Seminar_2_Statdisk_rmorel01_Seminar_2_Statdisk_Assignment_Rob_Morell

Seminar_2_Statdisk_rmorel01_Seminar_2_Statdisk_Assignment_Rob_Morell

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Excellent job Rob—this is laid out beautifully. You took a wrong turn to find the best predictive value though—see below. It is always good to provide a nice concise summary statement in problems such as this: The best predicted breaking distance for a 4,000lb car = 136.781 feet. See below for other feedback, esp for the last two questions. Score = 5 – 1 = 4 Seminar 2 Statdisk Assignment Name : Rob Morell 1) 2. Using the prediction procedure, find the best predicted braking distance for a car that weighs 4000 lbs. For #2, first we need to find r the linear correlation coefficient for the data set. r for the weight and braking distances = -0.0867286 Next we need to find the critical value, so refer to Table A-6. We have paired data and n = 32. Critical value = ±0.3493697 Does the linear correlation for the data support a linear correlation?
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Unformatted text preview: The absolute value of r = -0.0867286 is less than the critical value r = 0.3493697 Th ere is not a linear correlation between the weight and braking distance. If there is a linear correlation, then we input the necessary data into the prediction equation to find the best predicted value for braking distance for a vehicle weighing 4000 lbs. yhat = b0 + b1x If there is not a linear correlation, then we use the sample mean for y as the best predicted braking distance. You'll need to calculate that. 3. Just for practice, find the regression equation, letting the first variable be the predictor (x) variable. yhat = b0 + b1x Y = b0 + b1x y = 141.244 + -0.0012379*4000 y=141.244 + -4.9516 y=141.244 - 4.9516= 136.2924 y=136 .2924...
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Seminar_2_Statdisk_rmorel01_Seminar_2_Statdisk_Assignment_Rob_Morell

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