Applied Decision Methids Case Study

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Richard Joyal Module 3 Homework Chapter 4 Problems Applied decision methids 4-13: A:) [pic] B) Predict the final average of a student who made an 83 on the fist test: y = 0.7399x + 18.989 y = 0.7399(83) + 18.989 y = 80.40 C) Give values of r and r2 for this model. Interpret the value of r2 in the context of the problem: r2 = 0.8474 r = sqrt(0.8474) = 0.92054 84% of the variance in the second exam score is explained by the first exam score. 4-15: Yes there is a correlation between the first test grade and the final average which lead me to believe that the first test grade is usually a good indication of what the students over all performance will be and what their final average will be. 4-17: Accountants at the firm walker and walker believed that…show more content…
The model could also be divided into trips that do/do not include airfare (or perhaps trips over/under 100 miles, which could substitute). A driving trip might be expected to cost less than an air trip under most circumstances. 4-27: | | | | | |MPG (y) |Horsepower (x) |Weight (x) | | | |37 |66 |1797 | | | |34 |63 |2199 | | | |35 |90 |2404 | | | |32 |99 |2611 | | | |30 |63 |3236 | | | |28 |91 |2606 | | | |26 |94 |2580 | | | |26 |88 |2507 | | | |25 |124 |2922 | | | |22 |97 |2434 | | | |20 |114 |3248 | | | |21 |102 |2812 | | | |18 |114 |3382 | | | |18 |142 |3197 | | | |16 |153 |4380 | | | |16 |139 |4036 | | | | | | | | | |Correlation coefficient, R = |-0.8362 |-0.8020 | | | |Coefficient of determination, R^2 = |0.6992 |0.6431 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | The absolute value of R is greater for the variable being "Horse power" it appears that between one of the two factors, the MPG is more so strongly correlated or linked with the factor of the horse power of the automobile. This variable is also strongly affected by the

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