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

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 temperature are shown in the following table. Month Temp. Usage/1000 July 68621 Aug. 74 Sept.62 oct. 50 Nov.41 Dec. 30 Month Temp. Usage/1000 185.79 214.47 288.03 424.84 454.58 539.03 621.55 675.06 562.03 452.93 369.95 273.98 21 Jan Feb. 24 Mar Apr May 50 June 59 32 47 The following output is from Minitab Predictor Constant Temp 1.668 -3.80 0.003 9.20836 0.03377 272.64 0.00O Coef SE Coef -6.336 S= 1.94284 R-sq= 100.08 R-S (ad) = 100.0 Analys1s of Variance DE Souree Regression Residual Error 10 Total MS 1 280583 280583 74334.36 0.000 38 11 280621

Explanation / Answer

First enter the given data in minitab

Using minibab:

Step 1) Click on Stat>>>Regression >>>General regression

step 2) In response select Usage data column

In Model select Temperature data column

step 3) Click on prediction

Then then put 43 in the box "new observation for continuous predictor

Step 4) Click on Option and put confidence level = 99.0

Then click on OK and again click on OK

so we get the following output.

General Regression Analysis: usage versus temp

Regression Equation

usage = -6.3355 + 9.20836 temp

Coefficients

Term Coef SE Coef T P
Constant -6.33550 1.66765 -3.799 0.003
temp 9.20836 0.03377 272.643 0.000

Summary of Model

S = 1.94284 R-Sq = 99.99% R-Sq(adj) = 99.99%
PRESS = 53.5709 R-Sq(pred) = 99.98%

Analysis of Variance

Source DF Seq SS Adj SS Adj MS F P
Regression 1 280583 280583 280583 74334.4 0.000000
temp 1 280583 280583 280583 74334.4 0.000000
Error 10 38 38 4
Lack-of-Fit 9 36 36 4 3.0 0.424029
Pure Error 1 1 1 1
Total 11 280621

Fits and Diagnostics for Unusual Observations

Obs usage Fit SE Fit Residual St Resid
12 273.98 269.915 0.790638 4.06464 2.29035 R

R denotes an observation with a large standardized residual.

Predicted Values for New Observations

New Obs Fit SE Fit 99% CI 99% PI
1 389.624 0.573171 (387.808, 391.441) (383.204, 396.044)

Values of Predictors for New Observations

New Obs temp
1 43

See the output of heading "Predicted Values for New Observations"

a) Form the given output the mean pounds of steam(Fit) for given x = 43 is 389.624

b) 99% confidence interval for mean response is (387.808, 391.441)

c) b) 99% prediction interval for mean response is (383.204, 396.044)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.