Question

anna set up a lemonade stand on her block over the summer. she recorded each day’s high temperature and the number of cups of lemonade she sold for 10 days.

after plotting her results, anna noticed that the relationship between her two variables was fairly linear, so she used her data to calculate the following least squares regression equation for predicting lemonade sales, in cups, from the daily high temperature, in degrees fahrenheit:

$hat{y} = -34 + \frac{3}{5}x$

what is the residual if a day had a high temperature of 95 degrees and anna sold 21 cups of lemonade?

\boxed{} cups

show calculator

Explanation:

Step1: Recall the residual formula

The residual is calculated as the observed value minus the predicted value, i.e., \( \text{Residual} = y - \hat{y} \), where \( y \) is the observed value and \( \hat{y} \) is the predicted value from the regression equation.

Step2: Calculate the predicted value \( \hat{y} \)

Given the regression equation \( \hat{y} = -34 + \frac{3}{5}x \), and \( x = 95 \) (the high temperature).
First, calculate \( \frac{3}{5} \times 95 \):
\( \frac{3}{5} \times 95 = 3 \times 19 = 57 \) (since \( 95 \div 5 = 19 \))
Then, \( \hat{y} = -34 + 57 = 23 \)

Step3: Calculate the residual

The observed value \( y = 21 \) (cups sold).
Using the residual formula: \( \text{Residual} = y - \hat{y} = 21 - 23 = -2 \)

Answer:

-2