Questions: Would a linear or exponential model for the relationship between the number of employees and the number of customers be more appropriate?

Solution Steps

To determine whether a linear or exponential model is more appropriate for the relationship between the number of employees and the number of customers, we can plot the data points and visually inspect the trend. Additionally, we can fit both linear and exponential models to the data and compare their goodness of fit using metrics such as R-squared.

We need to determine whether a linear or exponential model better describes the relationship between the number of employees and the number of customers for the given data.

The given data is: \[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Number of employees} & 1 & 2 & 3 & 4 & 10 \\ \hline \text{Number of customers} & 4 & 8 & 13 & 17 & 39 \\ \hline \end{array} \]

A linear relationship can be described by the equation \( y = mx + b \), where \( y \) is the number of customers, \( x \) is the number of employees, \( m \) is the slope, and \( b \) is the y-intercept.

Let's calculate the differences between consecutive customer values to see if they are approximately constant: \[ \begin{align_} \Delta y_1 &= 8 - 4 = 4 \\ \Delta y_2 &= 13 - 8 = 5 \\ \Delta y_3 &= 17 - 13 = 4 \\ \Delta y_4 &= 39 - 17 = 22 \\ \end{align_} \] The differences are not constant, suggesting that a linear model may not be appropriate.

An exponential relationship can be described by the equation \( y = ab^x \), where \( y \) is the number of customers, \( x \) is the number of employees, \( a \) is a constant, and \( b \) is the base of the exponential function.

Let's calculate the ratios between consecutive customer values to see if they are approximately constant: \[ \begin{align_} \frac{y_2}{y_1} &= \frac{8}{4} = 2 \\ \frac{y_3}{y_2} &= \frac{13}{8} \approx 1.625 \\ \frac{y_4}{y_3} &= \frac{17}{13} \approx 1.3077 \\ \frac{y_5}{y_4} &= \frac{39}{17} \approx 2.2941 \\ \end{align_} \] The ratios are not constant, suggesting that an exponential model may not be appropriate either.

Since neither the differences nor the ratios are constant, neither a linear nor an exponential model perfectly fits the data. However, the differences between consecutive customer values are closer to being constant than the ratios, suggesting that a linear model might be a slightly better fit than an exponential model.

Final Answer