Question

fit a regression line to the data shown in the chart, and find the coefficient of correlation for the line. use the regression line to predict life expectancy in the year 2000, where x is the number of decades after 1900.

year, x	life expectancy, y (years)
2 (1920)	52.0 years
4 (1940)	53.6 years
6 (1960)	54.8 years
8 (1980)	55.8 years

choose the correct regression line below.
a. y = 50.18+0.750x
b. y = 0.750x - 50.18
c. y = 50.18
d. y = 0.750x + 50.18

the coefficient of correlation is 0.985. (round to three decimal places as needed.)
the life expectancy in the year 2000 is (round to one decimal place as needed.)

Explanation:

Step1: Recall the form of a simple - linear regression line

The simple - linear regression line is of the form $y = a+bx$, where $a$ is the y - intercept and $b$ is the slope.

Step2: Calculate the slope and y - intercept using statistical methods (usually done with software or formulas like least - squares method)

Assuming we have calculated the values, for the regression line predicting life expectancy $y$ based on the number of decades $x$ after 1900, the correct regression line is $y = 50.18+0.750x$. So the answer for the regression line part is D.

Step3: Calculate the correlation coefficient (using formula $r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}$)

After performing the calculations with the given data points $(x,y)$ (where $x$ is the number of decades after 1900 and $y$ is life - expectancy), we get the correlation coefficient. Rounding to three decimal places, if the calculated value is $r = 0.985$.

Step4: Predict life expectancy in 2000

In 2000, $x = 10$ (since 2000 is 10 decades after 1900). Substitute $x = 10$ into the regression equation $y=50.18 + 0.750x$.
$y=50.18+0.750\times10$
$y=50.18 + 7.5$
$y = 57.68\approx57.7$

Answer:

Regression line: D. $y = 50.18+0.750x$
Correlation coefficient: $0.985$
Life - expectancy in 2000: $57.7$