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Solving 3 Questions Relating to Power Model

1) A coworker is considering the use of a log linear (power) model using weight to estimate the cost of a missile engine. They have performed the following calculations in log space using natural logarithms. Select the corresponding unit space form of this power model equation.

Log Space b1 = 1.109438 b0 = -0.122932

Cost = 2.421347 (Weight)1.109438
Cost = 0.884324 (Weight)1.109438
Cost = 0.884324 + 1.109438 (Weight)
Cost = 1.109438 + 0.884324 (Weight)

2) You have calculated the following power model and associated unit space values:

(SEE ATTACHMENT FOR EQUATION)

Power equation because it has a higher standard error than the linear model.
Linear equation because it has a lower standard error than the power model.
Power equation because it has a lower standard error than the linear model.
Linear equation because it has a higher standard error than the power model.

3) Given a one independent variable linear equation that states cost in $K, and given the following information, calculate the STANDARD ERROR and determine its meaning.

SEE ATTACHMENT FOR EQUATION

If we used this equation, we could typically expect to be off by ± 36.39%.
If we used this equation, we could typically expect to be off by ± $36.39K.
If we used this equation, we could typically expect to be off by ± $42.01K.
If we used this equation, we could typically expect to be off by ± 42.01%.

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Solution Preview

1) Since we use power model, the equation will be cost=b0*weight^(b1) (please note the power sign ^).
Now b0=e^( -0.122932)=0.884324 , b1=1.109438.
So we choose second ...

Solution Summary

The solution gives detailed steps on solving 3 questions: first for setting up power model, second for distinguishing between power model and linear equation and third for calculating standard error.

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