Data Analysis: Stopping Distance In testing a new automobile braking system, engineers recorded

Data Analysis: Stopping Distance In testing a new automobile...

Key Concepts

Least Squares Regression

Least squares regression is a method used to determine the best fitting curve by minimizing the sum of the squares of the residuals, which are the differences between the observed values and those predicted by the model. This technique is fundamental in statistics for model fitting and is applicable in various regression analyses to ensure the most accurate prediction of outcomes based on the data.

Quadratic Regression Model

A quadratic regression model is a specific type of regression analysis where the relationship between the dependent variable and the independent variable is modeled as a second-degree polynomial. This model is particularly useful when the data shows curvature, enabling the inclusion of a squared term to capture non-linear trends that cannot be modeled by a simple linear regression.

Systems of Equations in Regression

When applying the least squares method to a quadratic model, a system of equations—often called the normal equations—is formed based on summations of the observed data. Solving this system is crucial for determining the coefficients of the quadratic function, which represent the constants that best fit the data according to the least squares criterion.

Graphical Representation of Models

Graphing both the regression model and the observed data points is an essential step in validating the model’s accuracy and visual intuition. By displaying the fitted quadratic curve alongside actual data, one can assess the model fit visually, identify outliers, and understand the overall trend in the dataset.

Extrapolation and Predictive Analysis

Extrapolation refers to using the regression model to predict values outside the range of the original data. While this can be a useful tool for prediction in new or unobserved conditions, it carries risks because the behavior of the model may not hold true beyond the range of the data on which it was based, and assumptions underlying the model may no longer be valid.

Transcript

00:01 All right.

00:02 Okay, so here we have a problem here about the final speed and the stopping distance.

00:11 So an idea, of course, the faster you're traveling, all right, and then if you immediately hit the brakes, of course, the further you will stop.

00:23 Right, i mean, okay, there's, of course, skidding involved here.

00:29 All right.

00:30 So, okay, so we have here a table.

00:33 That relates to speed in miles per hour.

00:36 So the speed is in miles per hour.

00:41 And we have here the stopping distance or the skidding distance, which is, of course, in feet.

00:48 All right.

00:49 So according to the table, if you are traveling at 30 months per hour and then suddenly you hit the break, you skid a total distance of 55 feet.

01:02 This is very, very important.

01:04 All right so that you know we drive you know very very safe okay so um using quadratic regression right you were able to come up with this three particular systems which we labeled as equations one two and three our objective is to figure out what would be the quadratic function that would model this particular data right so to begin with all right we're going to try and eliminate um c right first because c has the uh smallest coefficients right so we're going to use equations one and two to eliminate c first right over there and then later we'll use a different pair which is one and three eliminating c again all right so we'll start off with equation one and two three c plus 120 b plus 5 000 a is equal to 340.

02:13 This is equation 1.

02:15 Equation 2, on the other hand, is 120c plus 5 ,000b plus 216 ,000a is equal to 15 ,2102.

02:30 All right.

02:31 So the lcm of 3 and 120 is actually 120.

02:35 So since equation 2 has already 120 as it's coefficient.

02:41 Efficient for c, all we have to do is to multiply 40 to equation 1, but we also have to make sure it becomes opposite in sign.

02:50 So we're going to multiply instead negative 40.

02:54 All right.

02:55 So doing that, going that, we'll get a new equation 1, which is, of course, you know, just transform because everything, every coefficient is just multiplied by negative 40.

03:06 So we distribute negative 40 to all the coefficients here, including the constant.

03:10 We have negative 120c all right so negative 40 times 120 this is negative 48 followed by two zeros right and we have negative 40 times 5 ,000 we have negative 20 followed by 1 2 340s 1 1 2 3 and 4 0 0 0 and then finally we have negative 40 times 3 14 8 so for this one, we'll use a calculator.

03:47 All right.

03:48 Negative 40 times 348.

03:52 I think it's 348.

03:55 Yes.

03:56 And the answer we get is negative 13 ,920.

04:04 All right.

04:05 So now we have adjusted equation one.

04:07 Question two is as is, you don't need to do anything because it's coefficient.

04:13 Oops, this is plus, is already.

04:16 Oops, i forgot the b here.

04:18 Right don't forget the letters right as later you might get confused about them right so we need to make sure that those those variables are always present is equal to 15 ,2005 all right so now that we have this one all we have to do is of course to add this two together all right that way by adding them we should actually eliminate c so now we're one variable left so negative 4 ,800 plus 5 ,000, this is of course 200.

05:05 Negative 200 ,000 plus 216, this is of course plus 16 ,000.

05:12 All right.

05:14 And then finally we have negative 13 ,920 plus 15 ,250.

05:23 So here we already have the 13 ,920.

05:26 It's already negative also.

05:28 We're just going to add 15 ,220.

05:33 The answer we get is 1 ,330.

05:42 All right.

