Can the consumption of water in a city be predicted by temperature? The followin

Question

Can the consumption of water in a city be predicted by temperature? The following data represent a sample of a day’s water consumption and the high temperature for that day.

a. Develop a least squares regression line to predict the amount of water used in a day in a city by the high temperature for that day.
b. What would be the predicted water usage for a temperature of 100?
c. Evaluate the regression model by calculating se, by calculating r2, and by testing the slope. Let = .01.

*(Round your answer to 2 decimal places.)
**(Round the value of SSE to 4 decimal places. Round your answers to 3 decimal places.)


a. y =

+

x (Do not round the intermediate values. Round your answers to 4 decimal places.)

b. y =

(Do not round the intermediate values. Round your answers to 2 decimal places.)

c. se =

**

r2 =

*

Observed t =

*
The decision is to

.

Water Use
(millions of gallons) Temperature
(degrees Fahrenheit) 219 103 56 39 107 77 129 78 68 50 184 96 150 90 112 75

Explanation / Answer

a. To develop a least square regression equation from the given data, we will require x,y, xy and x2

The data with above table is given below

so lets say line equation is y^ = a +bx

Here n = 8

and a and b = line constants.

so by formulas

a = [(y) (x2 ) - (x) (xy)]/ [ n (x2 ) - (x)2 ]

all the values are given above and putting all these values from the table, we get

a = [ 1025 * 49584 - 608 * 86006] / [8 * 49584 - 6082 ]

a = -1468048/ 27008= -54.356

b = [ n(xy) - (x)((y)]/ [ n (x2 ) - (x)2 ]

b = [8 * 86006 - 1025 * 608] / [ 8 * 49584 - 6082 ]

b = 64848/ 27008 = 2.401

so y^ = -54.356 +2.401 x

b. for a temperatire of 100 degree the water use

y^(100) = -54.356 +2.401 * 100 =185.744 millions of

(c) SE = To calculate SE the table is here

here y' = y predicted by putting value in regression line

SE = (y-y')2/ (n-2) = 1919.831/ 6 =  319.97

R2 = [nxy - xy]2 / [n(x2 - (x)2] [n(y)2 - y2]

R2 = [ 8 * 86006 - 608 * 1025]2 / [8 * 49584 - 6082 ] [8 * 152711 - 10252 ]

R2 = 648482/ (27008 * 171063) = 0.9102

t = r sqrt [(n-2) / (1 -r2 )]

t = sqrt [0.9102 * 6/ (1 - 0.9102) ]

t = 7.798

so t > tcritical for dF = 6 and alpha = 0.01 (tcr = 3.143)

so we will reject the null hypothesis and regression is significant.

Temperature(Degree Fahreneit)(x) Water Use (Millions of gallons) (y) x^2 xy 39 56 1521 2184 50 68 2500 3400 77 107 5929 8239 75 112 5625 8400 78 129 6084 10062 90 150 8100 13500 96 184 9216 17664 103 219 10609 22557 608 1025 49584 86006

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