Answer
(a) A linear model is appropriate since the scatter plot can be approximately fit with a straight line. The scatter plot is shown in the image.
(b) $y = -0.000105x + 14.521$. The linear model (blue) is shown in the image.
(c) $y = -0.00009979x + 13.951$. The regression line (red) is shown in the image.
(d) Approximately 11.5 per 100 population.
(e) Approximately 6%.
(f) No
Work Step by Step
(a) The scatter plot is made by plotting points from the table.
(b) First, the slope is found as follows:
$m = \frac{8.2-14.1}{60000-4000} \approx -0.000105$
And the y-intercept is found as follows by inputting a pair of values into the equation:
$y=mx+b$
$14.1 = (-0.000105)(4000) + b$
$b \approx 14.521$
Thus the linear model equation is:
$y = -0.000105x + 14.521$
(c) Using technology, the regression line is found to be:
$y = -0.00009979x + 13.951$
(d) 25000 is input into the equation in (c):
$y = -0.00009979(25000) + 13.951 \approx 11.5$
(e) 80000 is input into the equation in (c):
$y = -0.00009979(80000) + 13.951 \approx 6$
The result of 6 is per 100 population, thus there is approximately a 6% chance.
(f) At 200000, the model is no longer appropriate since it results in negative values for the ulcer rate.
