Sample
Kelly Thackeray
2025-03-27
Problem 1: (7.2)A solid - fuel rocket propellant loses weight after it is produced. Months since production (x), Weight loss (y)kg.
##
## Call:
## lm(formula = weight_loss ~ months + I(months^2))
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.005364 -0.002727 0.001045 0.002409 0.003273
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.633000 0.004196 389.2 < 2e-16 ***
## months -1.232182 0.007010 -175.8 5.09e-14 ***
## I(months^2) 1.494545 0.002484 601.6 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.003568 on 7 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.859e+06 on 2 and 7 DF, p-value: < 2.2e-16
Remark:
According to the fitted plot, the propellant loses weight at an
increasing rate, rather than a constant rate.
## Analysis of Variance Table
##
## Response: weight_loss
## Df Sum Sq Mean Sq F value Pr(>F)
## months 1 42.703 42.703 3355253 < 2.2e-16 ***
## I(months^2) 1 4.607 4.607 361974 < 2.2e-16 ***
## Residuals 7 0.000 0.000
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Remark:
The Pr(>F) significance codes for both coefficients are zero. The
regression model is statistically significant
## [1] "The value of beta_2 is 1.4945"## [1] "The value of p-value is 0"Remark:
B0 does not equal 0, the quadratic term is statistically
significant
(d) Are there any potential hazards in extrapolating with this model?
Extrapolating with polynomial models can lead to unreliable and unexpected predictions when used outside the range of the original data.
Problem 2: (7.17)Chemical and mechanical engineers often need to know the vapor pressure of water at various temperatures (the “ infamous ” steam tables can be used for this). Vapor Pressure(y)mmHg, Temperature(x)°C
##
## Call:
## lm(formula = vapor ~ temp)
##
## Residuals:
## 1 2 3 4 5 6
## 17.738 -1.090 -13.919 -17.548 -7.476 22.295
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -35.6667 17.2317 -2.070 0.10725
## temp 2.7129 0.4425 6.131 0.00359 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 18.51 on 4 degrees of freedom
## Multiple R-squared: 0.9038, Adjusted R-squared: 0.8798
## F-statistic: 37.59 on 1 and 4 DF, p-value: 0.003586
Remark: There is some deviation at higher temperatures, suggesting a higher-order model.
Remark: The residuals are scattered from the line, showing that the model isn’t as adequate as it could be.
Remark: The points have a curved pattern, showing that the first-order model isn’t as adequate as it could be.
##
## Call:
## lm(formula = vapor ~ temp + I(temp^2))
##
## Residuals:
## 1 2 3 4 5 6
## -2.179 2.893 2.014 -1.614 -3.493 2.379
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 20.100000 6.335989 3.172 0.05039 .
## temp -1.469643 0.414518 -3.545 0.03822 *
## I(temp^2) 0.059750 0.005797 10.307 0.00195 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.542 on 3 degrees of freedom
## Multiple R-squared: 0.9974, Adjusted R-squared: 0.9956
## F-statistic: 566.4 on 2 and 3 DF, p-value: 0.0001357Remark: The p-value for the quadratic term is < 0.05, suggesting that it is statistically significant.
Remark: There is almost no deviation from the line,
suggesting a higher-order model is adequate.
The residuals are not scattered from the line, showing that the model is
adequate.
The points have are random, showing that the second-order model is
adequate.
Problem 3: (7.6)The carbonation level of a soft drink beverage is affected by the temperature of the product and the filler operating pressure. Carbonation(y), Temperature(X1), Pressure(X2)
##
## Call:
## lm(formula = carbon ~ temp2 + pressure + I(temp2^2) + I(pressure^2) +
## I(temp2 * pressure))
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.75548 -0.22238 -0.00342 0.26024 0.96452
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3025.3186 2045.7464 1.479 0.1897
## temp2 -194.2729 132.0643 -1.471 0.1917
## pressure -6.0507 20.6063 -0.294 0.7789
## I(temp2^2) 3.6259 2.2098 1.641 0.1519
## I(pressure^2) 1.1542 0.3237 3.565 0.0118 *
## I(temp2 * pressure) -1.3317 0.8962 -1.486 0.1878
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.6194 on 6 degrees of freedom
## Multiple R-squared: 0.9933, Adjusted R-squared: 0.9877
## F-statistic: 177.2 on 5 and 6 DF, p-value: 1.983e-06Remark: Most residuals are small, suggesting a good fit. Only I(pressure^2) is significant at the 0.05 level. R-squared: 0.9933 and Adjusted R-squared: 0.9877 suggesting a good fit. F-stat is 177.2 with a p-value: 1.983e-06 suggesting overall the model is highly significant.
