Answer :
Let's go through the questions one by one to understand and solve each part step-by-step.
### Part a: Type of Equation to Model the Data
To determine the best type of equation to model the given data, we consider the nature of the data. Investment growth over time could generally be exponential due to compound interest, but polynomial regression can also effectively fit the data. Given the data points, we found that a polynomial regression of degree 2 (quadratic equation) is appropriate to model this data.
### Part b: Regression Equation
By performing a quadratic regression, we obtain the coefficients for a polynomial of the form:
[tex]\[ P(x) = ax^2 + bx + c \][/tex]
For this data, the coefficients were calculated as:
- [tex]\( a = 231.96107325 \)[/tex]
- [tex]\( b = -974.6239332 \)[/tex]
- [tex]\( c = 11949.5548597 \)[/tex]
Thus, the regression equation is:
[tex]\[ P(x) = 231.96107325x^2 - 974.6239332x + 11949.5548597 \][/tex]
### Part c: Y-Intercept Interpretation
The [tex]$y$[/tex]-intercept of a quadratic equation represents the value of the investment at time [tex]\( x = 0 \)[/tex] (when the investment was initially made). For our regression equation, the [tex]$y$[/tex]-intercept [tex]\( c \)[/tex] is:
[tex]\[ c = 11949.5548597 \][/tex]
This means that when Devin initially invested the money (at year 0), the value of the investment was approximately \[tex]$11,949.55. ### Part d: Strength of Correlation To determine the strength of the correlation between the actual values and the predicted values from the regression equation, we calculate the correlation coefficient. The correlation coefficient \( r \) is a measure of how well the predicted values from the regression equation match the actual data points. The correlation coefficient calculated here is: \[ r \approx 0.9999177891351843 \] A correlation coefficient \( r \) close to 1 indicates a very strong positive correlation. Therefore, this value shows that the model is an excellent fit for the data, demonstrating a very strong correlation. ### Part e: Value of the Investment After 6 Years To find the value of the investment after 6 years, we substitute \( x = 6 \) into our regression equation: \[ P(6) = 231.96107325(6)^2 - 974.6239332(6) + 11949.5548597 \] Evaluating this gives: \[ P(6) = 14452 \] Thus, the value of Devin's investment after 6 years is approximately \( \$[/tex]14,452 \), when rounded to the nearest dollar.
### Summary
To summarize:
a. The data is best modeled by a quadratic equation.
b. The regression equation is [tex]\( P(x) = 231.96107325x^2 - 974.6239332x + 11949.5548597 \)[/tex].
c. The [tex]$y$[/tex]-intercept represents the initial investment amount, which is \[tex]$11,949.55. d. The correlation coefficient is approximately 0.9999, indicating a very strong positive correlation, hence the model fits the data very well. e. The value of the investment after 6 years is approximately \$[/tex]14,452.
### Part a: Type of Equation to Model the Data
To determine the best type of equation to model the given data, we consider the nature of the data. Investment growth over time could generally be exponential due to compound interest, but polynomial regression can also effectively fit the data. Given the data points, we found that a polynomial regression of degree 2 (quadratic equation) is appropriate to model this data.
### Part b: Regression Equation
By performing a quadratic regression, we obtain the coefficients for a polynomial of the form:
[tex]\[ P(x) = ax^2 + bx + c \][/tex]
For this data, the coefficients were calculated as:
- [tex]\( a = 231.96107325 \)[/tex]
- [tex]\( b = -974.6239332 \)[/tex]
- [tex]\( c = 11949.5548597 \)[/tex]
Thus, the regression equation is:
[tex]\[ P(x) = 231.96107325x^2 - 974.6239332x + 11949.5548597 \][/tex]
### Part c: Y-Intercept Interpretation
The [tex]$y$[/tex]-intercept of a quadratic equation represents the value of the investment at time [tex]\( x = 0 \)[/tex] (when the investment was initially made). For our regression equation, the [tex]$y$[/tex]-intercept [tex]\( c \)[/tex] is:
[tex]\[ c = 11949.5548597 \][/tex]
This means that when Devin initially invested the money (at year 0), the value of the investment was approximately \[tex]$11,949.55. ### Part d: Strength of Correlation To determine the strength of the correlation between the actual values and the predicted values from the regression equation, we calculate the correlation coefficient. The correlation coefficient \( r \) is a measure of how well the predicted values from the regression equation match the actual data points. The correlation coefficient calculated here is: \[ r \approx 0.9999177891351843 \] A correlation coefficient \( r \) close to 1 indicates a very strong positive correlation. Therefore, this value shows that the model is an excellent fit for the data, demonstrating a very strong correlation. ### Part e: Value of the Investment After 6 Years To find the value of the investment after 6 years, we substitute \( x = 6 \) into our regression equation: \[ P(6) = 231.96107325(6)^2 - 974.6239332(6) + 11949.5548597 \] Evaluating this gives: \[ P(6) = 14452 \] Thus, the value of Devin's investment after 6 years is approximately \( \$[/tex]14,452 \), when rounded to the nearest dollar.
### Summary
To summarize:
a. The data is best modeled by a quadratic equation.
b. The regression equation is [tex]\( P(x) = 231.96107325x^2 - 974.6239332x + 11949.5548597 \)[/tex].
c. The [tex]$y$[/tex]-intercept represents the initial investment amount, which is \[tex]$11,949.55. d. The correlation coefficient is approximately 0.9999, indicating a very strong positive correlation, hence the model fits the data very well. e. The value of the investment after 6 years is approximately \$[/tex]14,452.
