To find the linear regression equation for the given data, we need to first summarize the values and then perform the calculations.
Given data:
- Years since 2014 (x): 0, 1, 2, 3, 4, 5
- New Cases (y): 10471047, 10201020, 989989, 943943, 964964, 880880
Step 1: Calculate the means of x and y
\[ \bar{x} = \frac{0 + 1 + 2 + 3 + 4 + 5}{6} = \frac{15}{6} = 2.5 \]
\[ \bar{y} = \frac{10471047 + 10201020 + 989989 + 943943 + 964964 + 880880}{6} = \frac{11874143}{6} \approx 1979023.83 \]
Step 2: Calculate the slope (m)
Using the formulas:
\[ m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} \]
Where:
- \( n = 6 \)
- \( \sum x = 0 + 1 + 2 + 3 + 4 + 5 = 15 \)
- \( \sum y = 11874143 \)
- \( \sum xy = 0 \cdot 10471047 + 1 \cdot 10201020 + 2 \cdot 989989 + 3 \cdot 943943 + 4 \cdot 964964 + 5 \cdot 880880 \)
- \( \sum xy = 0 + 10201020 + 1979978 + 2831829 + 3859856 + 4404400 = 33399983 \)
- \( \sum x^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 0 + 1 + 4 + 9 + 16 + 25 = 55 \)
Now we can calculate \( m \):
\[ m = \frac{6(33399983) - (15)(11874143)}{6(55) - 15^2} \]
Calculating each term: \[ 6(33399983) = 200399898 \] \[ (15)(11874143) = 178111145 \] \[ 6(55) = 330 \] \[ 15^2 = 225 \]
Now substituting in:
\[ m = \frac{200399898 - 178111145}{330 - 225} = \frac{22288753}{105} \approx 212,851.2 \]
Step 3: Calculate the y-intercept (b)
Using the formula:
\[ b = \bar{y} - m\bar{x} \]
Substituting values:
\[ b = 1979023.83 - 212851.2 \cdot 2.5 \approx 1979023.83 - 532128 = 1445895.83 \]
Final Regression Equation
The linear regression equation can be written as:
\[ y = mx + b \]
Substituting \( m \) and \( b \):
\[ y = 212851.2x + 1445895.83 \]
Rounded to the nearest tenth, we have:
\[ y \approx 212851.2x + 1445895.8 \]
Step 4: Find the project number of new cases for 2026
For the year 2026, \( x = 12 \) (since \( x \) is years since 2014, 2026 is 12 years after 2014):
\[ y = 212851.2(12) + 1445895.8 \] \[ y \approx 2554214.4 + 1445895.8 \approx 4000109.2 \]
Final Answer
The projected number of new cases for 2026, rounded to the nearest whole number, is approximately:
\[ \boxed{4000110} \]
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