\begin{array}{ccccccccccccc} & 0.1064 & = & -1.4503 b \\ \Rightarrow & b & = & -0.0734 \end{array}$$

Substituting the value of $b$ in equation (1), we get

$$\begin{array}{ccccccccccccc} 4.8230 & = & 5 A & + & 3.6812 (-0.0734) \\ A & = & 1.0186 \end{array}$$

Now $$a = \log^{-1} A= \log^{-1} (1.0186) = 10^{1.0186} = 10.4376$$

Thus, the power curve of the best fit is $Y=10.4376 X^{-0.0734}.$

X <- c(6, 2, 10, 5, 8)
Y <- c(9, 11, 12, 8, 7)
data <- data.frame(X, Y)
data

# Create a scatterplot of Y versus X
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
)

# Fit a quadratic regression model to the data
model <- lm(log10(Y) ~ log10(X), data = data)

# Summarize the quadratic regression model
summary(model)

## 
## Call:
## lm(formula = log10(Y) ~ log10(X), data = data)
## 
## Residuals:
##         1         2         3         4         5 
## -0.007283  0.044852  0.133936 -0.064247 -0.107259 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)  1.01863    0.15679   6.497   0.0074 **
## log10(X)    -0.07339    0.20240  -0.363   0.7410   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.109 on 3 degrees of freedom
## Multiple R-squared:  0.04198,    Adjusted R-squared:  -0.2774 
## F-statistic: 0.1315 on 1 and 3 DF,  p-value: 0.741

# Calculate the value of a
a <- 10^((coef(model)[1]))

# Calculate the value of b
b <- coef(model)[2]

# Calculate the predicted Y for a range of X
predicted_values <- a*(seq(min(data$X), max(data$X), length.out = 5))^b
print(a)

## (Intercept) 
##    10.43836

print(b)

##    log10(X) 
## -0.07338735

# Plot the data points and the predicted Y values
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
)
lines(seq(min(data$X), 
          max(data$X), 
          length.out = 5), 
      predicted_values, col = "red"
)

6 Example 6

Fit the power curve $Y = a X^b$ to the following data:

$$\begin{array}{c|cccccccccccc} x & y \\ \hline 0.135 & 22000 \\ 0.67 & 13000 \\ 2.45 & 8000 \\ 75 & 2550 \\ 1775 & 900 \\ \hline \end{array}$$

Solution

Let the power curve be $Y=aX^b$, and the normal equations for estimating $a$ and $b$ are:

$$\begin{array}{ccccccccccccc} \displaystyle \sum u & = & n A & + & \displaystyle b\sum v \\ \displaystyle \sum uv & = & \displaystyle A \sum v & + & \displaystyle b \sum v^{2} \end{array}$$

where $u=\log y, v= \log x$, and $A=\log a$.

Note: Here we are using $\log$ with base 10.

To find the values of $a$ and $b$ from the above normal equations, we require $\displaystyle \sum u$, $\displaystyle \sum v$, $\displaystyle \sum uv$, and $\displaystyle \sum v^2$, which are calculated in the following table:

$$\begin{array}{ccccccccc} i & x & y & u=\log y & v=\log x & uv & v^2 \\ \hline 1 & 0.135 & 22000 & 4.342422681 & -0.869666232 & -3.776458368 & 0.756319354 \\ 2 & 0.67 & 13000 & 4.113943352 & -0.173925197 & -0.715518409 & 0.030249974 \\ 3 & 2.45 & 8000 & 3.903089987 & 0.389166084 & 1.518950247 & 0.151450241 \\ 4 & 75 & 2550 & 3.40654018 & 1.875061263 & 6.387471535 & 3.515854741 \\ 5 & 1775 & 900 & 2.954242509 & 3.249198357 & 9.598919909 & 10.55728997 \\ \hline n = 5& \displaystyle \sum x = 1853.255 & \displaystyle \sum y = 46450 & \displaystyle \sum u = 18.72023871 & \displaystyle \sum v= 4.469834276 & \displaystyle \sum uv = 13.01336491 & \displaystyle \sum v^2 = 15.01116428 \\ \hline \end{array}$$

