There is a strong evidence that the slope is not 0. Hence emitter drive-in time influences gain in a positive linear fashion. Hence the linear model is not adequate. A better model is a quadratic one using emitter drive-in time to explain the variability in gain. Hence, lack-of-fit test is insignificant and the linear model is adequate.
The lack-of-fit test is significant and the linear model does not appear to be the best model. If there were replicates, a lack-of-fit test could be performed.
The results do not change. The pure error test is not as sensitive because the loss of error degrees of freedom. It does not appear to be linear. Decision: Reject H0 ; the lack-of-fit test is significant.
Decision: Fail to reject H0 at level 0. Hence H0 cannot be rejected. The linear model is adequate at the level 0. The lack-of-fit test is insignificant. Solutions for Exercises in Chapter 11 c Observations 4, 9, 10, and 17 have the lowest standard errors of prediction. The R2 is only 0. There is still a pattern in the residuals.
The residuals appear to be more random. This model is the best of the three models attempted. Perhaps a better model could be found. Hence, the time it takes to run two miles has a significant influence on maximum oxygen uptake. Therefore, there is no lack of fit and the quadratic model fits the data well. Solutions for Exercises in Chapter 12 0. Again, fail to reject H0. Hence, Hence, we reject H0.
The regression explained by the model is significant. At level of 0. The partial f -test statistic is Solutions for Exercises in Chapter 12 Hence, the drive ratio variable is important.
Hence, variable x3 appears to be unimportant. Hence, the model with interaction and pure quadratic terms is better. There appears to be little advantage using the full model. Hence, the model without using x1 is very competitive. In comparing the three models, it appears that the model with x2 only is slightly better.
So, the difference between the estimates are smaller than one standard error of each. So, no significant difference in a van and an suv in terms of gas mileage performance. The company would prefer female customers. Perhaps other variables need to be considered. Hence only x1 remains in the model and the final model is the same one as in a. Solutions for Exercises in Chapter 12 c For the stepwise regression, after x1 is entered, no other variables are entered.
Hence the final model is still the same one as in a and b. Fail to reject H0. Both models appear to better than the full model. When the Cp method is used, the best model is model with the constant term. We do not appear to have the normality. The residual plot and a normal probability plot are given here. Using the s2 criterion, the model with x1 , x3 and x4 has the smallest value of 0. These two models are quite competitive.
However, the model with two variables has one less variable, and thus may be more appealing. Note that observations 2 and 14 are beyond the 2 standard deviation lines. Both of those observations may need to be checked. Pn The orthogonality is maintained with the use of interaction terms. There are no degrees of freedom available for computing the standard error.
The t-tests for each coefficient show that x12 and x22 may be eliminated. Therefore, both x12 and x22 may be dropped out from the model. Therefore, some other models may be explored. It is about Hence, both coefficients are significant. Chapter 13 One-Factor Experiments: General Decision: The treatment means do not differ significantly.
The mean number of hours of relief differ significantly. The amount of money spent on dog food differs with the shelf height of the display. However, this decision is very marginal since the P -value is very close to the significance level. Decision: There is significant difference in the mean sorption rate for the three solvents. The mean sorption for the solvent Chloroalkanes is the highest.
We know that it is significantly higher than the rate of Esters. Average specific activities differ. A, C, D A, D B is significantly different from the average of A, C, and D. C is significantly different from the average of A and D. A can not be shown to differ significantly from D. The mean nitrogen loss was significantly different for the three protein levels. Hence, the mean nitrogen loss for 29 grams of protein was different from the average of the two higher protein levels.
The treatment means are different. There is a significant difference in mean yield reduction for the 4 preselected blends. There is a significant difference in the average species count for the different procedures. There is a significant difference in mean pressure for the different angles.
Hence 8. Therefore, concentrations 1 and 2 are significantly different from the control. Solutions for Exercises in Chapter 13 The means are not all equal for the different financial leverages. Hence 5. Therefore, the mean rate of return are significantly higher for median and high financial leverage than for control.
The means are not all equal. Hence, for the first contrast, the test is significant and the for the second contrast, the test is insignificant. Decision: Do not reject H0 ; could not show that the varieties of potatoes differ in field. Decision: Reject H0 ; mean percent of foreign additives is not the same for all three brand of jam. The means are: Jam A: 2. Based on the means, Jam A appears to have the smallest amount of foreign additives.
Decision: Fail to reject H0 ; there is no significant evidence to conclude that courses are of different difficulty. Decision: Reject H0 ; the mean concentration is different at the different stations. Decision: Do not reject H0 ; the treatment means do not differ significantly. Decision: Reject H0 ; differences among the diets are significant.
Decision: Do not reject H0 ; cannot show that the analysts differ significantly. Decision: Reject H0 ; the mean weight losses are different for different treatments and the therapists had the greatest effect on the weight loss.
Decision: Reject H0 ; grades are affected by different professors. Decision: Do not reject H0 ; color additives could not be shown to have an effect on setting time.
Decision: Reject H0 ; color densities of fabric differ significantly for three levels of dyes. Hence, there is significant difference in flaws among three materials. Once the data is transformed, it is not necessary that the data would follow a normal distribution.
