Question

Use the least squares regression line of this data set to predict a value.
The manager of a ski resort in the Alps always worries there won't be enough snow to keep the resort open into the spring. She decided to see if there was a relationship between the temperature in January and the amount of snow in the spring.
For several years, she recorded the average temperature in January (in Celsius), x. On March 1, she also measured the depth of the snow at the bottom of a particular ski slope (in centimeters), y.
Average temperature (in Celsius) Snow depth (in centimeters)
–
7.98 69.87
–
6.10 55.76
–
3.17 62.68
–
2.85 56.56
–
2.04 56.65
4.16 37.21
The least squares regression line of this data set is:
y=
–
2.346x+49.424
What snow depth does this line predict for March 1st if the average temperature in January were 2.03 degrees Celsius?
Round your answer to the nearest thousandth.
centimeters

Answer 1

To predict the snow depth using the least squares regression line given by the equation:

\[ y = -2.346x + 49.424 \]

we substitute \( x = 2.03 \) (the average temperature in January).

Now, let's calculate:

\[
y = -2.346(2.03) + 49.424
\]

Calculating the multiplication first:

\[
-2.346 \times 2.03 \approx -4.76038
\]

Now, adding this to 49.424:

\[
y \approx 49.424 - 4.76038 \approx 44.66362
\]

Rounding this to the nearest thousandth, we get:

\[
y \approx 44.664
\]

Thus, the predicted snow depth for March 1st if the average temperature in January were 2.03 degrees Celsius is approximately **44.664 centimeters**.