Question

The local coffee shop keeps track of how much coffee they sell versus the temperature on that day, here are their figures:

Sales in Php

1796

1657

1435

1573

1476

1403

Temp in Celsius

21.2

22.7

24.2

26.6

29.5

30.9

Suppose you want to create a regression equation that will predict the total sales in Php using the temperature of the day in Celcius, what is the total sales for the day in Php if the temperature is 29.1? Round your answers to the nearest hundredths.

Answer 1

In order to compute the regression coefficients, the following table needs to be used:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 21.2 & 1796 & 38075.2 & 449.44 & 3225616 \\ \hdashline & 22.7 & 1657 & 37613.9 & 515.29 & 2745649 \\ \hdashline & 24.2 & 1435 & 34727.4 & 585.64 & 2059225 \\ \hdashline & 26.6 & 1573 & 41841.8 & 707.56 & 2474329 \\ \hdashline & 29.5 & 1476 & 43542 & 870.25 & 2178576 \\ \hdashline & 30.9 & 1403 & 43352.7 & 954.81 & 1968409 \\ \hdashline Sum= & 155.1 & 9340 & 239152.6 & 4082.99 & 14651804 \\ \hdashline \end{array}$$ $$\bar{X}=\dfrac{1}{n}\sum _{i}X_i=\dfrac{155.1}{6}$$

$$=25.85$$

$$\bar{Y}=\dfrac{1}{n}\sum _{i}Y_i=\dfrac{9340}{6}$$

$$=1556.666667$$

$$SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _{i}X_i)^2$$

$$=4082.99-\dfrac{155.1^2}{6}=73.655$$

$$SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _{i}Y_i)^2$$

$$=14651804-\dfrac{(9340)^2}{6}=112537.333333$$

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)$$

$$=239152.6-\dfrac{155.1(9340)}{6}=−2286.400000$$

$$b=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{239152.6-\dfrac{155.1(9340)}{6}}{4082.99-\dfrac{155.1^2}{6}}$$

$$=-31.042$$

$$a=\bar{Y}-b\bar{X}=2359.1029$$

Therefore, we find that the regression equation is:

$$Y =2359.1029 -31.042 X$$

$$X=29.1\\Y = 2359.1029 -31.042 (29.1)=1456$$

The total sales for the day in Php if the temperature is 29.1 is Php1456.