Substituting the values of $\displaystyle \sum u=18.72023871, \displaystyle \sum v=4.469834276,\sum uv= 13.01336491$ and $\displaystyle \sum v^2=15.01116428$ into the above normal equations, we obtain:

$$\begin{array}{ccccccccccccc} 1853.255 & = & 5 A & + & 4.469834276 b & & (1) \\ 13.01336491 & = & 4.469834276 A & + & 15.01116428 b & & (2) \end{array}$$

These two equations can be solved for $A$ and $b$ using Cramer’s rule:

$$A = \dfrac{\left| \begin{array}{cc} \color{red}{ \sum u} & \sum v \\ \color{red}{\sum uv} & \sum v^{2} \end{array} \right|} {\left| \begin{array}{cc} n & \sum v \\ \sum v & \sum v^{2} \end{array} \right|} = \dfrac{\left| \begin{array}{cc} \color{red}{18.72023871} & 4.469834276 \\ \color{red}{13.01336491} & 15.01116428 \end{array} \right|} {\left| \begin{array}{cc} 5 & 4.469834276\\ 4.469834276 & 15.01116428 \end{array} \right|} = 4.0461$$

Now $a = \log^{-1} (A) = \log^{-1} (4.0461) =10^{4.0461} = 11120$ and

$$b = \dfrac{\left| \begin{array}{cc} 5 & \color{red}{18.72023871} \\ 4.469834276 & \color{red}{13.01336491} \end{array} \right|} {\left| \begin{array}{cc} 5 & 4.469834276\\ 4.469834276 & 15.01116428 \end{array} \right|} = -0.3379$$

Thus, the power curve of the best fit is $Y=11120 X^{-0.3379}.$

# Create a data frame to store the data
Mass <- c(0.135, 0.67, 2.45, 75, 1775)
RPM <- c(22000, 13000, 8000, 2550, 900)
data <- data.frame(Mass, RPM)
data

# Fit a power regression model to the data
power_model <- lm(log10(RPM) ~ log10(Mass), data = data)
 
# Summarize the power regression model
summary(power_model)

## 
## Call:
## lm(formula = log10(RPM) ~ log10(Mass), data = data)
## 
## Residuals:
##         1         2         3         4         5 
##  0.002468  0.009070 -0.011523 -0.006010  0.005994 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  4.046107   0.005161   784.0 4.58e-09 ***
## log10(Mass) -0.337886   0.002979  -113.4 1.51e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.009886 on 3 degrees of freedom
## Multiple R-squared:  0.9998, Adjusted R-squared:  0.9997 
## F-statistic: 1.287e+04 on 1 and 3 DF,  p-value: 1.51e-06

# Calculate the value of a
a <- 10^((coef(model)[1]))

# Calculate the value of b
b <- coef(model)[2]

# Calculate the predicted Y for a range of X
predicted_values <- a*(seq(min(data$Mass), max(data$Mass), length.out = 5))^b
print(predicted_values)

## [1] 12.090791  6.673574  6.342661  6.156732  6.028124

print(a)

## (Intercept) 
##    10.43836

print(b)

##    log10(X) 
## -0.07338735

# Plot the data points and the predicted Y values
plot(
  data$Mass, 
  data$RPM, 
  main = "RPM vs. Mass", 
  xlab = "Mass", 
  ylab = "RPM"
)
lines(seq(min(data$Mass), 
          max(data$Mass), 
          length.out = 5), 
      predicted_values, col = "red"
)

7 Example 7

Fit the exponential curve $Y=ab^X$ to the following data:

$$\begin{array}{c|cccccccccccc} x & y \\ \hline 2 & 1 \\ 4 & 3 \\ 6 & 6 \\ 8 & 12 \\ 10 & 24 \\ \hline \end{array}$$

Solution

Let the exponential curve be $Y=ab^X$, and the normal equations for estimating $a$ and $b$ are

$$\begin{array}{ccccccccccccc} \displaystyle \sum u & = & n A & + & \displaystyle B \sum x \\ \displaystyle \sum ux & = & \displaystyle A \sum x & + & \displaystyle B \sum x^{2} \end{array}$$

where $a=\log^{-1}(A)= 10^{A}$ and $b= \log^{-1}(B)=10^{B}$.

Note: Here we are using $\log$ with base $10$.

