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Stat 401B Lab 6: Solution Fall 2008
- What is the relationship between the distance walked and the time it takes to deliver soft drinks to vending machines? An industrial engineer collected 20 observations on the delivery time (in minutes), number of cases delivered and distance walked (in feet). Delivery time includes time to unload the cases from the delivery truck, walk to the machine and load the machine with the soft drinks. The data are given below.
Cases Distance Time Cases Distance Time 7 560 16.68 4 255 13. 3 220 11.50 6 462 19. 3 340 12.03 9 448 24. 4 80 14.88 6 200 15. 6 150 13.75 7 132 19. 7 330 18.11 3 36 9. 2 110 8.00 10 140 17. 7 210 17.83 9 450 18. 5 605 21.50 8 635 19. 10 215 21.00 4 150 10.
a) Fit a simple linear model with Time as your response variable and Distance as your explanatory variable. Use the JMP output to answer the following questions. Be sure to turn in JMP output with your answers.
- What is the least squares regression equation for this model?
Predicted Time = 12.06 + 0.0144Distance*
- Predict the delivery time for 6 cases and 200 feet.
Predicted Time = 12.06 + 0.0144(200) = 12.06 + 2.88 = 14.94 minutes
- Give an interpretation of the estimated slope within the context of the problem.
For each additional foot the delivery person walks, the time increases by 0. minutes, on average.
- Give and interpretation of the estimated intercept within the context of the problem.
If the delivery person does not have to walk any distance (the machine is right next to the truck) the average time is 12.06 minutes.
- How much of the variation in Time is explained by the linear relationship with Distance?
R^2 = 0.362. Only 36.2% of the variation in Time is explained by the linear relationship with Distance.
- Is Distance a statistically significant variable for predicting Time? Support your answer statistically.
Yes. F = 10.2335 or t = 3.20, with P-value = 0.0050. The small P-value indicates that the Distance is a statistically significant variable for predicting Time.
b) Fit a multiple regression model with Time as your response variable and Distance and Cases as explanatory variables. Use the JMP output to answer the following questions. Be sure to turn in JMP output with your answers.
- What is the least squares regression equation for this model?
Predicted Time = 6.30 + 1.221Cases + 0.0089Distance**
- Predict the delivery time for 6 cases and 200 fett. How does this compare to the prediction in a)?
Predicted Time = 6.30 + 1.221(6) + 0.0089(200) = 6.30 + 7.326 + 1.78 = 15. minutes. This predicted time is about a half a minute longer than the prediction in a).
- Give an interpretation of the estimated slope for Distance within the context of the problem.
Holding the number of cases constant, the average delivery time increases by 0. minutes for each additional foot of distance.
- Give an interpretation of the estimated slope for Cases within the context of the problem.
Holding the distance constant, the average delivery time increases by 1.221 minutes for each additional case.
- Why is there not an interpretation of the estimated intercept within the context of the problem for this model?
The estimated intercept gives the predicted value of time when the distance is zero (the truck is next to the machine) and there are zero cases delivered. Distance being zero makes sense but cases being zero does not. If there are no cases to deliver there is no delivery.
- Does Cases add significantly to the model with Distance? Support your answer statistically.
Yes. F = 33.2253 or t = 5.76 with a P-value < 0.0001. The small P-value indicates that Cases adds significantly to the model with Distance.
c) Fit a multiple regression model with Time as your response variable and Distance, Cases, and Distance*Cases as explanatory variables. Be careful! Remember to turn off the Center Polynomial option before you fit the model. Use the JMP output to answer the following questions. Be sure to turn in JMP output with your answers.
- What is the least squares regression equation for this model?
Predicted Time = 3.99 + 1.619Cases + 0.0199Distance – 0.00173CasesDistance**
- Predict the delivery time for 6 cases and 200 feet. How does this compare to the prediction in b)?
Predicted time = 3.99 + 1.619(6) + 0.0199(200) – 0.00173(6)(200) = 3.99 + 9.714 + 3.98 – 2.076 = 15.61 minutes. This predicted time is 0.20 minutes longer than in b).
e) For the “best” model for these data, compute the residuals. Plot the residuals versus Distance. Plot the residuals versus Cases. Analyze the distribution of the residuals. Use the JMP output to answer the following questions. Be sure to turn in JMP output with your answers.
Note: The “best” model would be the model that contains only Distance and Cases.
- What do the plots of residuals versus the explanatory variables tell you about the adequacy of the model you have chosen as “best”? What do they tell you about the equal standard deviation condition? Be sure to support your answers by referring to the plots.
In plot of residuals versus distance is a bit strange. There appears to be a butterfly type pattern, e.g. wide, narrow, wide spread. The linear relationship is probably ok but the equal standard deviation may be somewhat in doubt.
The plot of residuals versus cases is a random scatter and so the linear relationship with cases is the best we can do. Also the spread around the zero line is constant for different values of cases indicating that the equal standard deviation condition is met.
- What does the analysis of the distribution of residuals tell you about the conditions of identically and normally distributed residuals? Be sure to support your answers by referring to the analysis.
There are not many residuals to analyze and so the histogram and box plot may not be the most informative regarding a normal distribution. The histogram does show a possibly bi-modal distribution which would cast doubt on the identically distributed condition. The box plot is fairly symmetric although the “mound” is to the left of zero and the mean is pulled slightly right, indicating a slight skew to the right. The normal quantile plot shows points on the diagonal “normal model” line, then above then below. The normal distribution condition may not be met exactly. In general, there are problems with the identically distributed and normally distributed error conditions. However, all of the P-values are quite extreme so the no interaction model would still be the “best.”
