The efficiency for a steel specimen immersed in a phosphating tank is the weight

Question

The efficiency for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal loss (both in mg/ft^2). An article gave the accompanying data on tank temperature (x) and efficiency ratio (y). Determine the equation of the estimated regression line. (Round all numerical values to five decimal places.) Calculate a point estimate for true average efficiency ratio when tank temperature is 185. (Round your answer to four decimal places.) Calculate the values of the residuals from the least squares line for the four observations for which temperature is 185. (Round your answers to four decimal places.) Why do they not all have the same sign? These residuals do not all have the same sign because in the cases of the first two pairs of observations, the observed efficiency ratios were smaller than the predicted value. In the cases of the last two pairs of observations, the observed efficiency ratios were larger than the predicted value. These residuals do not all have the same sign because in the case of the second pair of observations, the observed efficiency ratio was equal to the predicted value. In the cases of the other pairs of observations, the observed efficiency ratios were larger than the predicted value. These residuals do not all have the same sign because in the cases of the first two pairs of observations, the observed efficiency ratios were larger than the predicted value. In the cases of the last two pairs of observations, the observed efficiency ratios were smaller than the predicted value. These residuals do not all have the same sign because in the case of the third pair of observations, the observed efficiency ratio was equal to the predicted value. In the cases of the other pairs of observations, the observed efficiency ratios were smaller than the predicted value. What proportion of the observed variation in efficiency ratio can be attributed to the simple linear regression relationship between the two variables? (Round your answer to four decimal places.)

Explanation / Answer

> tr <- read.csv("clipboard",header=TRUE,sep=" ")
> head(tr)
Temperature Ratio
1 173 0.84
2 175 1.21
3 176 1.36
4 177 1.01
5 177 1.09
6 178 1.10
> ratio_lm <- lm(Ratio~Temperature-1,data=tr)
> summary(ratio_lm)

Call:
lm(formula = Ratio ~ Temperature - 1, data = tr)

Residuals:
Min 1Q Median 3Q Max
-0.9000 -0.5254 -0.1591 0.2979 1.4161

Coefficients:
Estimate Std. Error t value Pr(>|t|)
Temperature 0.0091304 0.0007026 13 4.43e-12 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.6283 on 23 degrees of freedom
Multiple R-squared: 0.8801, Adjusted R-squared: 0.8749
F-statistic: 168.9 on 1 and 23 DF, p-value: 4.432e-12

> predict(ratio_lm,newdata=data.frame(Temperature=1.85))
1
0.01689132

> residuals(ratio_lm)
1 2 3 4 5 6 7
-0.73956635 -0.38782723 -0.24695767 -0.60608811 -0.52608811 -0.52521855 -0.53434899
8 9 10 11 12 13 14
0.05652057 -0.26087076 -0.01087076 -0.16087076 0.49912924 0.43912924 -0.90000120
15 16 17 18 19 20 21
-0.23000120 -0.84913164 0.20086836 0.25086836 0.93086836 -0.15739252 0.75260748
22 23 24
1.18347703 0.16434659 1.41608571

(a) Y = 0.00913 *  (Temperature)

(b) A point estimate when temperature is 185 is 0.0169

(c) The residuals are

-0.8491 0.2009 0.2509 0.9309

(d) Since "Multiple R-squared: 0.8801", 88.01% of the observed variation in efficiency ratio can be attributed to the simple linear regression model.

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