To find the values of $a$ and $b$ from the above normal equations, we require $\displaystyle \sum u$, $\displaystyle \sum x, \displaystyle \sum ux$ and $\displaystyle \sum x^{2}$, which are being calculated in the following table:

$$\begin{array}{ccccccccc} i & x & y & u=\log y & ux & x^2 \\ \hline 1 & 2 & 1 & 0 & 0 & 4 \\ 2 & 4 & 3 & 0.477121255 & 1.908485019 & 16 \\ 3 & 6 & 6 & 0.77815125 & 4.668907502 & 36 \\ 4 & 8 & 12 & 1.079181246 & 8.633449968 & 64 \\ 5 & 10 & 24 & 1.380211242 & 13.80211242 & 100 \\ \hline n = 5& \displaystyle \sum x = 30 & \displaystyle \sum y = 46 & \displaystyle \sum u = 3.714664993 & \displaystyle \sum ux = 29.01295495 & \displaystyle \sum x^2 = 220 \\ \hline \end{array}$$

Substituting the values of $\displaystyle \sum u , \displaystyle \sum x, \sum ux$ and $\displaystyle \sum x^{2}$ in the above normal equations, we get

$$\begin{array}{ccccccccccccc} 3.7147 & = & 5 A & + & 30 B & & (1) \\ 29.013 & = & 30 A & + & 220 B & & (2) \end{array}$$

Now, we solve the equation (1) and equation (2).

Multiplying equation (1) by $6$, i.e.,

$$\begin{array}{ccccccccccccc} 3.7147 & = & 5 A &+& 30 B & & (1) \times 6 \\ \end{array}$$

we have

$$\begin{array}{ccccccccccccc} 22.2882 & = & 30 A & + & 180 B & & (3) \\ 29.013 & = & 30 A & + & 220 B & & (2) \end{array}$$

Subtracting equation (2) from equation (3), we have

$$\begin{array}{ccccccccccccc} & 6.7248 & = & 40 B \\ \Rightarrow & B & = & 0.1681 \end{array}$$

Substituting the value of $B$ in equation (1), we get

$$A=-0.26566$$

Now $$a = \log^{-1} A= \log^{-1} (-0.26566) = 10^{-0.26566} = 0.5424$$

and

$$b = \log^{-1} B= \log^{-1} (0.1681) = 10^{0.1681} = 1.4727$$

Thus, the exponential curve of the best fit is $Y=0.5424(1.4727)^{X}.$

X <- c(2, 4, 6, 8, 10)
Y <- c(1, 3, 6, 12, 24)
data <- data.frame(X, Y)
data

# Create a scatterplot of Y versus X
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
)

# Fit a quadratic regression model to the data
model <- lm(log10(Y) ~ X , data = data)

# Summarize the quadratic regression model
summary(model)

## 
## Call:
## lm(formula = log10(Y) ~ X, data = data)
## 
## Residuals:
##          1          2          3          4          5 
## -7.044e-02  7.044e-02  3.522e-02  3.469e-17 -3.522e-02 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.26581    0.06744  -3.942 0.029106 *  
## X            0.16812    0.01017  16.537 0.000481 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.0643 on 3 degrees of freedom
## Multiple R-squared:  0.9891, Adjusted R-squared:  0.9855 
## F-statistic: 273.5 on 1 and 3 DF,  p-value: 0.0004813

# Calculate the value of a
a <- 10^((coef(model)[1]))

# Calculate the value of b
b <- 10^(coef(model)[2])

# Calculate the predicted Y for a range of X
predicted_values <- a*b^(seq(min(data$X), max(data$X), length.out = 5))
print(predicted_values)

## [1]  1.176079  2.550849  5.532647 12.000000 26.027323

print(a)

## (Intercept) 
##   0.5422359

print(b)

##        X 
## 1.472733

# Plot the data points and the predicted Y values
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
)
lines(seq(min(data$X), 
          max(data$X), 
          length.out = 5), 
      predicted_values, col = "red"
)

8 Example 8

Fit an exponential curve $Y = a e^{bX}$ to the following data:

$$\begin{array}{c|cccccccccccc} x & y \\ \hline 0 & 1481.66 \\ 2 & 2333.32 \\ 4 & 3501.39 \\ 6 & 5938.71 \\ 8 & 8838.60 \\ \hline \end{array}$$

Solution

Let the exponential curve be $Y=a e^{bX}$, and the normal equations for estimating $a$ and $b$ are

$$\begin{array}{ccccccccccccc} \displaystyle \sum u & = & n A & + & \displaystyle B \sum x \\ \displaystyle \sum ux & = & \displaystyle A \sum x & + & \displaystyle B \sum x^{2} \end{array}$$

where $a=\log^{-1}(A)$ and $b= \dfrac{B}{\log e}$.

