15.15. Four different formulations of an industrial glue are being tested. The t
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Question
15.15. Four different formulations of an industrial glue are being tested. The tensile strength of the glue when it is applied to join parts is also related to the application thickness. Five observations on strength (y) in pounds and thickness (x) in 0.01 inches are obtained for each formulation. The data are shown in the following table. Analyze these data and draw appropriate conclusions. Glue Formulation ox 46.5 13 48.7 12 46.3 15 44.7 16 45.9 14 49.0 10 47.1 14 43.0 15 49.8 12 50.1 11 48.9 11 51.0 10 46.1 12 48.5 12 48.2 11 48.1 12 44.3 14 45.2 14 50.3 10 48.6 11Explanation / Answer
Output by using Minitab,
Here we use ANCOVA, to test whether treatment means i.e. effect of different types of glue are same or different. And second hypothesis is to test whether means of covariates (x's) are significantly different or not.
Procedure:
1. Enter values of x and y for each treatment (glue).
2. Stat -> ANOVA -> General linear model -> fit general linear model.
3. we get General linear model window. In that window fill following information:
Responses: y
Factors: treatment(glue)
covariates: x.
General Linear Model: Y versus X, treatment
Method
Factor coding (-1, 0, +1)
Factor Information
Factor Type Levels Values
treatment Fixed 4 1, 2, 3, 4
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
X 1 59.566 59.5658 42.62 0.000
treatment 3 1.771 0.5903 0.42 0.740
Error 15 20.962 1.3975
Lack-of-Fit 11 12.572 1.1429 0.54 0.808
Pure Error 4 8.390 2.0975
Total 19 91.585
Model Summary
S R-sq R-sq(adj) R-sq(pred)
1.18215 77.11% 71.01% 58.67%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 60.09 1.94 30.91 0.000
X -1.010 0.155 -6.53 0.000 1.08
treatment
1 -0.440 0.466 -0.94 0.360 1.55
2 0.129 0.469 0.27 0.788 1.57
3 0.393 0.459 0.85 0.406 1.51
Regression Equation
treatment
1 Y = 59.65 - 1.010 X
2 Y = 60.22 - 1.010 X
3 Y = 60.48 - 1.010 X
4 Y = 60.01 - 1.010 X
Interpretation:
H0:t1=t2=t3=t4 Vs H1:at least one of them is different
H0:X1=X2=X3=X4 Vs H1: at least one of them is not equal
Treatment>0.05
From ANOVA we see that treatments means are not significant.here we accept null hypothesis for treatments. For treatment all the means are equal and we accept the null hypothesis.
X(Cov)<0.05 therefore we reject the null hypothesis and conclude that average of the dependent variable is not same for all the groups. Difference between means are statistically significant.
From R-sq also we see that this model is explains only 77.11% variation.
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