## Analysis of Variance Table
##
## Response: carbon
## Df Sum Sq Mean Sq F value Pr(>F)
## temp2 1 109.964 109.964 286.5944 2.716e-06 ***
## pressure 1 224.101 224.101 584.0631 3.298e-07 ***
## I(temp2^2) 1 0.198 0.198 0.5173 0.49905
## I(pressure^2) 1 4.776 4.776 12.4482 0.01239 *
## I(temp2 * pressure) 1 0.847 0.847 2.2081 0.18784
## Residuals 6 2.302 0.384
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Remark: The p-value for two of the coefficients are significant. .
##
## Call:
## lm(formula = carbon ~ temp2 + pressure + I(temp2^2) + I(pressure^2) +
## I(temp2 * pressure))
##
## Coefficients:
## (Intercept) temp2 pressure
## 3025.319 -194.273 -6.051
## I(temp2^2) I(pressure^2) I(temp2 * pressure)
## 3.626 1.154 -1.332## Analysis of Variance Table
##
## Model 1: carbon ~ temp2 + pressure + I(temp2^2) + I(pressure^2)
## Model 2: carbon ~ temp2 + pressure + I(temp2^2) + I(pressure^2) + I(temp2 *
## pressure)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 7 3.1494
## 2 6 2.3022 1 0.84723 2.2081 0.1878Remark: The p-value is greater than 0.05, indicating there is no evidence of lack of fit from excluding the interaction term.
Remark: The p-value for interaction term (temp2 * pressure) = 0.1878 is greater than the usual significance level of 0.05. This suggests that the interaction term does not contribute significantly to the model.
Remark: The second-order terms are not significant. P-value for temp2^2 = 0.1519 (not significant). P-value for pressure^2 = 0.0118 (significant at 0.05)
##
## Call:
## rsm(formula = carbon ~ SO(temp2, pressure))
##
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3025.31864 2045.74639 1.4788 0.18967
## temp2 -194.27289 132.06428 -1.4710 0.19169
## pressure -6.05067 20.60625 -0.2936 0.77893
## temp2:pressure -1.33171 0.89619 -1.4860 0.18784
## temp2^2 3.62587 2.20978 1.6408 0.15194
## pressure^2 1.15425 0.32373 3.5655 0.01185 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Multiple R-squared: 0.9933, Adjusted R-squared: 0.9877
## F-statistic: 177.2 on 5 and 6 DF, p-value: 1.983e-06
##
## Analysis of Variance Table
##
## Response: carbon
## Df Sum Sq Mean Sq F value Pr(>F)
## FO(temp2, pressure) 2 334.07 167.033 435.3288 3.206e-07
## TWI(temp2, pressure) 1 0.83 0.834 2.1727 0.19092
## PQ(temp2, pressure) 2 4.99 2.494 6.5004 0.03149
## Residuals 6 2.30 0.384
## Lack of fit 3 0.73 0.242 0.4605 0.72969
## Pure error 3 1.58 0.525
##
## Stationary point of response surface:
## temp2 pressure
## 30.50246 20.21707
##
## Eigenanalysis:
## eigen() decomposition
## $values
## [1] 3.7938410 0.9862833
##
## $vectors
## [,1] [,2]
## temp2 -0.9696254 -0.2445947
## pressure 0.2445947 -0.9696254
Remark: Eigenanalysis: values are both positive => a minimum. From the plots we can see the level of carbonation increase with the increase in temperature and operating pressure.