Question

Manage.an outdoor coffee stand in Coast City are examing the relationship between (hot) coffee sales and daily temperature, hoping to be able to predict a day's total coffee sales from the maximum temperature that day. The bivariate data values for the coffee sales (denoted by y, in dollars) and the maximum temperature (denoted by x, in degrees Fahrenheit) for each of fifteen randomly selected days during the past year are given below. These data are plotted in the scatter plot in Figure 1. Also given are the products of the temperature values and coffee sales values for each of the fifteen days. (These products, written in the column labelled "xy," may aid in calculations.) Temperature, Coffee sales, y (in degrees Fahrenheit) 73.8 83.0 56.9 71.7 43.0 46.5 39.6 56.7 74.6 63.4 47.1 68.8 52.5 75.5 39.7 xy (in dollars) 1661.7 1533.6 1620.2 1953.7 1739.8 1954.7 2285.4 1956.1 2046.8 1829.8 2170.2 1743.6 2218.8 1518.6 1932.9 122633.46 127288.8 92189,38 40080.29 74811.4 90893.55 90501.84 10910.87 152691.28 116009.32 102216.42 119959.68 116487 114654.3 76736.13 2200 1800 1400 40 50 60 70 80 90 Figure 1 Answer the following. Carry your intermediate computations to at least four decimal places, and round your answer as specified below. (If necessary, consult a list of formulas.) What is the value of the sample correlation coefficient for these data? Round your answer to at least three decimal places.

Answer 1

The Sample Correlation coefficient is calculated as

r = ∑((X - M_y)(Y - M_x)) / √((SS_x)(SS_y))

Where,

X: X Values
Y: Y Values
M_x: Mean of X Values
M_y: Mean of Y Values
X - M_x & Y - M_y: Deviation scores
(X - M_x)² & (Y - M_y)²: Deviation Squared
(X - M_x)(Y - M_y): Product of Deviation Scores ,

Now.

X Values
∑ = 892.8
Mean = 59.52
∑(X - M_x)² = SS_x = 2937.944

Y Values
∑ = 28165.9
Mean = 1877.727
∑(Y - M_y)² = SS_y = 817231.649

X and Y Combined
N = 15
∑(X - M_x)(Y - M_y) = -28370.648

By formula

r = -28370.648 / √((2937.944)(817231.649)) = -0.579