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An analyst would like to predict water consumption in a city based on daily temperature. She has gathered the following data for a random sample of n= 7 days.

Temperature ( C) 40 5 25 15 10 30 35
Water Consumption(million gallons) 225 25 150 100 75 125 175

Using the following sums and sums of squares and cross-products

Summation of x= 160 Summation of y= 875 SS(x)= 1042.857 SS(y)= 26,250 SS(xy)= 5,000

A) Compute the regression line to explain "water consumption" in terms of "temperature".
B) Test, at the alpha= 0.01 level of significance, whether temperature is a significant predictor of water consumption.
C) Compute the coefficient of determination of the regression model.
D) Find a 99% confidence interval for water consumption on a particular day when the temperature is 28ºC.

https://brainmass.com/statistics/regression-analysis/compute-regression-line-explain-water-consumption-16457

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An analyst would like to predict water consumption in a city based on daily temperature. She has gathered the following data for a random sample of n= 7 days.

Temperature ( C) 40 5 25 15 10 30 35
Water Consumption(million gallons) 225 25 150 100 75 125 175

Using the following sums and sums of squares and cross-products

Σx= 160 Σy= 875 SS(x)= 1042.857 SS(y)= 26,250 SS(xy)= 5,000

n= 7
Σx= 160
Σy= 875
SS(x)= 1042.857
SS(y)= 26250
SS(xy)= 5000

See the answer on the following pages
A) Compute the regression line to explain "water consumption" in terms of "temperature".

Regression equation: Y=a+bX
b=SS(xy)/ SS(x)= 4.7945 =5000/1042.857

ΣY=na+bΣX
or 875=7*a + 4.7945*160
or a= 15.4114 =(875-4.7945*160)/7

or Y= 15.4114 + 4.7945 * X

where
Y= Water Consumption(million gallons)
X= Temperature ( C)

B) Test, at the α= 0.01 level of significance, whether temperature is a significant predictor of water consumption.

n= 7
k= 1
Degrees of freedom=n-k-1
n-k-1= 5 degrees of freedom

We will do the F test to see whether regression as a whole is significant

Computed F value
F=((SS regression/k)/(SS error/(n-k-1))= 52.6317 =(23972.606/1)/(2277.394/5)

F value for F for significance level=α= 0.01
degrees of freedom in numerator= 1
and degrees of freedom in numerator= 5
is 16.2581

Since the computed F value= 52.6317 is very much greater than 16.2581

The regression as a whole is significant

Alternatively
P value associated with F= 52.6317 is 0.00078

Since this probability value = 0.00078 is much smaller than 0.01 , the level of significance of the test

The regression as a whole is significant

C) Compute the coefficient of determination of the regression model.

Coefficient of determination r 2
= 1-( SS error/ SS total)= 0.9132 =1-( 2277.394 / 26250 )

D) Find a 99% confidence interval for water consumption on a particular day when the temperature is 28ºC.

or Y= 15.4114 + 4.7945 * X
X= 28 degrees C
Y predicted= 149.66 degrees C =15.4114 + 4.7945 * 28

To calculate the confidence interval we have to calculate the standard error of estimate and read from the tables the t value (from Student's T distribution) for the degrees of freedom and significance level

standard error of estimate Se

standard error= Se=√{SS error /(n-k-1)}= =sqrt( 2277.394 / 5 )= 21.3419

Significance level α= 0.01
degrees of freedom 5
t value= 4.0321 corresponding to 5 degrees of freedom and 0.01 significance level

Upper acceptance limit= Y predicted +t* Se = 235.71 =149.66+4.0321*21.3419
Lower acceptance limit= Y predicted -t* Se = 63.610 =149.66-4.0321*21.3419

99 % Confidence interval is between 63.610 and 235.71 degrees C

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