Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

The number of pounds of steam used per month by a chemical plant is thought to b

ID: 3218659 • Letter: T

Question

The number of pounds of steam used per month by a chemical plant is thought to be related to the average ambient temperature (in F) for that month. The past year's usage and temp are shown in the the following table: Month Temp Usage/1000 Jan. 21 185.79 Feb 24 214.47 Mar 32 288.03 Apr 47 424.84 May 50 454.58 June 59 539.03 July 68 621.55 Aug 74 675.06 Sept 62 562.03 Oct 50 452.93 Nov 41 369.95 Dec 30 273.98

a.) Assuming that a simple linear regression model is appropriate, average temp fit the regression model relating steam usage

(y) to the average temperature (x). What is the estimate of ?2?Graph the regression line.

b.) what is the estimate of expected steam usage average temp is 55 degrees F.

c.) What change in mean steam usage is expected when the monthly average temp changes by 1 degree F?

d.) Suppose the monthly average temp is 42 degrees F. Calculate thefitted value of y and the corresponding residua

in this question, monthly average temperature is 42 F

and there more requirements in (e,f,g,h)

please help

thank you

Explanation / Answer

a)   Coefficients(a)
Unstandardized Coefficients       Standardized Coefficients
Model B Std. Error Beta t   Sig.
1   (Constant)   -6.336   1.668 -3.799   .003
X 9.208   .034 1.000 272.643   .000
a. Dependent Variable: Y

therefore the regression model of the problem is Y = -6.336+9.208X

b) The estimate of expected steam usage average temp is 55 degrees F

Y = -6.336+9.208(55) = 500.104

d) The monthly average temp is 42 degrees F , then Y = 6.336+9.208(42)= 380.4

f) the standard errors of the slope and intercept = 0.034

Ho:b1=0 versus H1:b1 0

test statistic t = b1 / SE follows t(n-2) = 9.208/0.034 = 270.82

from t-table the critical value of t at 1% level with n-2 = 12-2 = 10 df is 3.167

therefore calculated value > table value (270.82>3.167)

therefore H0 is rejected i.e we may conclude that H1:b1 0

p-value : 0.00001

the result is signficant at p(0.00001)<0.01