STA 106 Discussion 7
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STA 106 Discussion 7
Problem I
1. The weight of dried plants (measuring plant mass) was measured, and let Factor A be fertilizer type (Con = control, T1 = treatment 1, T2 = treatment 2), and Factor B be location (G = Greenhouse, O = outside). The regression formation of the ANOVA model is:
= 5.26 + (-0.11)XA,T 1 + (0.07)XA,T2 + (-0.46)XB,O + (-0.52)XA,T 1XB,O + (0.85)XA,T2XB,O You may assume alphabetical ordering for the allocation of i = 1, 2, 3, and similarly for j = 1, 2.
(a) What combination of estimated βi ’s would result in the estimate of µ32 ?
Based on alphabetical ordering of R, i = 3 implies T2, and j = 2 implies outside location Thus, = 5.26 + (-0.11)0 + (0.07)1 + (-0.46)1 + (-0.52)(0)(1) + (0.85)(1)(1) = 5.72 = 32 In otherwords, the estimated value of µ32 is : βˆ0 + βˆ2 + βˆ3 + βˆ5 .
(b) Interpret the value of βˆ2 in terms of the problem.
11 = βˆ0 , and 31 = βˆ2 + βˆ0 , so that βˆ2 = 31 - 11 .
Thus, for plants in a green house, the estimated average difference in dry weight for treatment 2 plants vs. the control is: 0.07.
(c) What contrast does βˆ5 estimate?
We have the following estimates for all combinations of treatment means:
Thus, βˆ5 = 32 - 12 - 31 + 11 = (32 - 12 ) - (11 + 31 ) is the estimate of the contrast.
Indices |
XA,T 1 |
XA,T2 |
XB,O |
XA,T 1XB,O |
XA,T2XB,T2 |
Model |
Estimates |
|||||||
i i i i i i |
= = = = = = |
1 2 3, 1 2 3, |
j j j j j j |
= = = = = = |
1 1 1 2 2 2 |
0 1 0 0 1 0 |
0 0 1 0 0 1 |
0 0 0 1 1 1 |
0 0 0 0 1 0 |
0 0 0 0 0 1 |
βˆ0 βˆ0 βˆ0 βˆ0 βˆ0 βˆ0 |
+ + + + + |
βˆ1 βˆ2 βˆ3 βˆ1 + βˆ3 + βˆ4 βˆ2 + βˆ3 + βˆ6 |
µ 11 µ21 µ31 µ 12 µ22 µ32 |
(d) What is the estimated average dry weight of a plant that was in a green house and had treatment 1? This is 11 = βˆ0 = 5.26.
(e) Estimate the value of µ 11 - µ22 .
This is βˆ0 - (βˆ0 + βˆ1 + βˆ3 + βˆ4 ) = -βˆ1 - βˆ3 - βˆ4 = -(-0.11) - (-0.46) - (-0.52) = 1.09
True/False
For each of the following questions indicate true or false, then fully explain your answer. You may use examples to illustrate
your answer.
(I) In the “regression formation” of Two Factor ANOVA, the value of the estimated intercept βˆ0 is the estimated value of a treatment mean.
TRUE. The intercept represents the average value when all X indicator variables are 0, which would be the
average value of Y at a specific combination of groups (or factors).
(II) When testing if a “reduced model” is a better fit than a “full model”, the smaller the p-value the more evidence there is to suggest the “reduced model” is a statistically better fit.
FALSE. When the p-value is small, we reject the null hypothesis, which is the hypothesis that the reduced model is a better fit. The null is always referencing the smaller model.
(III) If you used Y = ,Y as the “best” transformation for your data set and found a confidence interval for µ 11 (using the transformed data) was (2, 4), the confidence interval for µ 11 in terms of the original units would be (22 , 42 ).
FALSE. You cannot square the sample means to get back to the original units, since after we have transformed the data we have found the average of the square root of Y, and squaring that will result in something strange (and certainly not the average of the original Y).
(IV) The value of a partial R2 does not depend on the units of Y .
TRUE. It is the proportion of reduction in error, and proportions do not have any units. You can also see this
through the equation , which shows that the units of SSE would cancel.
2022-05-25