Determine the exponential regression of the data below using either a calculator or spreadsheet program. Round the values in the regression equation to two decimals. Then, estimate the [tex]x[/tex] value when the [tex]y[/tex] value is 5.2. Round your answer to one decimal place.

[tex](-4.7,10.7),(-7.8,20.6),(-10.5,30.2),(-15.6,41),(-20.8,56.1),(-22,65.1)[/tex]



Answer :

Certainly! Let's go through the steps required to find the exponential regression equation and estimate the [tex]\( x \)[/tex] value for a given [tex]\( y \)[/tex] value. Here's the detailed solution:

### 1. Gather the Data Points

We are given the following data points:

[tex]\[ (-4.7, 10.7), (-7.8, 20.6), (-10.5, 30.2), (-15.6, 41), (-20.8, 56.1), (-22, 65.1) \][/tex]

### 2. Transform the [tex]\( y \)[/tex]-Values

We will apply a logarithmic transformation to the [tex]\( y \)[/tex]-values to linearize the exponential relationship.

[tex]\[ y = a \cdot e^{bx} \Rightarrow \ln(y) = \ln(a) + bx \][/tex]

### 3. Perform Linear Regression on Transformed Data

Using the transformed data [tex]\((x, \ln(y))\)[/tex], we perform linear regression to find the coefficients [tex]\( \ln(a) \)[/tex] and [tex]\( b \)[/tex]. Linear regression will give us a linear equation of the form:

[tex]\[ \ln(y) = bx + \ln(a) \][/tex]

From this, we can determine [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:

[tex]\[ a = e^{\ln(a)} \][/tex]

### 4. Round the Coefficients

The coefficients [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are rounded to two decimal places:

[tex]\[ a = 9.03 \quad \text{and} \quad b = -0.09 \][/tex]

Therefore, the estimated exponential regression equation becomes:

[tex]\[ y = 9.03 \cdot e^{-0.09x} \][/tex]

### 5. Estimate [tex]\( x \)[/tex] for a Given [tex]\( y \)[/tex]-Value

To estimate the [tex]\( x \)[/tex] value when [tex]\( y \)[/tex] is 5.2, we substitute [tex]\( y = 5.2 \)[/tex] into the regression equation and solve for [tex]\( x \)[/tex]:

[tex]\[ 5.2 = 9.03 \cdot e^{-0.09x} \][/tex]

First, isolate the exponential term:

[tex]\[ \frac{5.2}{9.03} = e^{-0.09x} \][/tex]

Then, take the natural logarithm of both sides:

[tex]\[ \ln\left(\frac{5.2}{9.03}\right) = -0.09x \][/tex]

Solve for [tex]\( x \)[/tex]:

[tex]\[ x = \frac{\ln\left(\frac{5.2}{9.03}\right)}{-0.09} \][/tex]

After computing, we get:

[tex]\[ x \approx 6.0 \][/tex]

Rounding to one decimal place, the estimated [tex]\( x \)[/tex] value is:

[tex]\[ x = 6.0 \][/tex]

### Conclusion

Thus, the exponential regression equation is:

[tex]\[ y = 9.03 \cdot e^{-0.09x} \][/tex]

When the [tex]\( y \)[/tex] value is 5.2, the estimated [tex]\( x \)[/tex] value is:

[tex]\[ x \approx 6.0 \][/tex]