Question

At a famous ice cream parlor on the island, ice cream sales vary with the ambient temperature that day as shown in the following table. a.Determine the linear regression that best fits these data, send the Excel table, show the entire procedure (25 pts.) b. How much will the sale of ice cream be when the ambient temperature is 88 degrees? Explain. (5pts) c. Predict the temperature at which the sale of ice cream would be $ 650 (5pts) 71 Ambient temperature vs ice cream sales Temperature in degrees Ice cream dollar sales 66 $185 67 $215 68 $332 $325 72 $408 175 $421 80 $406 83 $412 86 $522 91 $445 92 $544 92 $614

Answer 1

solution:

a)

let X be the temperature as explanatory variable and Y (sales) be the dependent variable.

so first of all calculating the mean of both variables and make the necessary computation as shown in the table below:

dy= temperature sale in dx = (x) dollor (Y (X-mean) (Y-mean dx^2 dy^2 dx dy 66 185 - 12.58 -217.42 158.34 47270.01 2735.83 67 215 -11.58 -187.42 134.17 35125.01 2170.91 68 332 -10.58 -70.42 112.01 4958.51 745.24 71 325 -7.58 - 77.42 57.51 5993.34 587.08 72 408 -6.58 5.58 43.34 31.17 -36.76 75 421 -3.58 18.58 12.84 345.34 -66.59 80 406 1.42 3.58 2.01 12.84 5.08 83 412 4.42 9.58 19.51 91.84 42.33 86 522 7.42 119.58 55.01 14300.17 886.91 91 445 12.42 42.58 154.17 1813.34 528.74 92 544 13.42 141.58 180.01 20045.84 1899.58 92 614 13.42 211.58 180.01 44767.51 2838.74 943 4829 0.00 0.00 1108.92 174754.92 12337.08 total

so mean of X =

Χ. ΣΧ 943 12 78.583 n

mean of Y =

Υ ΣΥ 4829 402.42 n 12

regression equation of Y on X as follows:

$(Y-\bar Y) = \frac{\sum dxdy}{\sum dx^2}(X-\bar X)$

$(Y-402.42) = \frac{12337.08}{1108.92}(X-78.583)$

$\hat Y=11.13 X-471.84$

b)

we have to find the sale of ice cream when when th temperature is 88 degree.

so x = 88

putting the value of X in equation we get

$\hat Y=11.13 (88)-471.84$

$\hat Y=507.6$

so predicted sale at temperature 88 degree is $ 507.6

c)

here we have to predict the temperature with the given sale so we have to find the regreesion line of X on Y

the regression line of X on Y as follows

$(X-\bar X)=\frac{\sum dxdy}{\sum dy^2}(Y-\bar Y)$

$(X-78.583)=\frac{12337.08}{174754.92}(Y-402.42)$

$\hat X=0.071Y+50.17$

so if Y = 650

predicted value of X will be

$\hat X=0.071(650)+50.17 = 96.32$

so predicted temperature will be 96.32 degree.