Answer :
To solve the problem and find the exponent in the power regression model, we need to follow these steps:
### Step-by-Step Solution
1. Rescale the Year Data:
- According to the problem, we need to rescale the data such that the year 2003 corresponds to [tex]\( x = 1 \)[/tex].
- Therefore, the rescaled years will be:
- 2003 → [tex]\( x = 1 \)[/tex]
- 2004 → [tex]\( x = 2 \)[/tex]
- 2005 → [tex]\( x = 3 \)[/tex]
- 2006 → [tex]\( x = 4 \)[/tex]
- 2007 → [tex]\( x = 5 \)[/tex]
- 2008 → [tex]\( x = 6 \)[/tex]
2. Profits Data:
- The profits for the respective years (already given in the problem) are:
- 2003: 31.3 million
- 2004: 32.7 million
- 2005: 31.8 million
- 2006: 33.7 million
- 2007: 35.9 million
- 2008: 36.1 million
3. Convert Data for Power Regression:
- In power regression, we often deal with the natural logarithm of both the independent and dependent variables. This helps in transforming the relationship into a linear model.
- So, take the natural logarithm of the rescaled years and profit values.
4. Perform Linear Regression on Log-Log Data:
- Using the logarithmic values of both the rescaled years and the profits, perform a linear regression. This approach transforms a power model of the form [tex]\( y = ax^b \)[/tex] into a linear form [tex]\( \log(y) = \log(a) + b\log(x) \)[/tex].
- The slope of this line, obtained through linear regression on the log-transformed data, gives us the exponent [tex]\( b \)[/tex] in the power law.
5. Find the Exponent:
- From the regression calculation, we find that the exponent, which represents the slope of the line in the log-log space, is approximately [tex]\( 0.0789 \)[/tex].
6. Conclusion:
- Comparing this result with the given options, the exponent in the power regression model that best fits this data is 0.0789.
Therefore, among the options provided, the correct choice is 0.0789.
### Step-by-Step Solution
1. Rescale the Year Data:
- According to the problem, we need to rescale the data such that the year 2003 corresponds to [tex]\( x = 1 \)[/tex].
- Therefore, the rescaled years will be:
- 2003 → [tex]\( x = 1 \)[/tex]
- 2004 → [tex]\( x = 2 \)[/tex]
- 2005 → [tex]\( x = 3 \)[/tex]
- 2006 → [tex]\( x = 4 \)[/tex]
- 2007 → [tex]\( x = 5 \)[/tex]
- 2008 → [tex]\( x = 6 \)[/tex]
2. Profits Data:
- The profits for the respective years (already given in the problem) are:
- 2003: 31.3 million
- 2004: 32.7 million
- 2005: 31.8 million
- 2006: 33.7 million
- 2007: 35.9 million
- 2008: 36.1 million
3. Convert Data for Power Regression:
- In power regression, we often deal with the natural logarithm of both the independent and dependent variables. This helps in transforming the relationship into a linear model.
- So, take the natural logarithm of the rescaled years and profit values.
4. Perform Linear Regression on Log-Log Data:
- Using the logarithmic values of both the rescaled years and the profits, perform a linear regression. This approach transforms a power model of the form [tex]\( y = ax^b \)[/tex] into a linear form [tex]\( \log(y) = \log(a) + b\log(x) \)[/tex].
- The slope of this line, obtained through linear regression on the log-transformed data, gives us the exponent [tex]\( b \)[/tex] in the power law.
5. Find the Exponent:
- From the regression calculation, we find that the exponent, which represents the slope of the line in the log-log space, is approximately [tex]\( 0.0789 \)[/tex].
6. Conclusion:
- Comparing this result with the given options, the exponent in the power regression model that best fits this data is 0.0789.
Therefore, among the options provided, the correct choice is 0.0789.