MATH 220--Assignment #7--Due Thursday, February 12, 2015
Question 1: (R/Rcmdr)
Wild bears were anesthetized, and their bodies were measured and weighed. One goal of the study was to make a table (or perhaps a set of tables) for hunters, so they could estimate the weight of a bear based on other measurements. We are particularly interested in predicting the weight of a bear from its length. The data file of weights and lemgths of 143 bears is provided separately.
- What are the response and the explanatory variables.
(Ans): The response variable is weight of bears and explanatory variable is length of bears .
- describe the relationship between weight and length of a bear. Is there an overall pattern? Do you see any deviation from that pattern?
Ans: Linear relationship exists between weight and length of bears
Yes , we observed there deviation exists .
- the correlation coefficient R between weight and length. What would happen to the value of R if the scales were transformed in kilograms for the weight variable and in meters for the length?
Ans: The co-relation co-efficient R between weight and length is 0.8746451
There will be no change to the value of R if the scales were transformed in kilograms for the weight variable and in meters for the length.
- the equation of the regression line for predicting the response from the explanatory variable. In simple language, what is the slope of the line telling us?
Answer: The equation of regression line is Weight = -441.3882 + 10.3382 *length
Slope of the regression line is 10.3382. It Says if there is one unit change in length either increase or decrease then weight will increase or decrease by 10.3382 units.
- the percent of variation in the weight variable that is explained by the regression line. Does the regression model provide a good fit?
Answer: The percent variation in the weight variable that is explained by the regression line is R^2 = ( 0.8746452)^2 x 100% = 76.5%
The regression line does not provide a good fit . For a well fitted regression line R^2 should be close to 100%
- Use the equation of the regression line to predict the weight of a bear if its length is 6 feet. How confident are you that the prediction is quite accurate? (Think percent of variation).
Answer: The predicted value of weight = -441.3882 + 10.3382 x 72 = 302.9622
I am 76.5% confident that the prediction is quite accurate.
- Question 2: (
This problem is a continuation of question 1 (same response and same explanatory variable).
- a residual plot from the regression of weight against length. The residuals exhibit a fairly definite pattern. Describe that pattern.
Recall from the class discussion that presence of a pattern in the residual plot essentially implies there is room to improve the mathematical model. The main problem in this situation is the strong skewness exhibited by the weight variable (you can check that by making a histogram of the weight variable). The usual remedy in this case is to transform the weight variable to the logarithmic scale. Let’s do that!
Answer : The residual plot from the regression of weight against length shows linear pattern .
- the percent of variation in the logten(Weight) variable that is explained by the regression line. Does the regression model provide a good fit?
Answer: The percent of variation in the logten(weight) variable that is explained by the regression line is (0.9377764)^2 x 100% = 87.94% . The model is not good fit . R^2 value should be close to 100% to be good fit .
- Does the model provide a good fit? Have we markedly improved on the model in question 2 by shifting to the logarithmic scale? Refer in particular to the correlation coefficients, the percent of variation, and the residual plots when comparing the models. For fun, recalculate the fitted value of the weight of a bear that is 6 feet in length.
Answer: The Model does not provide a good fit. Yes we have markedly improved on the model by shifting to the logarithmic scale.
Here R^2 = 87.94% . So 87.945 value of weight can be explained by this model.
Here residual plot showing points are scattered randomly.
The predicted value is :
log 10 weight = 0.597 – 0.026 x 72 = -1.275
So weight = 10 ^ -1.275 = 0.0531