Note: Here we are using $\log$ with base $10$.

To find the values of $a$ and $b$ from the above normal equations, we require $\displaystyle \sum u, \sum x, \displaystyle \sum ux$ and $\displaystyle \sum x^{2}$, which are being calculated in the following table:

$$\begin{array}{ccccccccc} i & x & y & u=\log y & ux & x^2 \\ \hline 1 & 0 & 1481.66 & 3.170748557 & 0 & 0 \\ 2 & 2 & 2333.32 & 3.367974304 & 6.735948607 & 4 \\ 3 & 4 & 3501.39 & 3.544240487 & 14.17696195 & 16 \\ 4 & 6 & 5938.71 & 3.773692118 & 22.64215271 & 36 \\ 5 & 8 & 8838.6 & 3.94638348 & 31.57106784 & 64 \\ \hline n=5 & \displaystyle \sum x = 20 & \displaystyle \sum y = 22093.68 & \displaystyle \sum u = 17.80303895 & \displaystyle \sum ux = 75.1261311 & \displaystyle \sum x^2 = 120 \\ \hline \end{array}$$

Substituting the values of $\displaystyle \sum u , \sum x , \sum ux$ and $\displaystyle \sum x^{2}$ in the above normal equations, we get

$$\begin{array}{ccccccccccccc} 17.80303895 & = & 5 A & + & 20 B & (1) \\ 75.1261311 & = & 20 A & + & 120 B & (2) \end{array}$$

By solving these equations as simultaneous equations, we get

$$\begin{array}{ccccccccccccc} A & =& 3.1692 \\ B & = & 0.09784 \end{array}$$

Now $$a = \log^{-1} A= \log^{-1} (3.1692) = 10^{3.1692} = 1476.4$$

and

$$b = \dfrac{0.09784}{\log 2.71828} = \dfrac{0.09784}{0.43429} = 0.2253$$

Thus, the exponential curve of the best fit is $Y=1476.4 \cdot e^{0.2253X}.$

# Create a data frame to store the data
data <- data.frame(
  X = c(0, 2, 4, 6, 8),
  Y = c(1481.66, 2333.32, 3501.39, 5938.71, 8838.60)
)
data

# Create a scatterplot of Y versus X
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
)

# Fit a quadratic regression model to the data
model <- lm(log(Y) ~ X , data = data)

# Summarize the quadratic regression model
summary(model)

## 
## Call:
## lm(formula = log(Y) ~ X, data = data)
## 
## Residuals:
##         1         2         3         4         5 
##  0.003542  0.007058 -0.037687  0.040032 -0.012945 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 7.297376   0.025506  286.10 9.42e-08 ***
## X           0.225307   0.005206   43.27 2.72e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.03293 on 3 degrees of freedom
## Multiple R-squared:  0.9984, Adjusted R-squared:  0.9979 
## F-statistic:  1873 on 1 and 3 DF,  p-value: 2.716e-05

# Calculate the value of a
a <- exp(coef(model)[1])

# Calculate the value of b
b <- coef(model)[2]

# Calculate the predicted Y for a range of X
predicted_values <- a*exp(seq(min(data$X), max(data$X), length.out = 5)*b)
print(predicted_values)

## [1] 1476.421 2316.909 3635.865 5705.669 8953.757

print(a)

## (Intercept) 
##    1476.421

print(b)

##         X 
## 0.2253065

# Plot the data points and the predicted Y values
plot(
  data$X, 
  data$Y, 
  main = "Y vs. X", 
  xlab = "X", 
  ylab = "Y"
)
lines(seq(min(data$X), 
          max(data$X), 
          length.out = 5), 
      predicted_values, col = "red"
)

9 Exercise

10 Solution