So, it is likely that the transformed data are normally distributed. Decision: Reject H0 ; operators are different. Decision: Not able to show a significant difference in the ran- dom treatments at 0. The g vector is the same as in part a with the exception that T. Thus one obtains k X T2 i. Hence, the variance component of pour is not significantly different from 0. Hence, the loom variance component is significantly different from 0 at level 0. Decision: Do not reject H0 ; there is insufficient evidence to claim that the concentration levels of Garlon would impact the heights of shoots.
Conclusion: do not reject homogeneous variance assumption. Decision: Reject H0 ; different rations have an effect on the daily milk production. Since this is a one-sided test, we find d0. Hence, significantly higher yields are obtained with the catalysts than with no catalyst. Decision: Do not reject H0 ; there is no sufficient evidence to conclude that the variances are not equal.
Decision: Reject H0 ; the laboratory means are significantly different. Therefore, [ 0. Decision: Do not reject H0 ; the variances are not significantly different. Critical region: reject H0 when 9 0. Decision: Reject H0 ; the variances are significantly different. Decision: Reject H0 ; diets do have a significant effect on mean percent dry matter. Decision: Reject H0 ; zinc is significantly different among the diets.
Materials 1 and 3 have better results with machine 1 but material 2 has better results with machine 2. The residual and normal probability plots are shown here: 0. Decision: Reject H0 ; the mean ozone levels differ significantly across the locations. Interaction is not significant, while both main effects, environment and strain, are all significant. Coating and humidity do not interact, while both main effects are all significant. An interaction plot is given here.
Hence, if 0. Although the main effects of speed showed insignificance, we might not make such a conclusion since its effects might be masked by significant interaction. Solutions for Exercises in Chapter 14 b In the graph shown, we claim that the cutting speed that results in the longest life of the machine tool depends on the tool geometry, although the variability of the life is greater with tool geometry at level 1.
Hence, a high cutting speed has longer life for tool geometry 1. Hence, a low cutting speed has longer life for tool geometry 2. For the above detailed analysis, we note that the standard deviations for the mean life are much higher at tool geometry 1. Hence, the results on the main effects can be considered meaningful to the scien- tist.
In this case, usually the degrees of freedom of errors are small. If we compare the mean differences of the method within the overall ANOVA model, we obtain the P -values for testing the differences of the methods at Lab 1 and 7 as 0. Hence, methods are no difference in Lab 1 and are significantly different in Lab 7. Similar results may be found in the interaction plot in d. There is a significant time effect.
There is a significant copper effect. The interaction plot is show here. There seems no interaction effect. Also, time 2 has lower magnesium uptake than time 1. All the main effects are significant. They are all significant at a level larger 2 than 0. Model in b would be more appropriate. However, due to significant interactions men- tioned in a , the insignificance of A and C cannot be counted.
Hence, although the overall test on factor C is insignificant, it is misleading since the significance of the effect C is masked by the significant interaction between A and C. Hence at level of 0. All these are significant. Hence, they are all signif- icant. On the other hand, the Stress main effect is strongly significant as well. However, both other main effects, Coating and Humidity, cannot be claimed as insignificant, since they are all part of the two significant interactions.
It appears to work best with medium humidity and an uncoated surface. The insignificance of the main effect B may not be valid due to the significant BC interaction.
Also, using factor A at level 1 is the best. Hence, the interaction is at least marginally significant. The tests of the method effect for different type of gold yields the P -values as 0. Here is an interaction plot. So, three brands averaged across the other two factore are significantly different. There is no significant interaction variance component.
Therefore, 2 observations for each treatment combination are sufficient. There does not appear to be an effect due to the inspector. A simple linear model seems suitable for the data. Therefore, students scoring below 59 should be denied admission. Solutions for Exercises in Chapter 11 d Residuals appear to be random as desired.
It appears that attending professional meetings would not result in publishing more papers. Now Now 3. Solutions for Exercises in Chapter 11 Solutions for Exercises in Chapter 11 80 60 y 40 20 5 10 15 20 25 30 x Level of significance: 0. Hence, lack-of-fit test is insignificant and the linear model is adequate. There is a strong evidence that the slope is not 0.
Hence emitter drive-in time influences gain in a positive linear fashion. Hence the linear model is not adequate. A better model is a quadratic one using emitter drive-in time to explain the variability in gain. Since T1.
The lack-of-fit test is significant and the linear model does not appear to be the best model. If there were replicates, a lack-of-fit test could be performed.
The results do not change. The pure error test is not as sensitive because the loss of error degrees of freedom. It does not appear to be linear. Decision: Reject H0 ; the lack-of-fit test is significant. There does not seem to be a strong linear relationship present. Decision: Fail to reject H0 at level 0. Hence H0 cannot be rejected. The linear model is adequate at the level 0.
The lack-of-fit test is insignificant. The R2 is only 0. There is still a pattern in the residuals. The residuals appear to be more random. This model is the best of the three models attempted.
