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Question 1037977: Hi there,
I'm really stumped on this and could really use your help!
The line of best fit, the regression line, is y = 0.0344x – 10.107, where x = year and y = average Washington, DC temperature, in degrees. The value of r2 is 0.6242.
(a) Use the regression line to estimate the average Washington, DC temperature in 1890, to the nearest tenth of a degree. Show some work.
(b) Use the regression line to predict the average Washington, DC temperature in 2018, to the nearest tenth of a degree. Show some work.62
(c) In what year (to the nearest year) does the regression line predict an average Washington, DC temperature of 62.0 degrees? Show work, solving an appropriate equation.
(d) What is the slope of the regression line and what are the units of measurement? In a sentence, interpret what the slope is telling us, in the context of this real-world application.
(e) What is the value of the correlation coefficient, r? Also, interpret its value: Looking at the graph and the size of r, do you judge the strength of the linear relationship to be very strong, moderately strong, somewhat weak, or very weak?
Please help!
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! y = 0.0344x – 10.107 Put in the year for x
for 1890 (x), 54.9 degrees.
for 2018 (x), 59.3 degrees
set y=62
(62+10.107)/0.0344 = year
2096.13, or early 2096.
x is year
y is temperature
slope is change in y over change in x
the units are temperature/year, so the slope is the change in temperature per year. One could interpret this as the average temperature is rising at 0.03 degrees per year. Parenthetically, a year a degree above normal (which would occur in 33 years) is noticeable.
The correlation coefficient is strong, since we define that as being greater than 0.7.
r^2=0.6242
r is the sqrt (0.6242) or 0.79.
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