Using the line of best-fit, it is found that:
- The linear regression model is [tex]y = -25.31x + 1000[/tex]
- The estimate is that the number of cases would reach 543 in the year of 2026.
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The line of best-fit is given by:
[tex]y = bx + a[/tex]
[tex]b = \frac{\sum (x - \overline{x})(y - \overline{y})}{\sum (x - \overline{x})^2}[/tex]
- After the slope is found, using the means for x and y, the coefficient a is found.
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The means are given by:
[tex]\overline{x} = \frac{0 + 1 + 2 + 3 + 4 + 5}{6} = 2.5[/tex]
[tex]\overline{y} = \frac{1015 + 960 + 950 + 902 + 929 + 866}{6} = 937[/tex]
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The sums are:
[tex]\sum (x - \overline{x}) = (0 - 2.5) + (1 - 2.5) + ... + (5 - 2.5)[/tex]
[tex]\sum (y - \overline{y}) = (1015 - 937) + ... + (866 - 937)[/tex]
Using a calculator:
[tex]\sum (x - \overline{x})(y - \overline{y}) = -443[/tex]
[tex]\sum (x - \overline{x})^2 = 17.5[/tex]
Thus, the slope is:
[tex]b = -\frac{443}{17.5} = -25.31[/tex]
And
[tex]y = -25.31x + a[/tex]
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Using the means to find a:
[tex]y = -25.31x + a[/tex]
[tex]937 = -25.31(2.5) + a[/tex]
[tex]a = 1000[/tex]
Thus, the linear regression model is:
[tex]y = -25.31x + 1000[/tex]
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The number of cases would reach 543 in x years after 2008, and x is found when y = 543. Thus:
[tex]y = -25.31x + 1000[/tex]
[tex]543 = -25.31x + 1000[/tex]
[tex]25.31x = 1000 - 543[/tex]
[tex]x = \frac{1000 - 543}{25.31}[/tex]
[tex]x = 18.1[/tex]
2008 + 18 = 2026
The estimate is that the number of cases would reach 543 in the year of 2026.
A similar problem is given at https://brainly.com/question/21826199