Perhaps a better model could be found. Hence, the time it takes to run two miles has a significant influence on maximum oxygen uptake. Solutions for Exercises in Chapter 11 8 6 4 residual 2 0 -2 -4 Time Therefore, there is no lack of fit and the quadratic model fits the data well.
Solutions for Exercises in Chapter 12 Again, fail to reject H0. Hence, M SR Hence, we reject H0. The regression explained by the model is significant. At level of 0. The partial f -test statistic is Hence, the drive ratio variable is important.
Hence, variable x3 appears to be unimpor- tant. Hence, the model with interaction and pure quadratic terms is better. Hence, the model without using x1 is very competitive. In comparing the three models, it appears that the model with x2 only is slightly better. The company would prefer female customers. Perhaps other variables need to be considered. Hence only x1 remains in the model and the final model is the same one as in a.
Hence the final model is still the same one as in a and b. Fail to reject H0. The residual plot and a normal probability plot are given here. When the Cp method is used, the best model is model with the constant term. We do not appear to have the normality. Both models appear to better than the full model. Using the s2 criterion, the model with x1 , x3 and x4 has the smallest value of 0. These two models are quite competitive. However, the model with two variables has one less variable, and thus may be more appealing.
Note that observations 2 and 14 are beyond the 2 standard deviation lines. Both of those observations may need to be checked. The orthogonality is maintained with the use of interaction terms. There are no degrees of freedom available for computing the standard error. The two-way interactions are not significant. Therefore, both x12 and x22 may be dropped out from the model.
Therefore, some other models may be explored. It is about Chapter 13 One-Factor Experiments: General However, this decision is very marginal since the P -value is very close to the significance level. The mean sorption for the solvent Chloroalkanes is the highest. We know that it is significantly higher than the rate of Esters. Therefore, [ 0. Now, Critical region: reject H0 when 9 0. A, C, D A, D Hence 8. Hence 5. Therefore, the mean rate of return are significantly higher for median and high financial leverage than for control.
The means are not all equal. Decision: Reject H0 ; mean percent of foreign additives is not the same for all three brand of jam. The means are: Jam A: 2. Based on the means, Jam A appears to have the smallest amount of foreign additives. Once the data is transformed, it is not necessary that the data would follow a normal distribution. So, it is likely that the transformed data are normally distributed. Solutions for Exercises in Chapter 13 Conclusion: do not reject homogeneous variance assumption.
Since this is a one-sided test, we find d0. Hence, significantly higher yields are obtained with the catalysts than with no catalyst. Decision: Reject H0 ; the laboratory means are significantly dif- ferent. Solutions for Exercises in Chapter 13 0. Materials 1 and 3 have better results with machine 1 but material 2 has better results with machine 2. An interaction plot is given here. Solutions for Exercises in Chapter 14 Hence, if 0. Solutions for Exercises in Chapter 14 speed and Hence, a high cutting speed has longer life for tool geometry 1.
Hence, a low cutting speed has longer life for tool geometry 2. For the above detailed analysis, we note that the standard deviations for the mean life are much higher at tool geometry 1. Using partial data such as we did here as above, the degrees of freedom in errors are often smaller 2 in both cases discussed here.
Similar results may be found in the interaction plot in d. The interaction plot is shown here. Also, time 2 has lower magnesium uptake than time 1. They are all significant at a level larger than 0. Model in b would be more appropriate. So, using traditional 0. Hence at level of 0. All these are significant. However, due to significant interactions mentioned in a , the insignificance of A and C cannot be counted.
Also, using factor A at level 1 is the best. Hence, they are all significant. It appears to work best with medium humidity and an uncoated surface. Hence, the inter- action is at least marginally significant. Here is an interaction plot. There is no significant interaction variance component. An interaction plot is given. Solutions for Exercises in Chapter 14 b No. Hence, there is no significant interaction when only A and B are considered.
Therefore, the contrast is significant. However, the general pat- terns of the gel generated are pretty similar for the two Solvent levels. It might be useful to try some additional runs at this combination.
It would be better to transform the data to get at least stable variance. To check on whether the assumption of the standard analysis of variance is violated, residual analysis may used to do diagnostics.
Chapter 15 2k Factorial Experiments and Fractions Interaction plots are given. Solutions for Exercises in Chapter 15 Averaged across Feed Rate a high concentration of reagent yields significantly higher viscosity, and averaged across concentration a low level of Feed Rate yields a higher level of viscosity. Here is the interaction plot of AB. AB Interaction 1 B Variable Degrees of Freedom Estimate f P -value x1 1 5. Hence Factor B, Tool Geometry, seems more significant than the other two factors.
There are possible quadratic terms missing in the model. So x2 , x3 , x1 x2 and x1 x3 are important in the model. It is significant. Since AD and ABC are confounded with blocks there are only 2 degrees of freedom for error from the unconfounded interactions. Comparatively factors A and C are more significant than the other two.
Note that the degrees of freedom on the error term is only 3, the test is not very powerful. There are total 7 two-factor interactions that can be estimated. So, factor C, the amount of grain refiner, appears to be most important. The following are two interaction plots.
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