05:42 Notice, all of this has a gcf of 10.

05:47 So what we're going to do is we're going to divide everything by 10, just to make sure that the coefficients, of course, are as low as possible.

05:56 So 200 divided by 10 is 20, 16 ,000 divided by 10 is 1 ,600.

06:06 And of course, 1 ,330 divided by 10 is 100.

06:11 All right.

06:11 Now we have our equation 4, right? this time without any c present.

06:18 Okay.

06:19 So now we're going to repeat, all right, this process.

06:22 This time we're going to pick a different pair, as mentioned earlier.

06:25 So we're going to go with equation 1 and 3 together, eliminating c.

06:30 Right so that we can come up with a new equation without any c all right so we're going to of course first begin the process by writing down equation one which is 3c plus 120b plus 5 000 a is equal to 300 and 4 to 8 all right so now we have equation one equation three on the other hand is 5 ,000 c plus 216 b plus 9 ,620 ,000 a is equal to 687 ,500.

07:24 All right.

07:26 So now we have our equations 1 and 3 together.

07:30 And the goal is to eliminate c.

07:32 All right.

07:33 So the lcm of 3 and 5 ,000 can be obtained.

07:36 Okay but just simply multiplying three in five thousand because they don't have any common factors all right so therefore the lcm which is the product of the two numbers so for three we need to multiply that by 5 ,000 and for 5 ,000 we need to multiply that by negative 3 all right now doing that right doing that we will have so 5 ,000 times 3 this is 15 right then followed by 3 zeros all right.

08:10 Okay, 5 ,000 times 120.

08:12 So 5 times 12 is 60 followed by 1, 2, 3, 4 0 ,000, 1, 2, 3, and 4 zeros.

08:23 All right.

08:23 Then we have 5 ,000 times 5 ,000.

08:26 So 5 times 5 is 25 followed by 6 zeros 8.

08:36 All right.

08:37 And finally we have 5 ,000 times 3148.

08:41 So we can use the calculator for that.

08:42 5 ,000 times 300 and 4.

08:48 And we have 1 ,740.

08:58 Okay.

08:59 So now we're finished with adjusting equation 1 so that the coefficient for c is 15 ,000.

09:05 So we're going to do the same process for equation 3 by distributing negative 3 to all its coefficients.

09:12 So negative 3 times 5 ,000.

09:15 This is of course negative 15 ,000.

09:17 All right three times negative three times two hundred sixteen thousand so negative three times two hundred sixteen thousand right we have negative six forty eight thousand all right and then we have negative three times nine million six hundred twenty thousand right negative three times nine million six hundred twenty thousand right so we have here negative 28 million 860 ,000 negative oops i forgot so with big numbers you know you have to always 8 million 860 ,000 all right and then finally we have negative three times 687 ,500 so negative 3 times 687 ,500 so negative 3 times 600 887 ,500 which we have 2 ,062 ,500 negative of course.

10:44 2 ,062 ,500.

10:46 All right.

10:49 Okay, so now that we have this one, all we have to do is, of course, to add them.

10:53 So we can, of course, eliminate the c.

10:56 All right, so we have here 600 ,000 b minus 6004 ,8 ,000 b.

11:02 Right? we have negative 48 ,000.

11:07 Thousand b all right and then here we have 25 million minus 28 million 860 ,000 we have here negative 3 million 860 ,000 a right and then we have 1 ,740 ,000 minus 2 million 62 ,500 so we have here already the negative 2 million 62 ,400 so we have here already the negative 2 million 62 ,400 a 500 so all we need to do is to add add the 1 ,740 ,000 right here.

11:45 So we add 1 ,740 ,000.

11:49 All right, so that should be it.

11:52 All right, we get negative 32 ,500.

12:01 322 ,500.

12:04 All right, okay.

12:07 So here, all right, we notice that there's actually some zeros here.

12:11 All right, but this one since it ends with actually e5, which means besides the two zeros, you could actually divide it with five followed by two zeros.

12:21 That would be a convenient factor to use in order to reduce the numbers, all right, and the coefficient.

12:29 So we're going to divide the 500.

12:30 We're going to keep the negative because we need actually one of the two equations later, all right, to have a negative coefficient.

12:39 All right.

12:40 So we have 48 ,000 divided by 500.

12:43 So we have negative 48 ,000 divide by 500.

12:51 That gives us negative 96.

12:54 So here we now have negative 96b.

12:57 All right.

12:58 What about negative 3 ,860 ,000 divided by 500? so we have negative 3 ,860 ,000.

13:10 Right.

13:11 It's not right.

13:13 Yep.

13:16 Right.

13:16 And then we divide this.

13:18 By 500.

13:20 All right, we get negative 7 ,720.

13:27 7 ,720.

13:31 And then finally we have negative 32 ,000 divided by 5003 ,3, 222 ,500.

13:42 Then we divide this by 500.

13:44 And we get negative 645.

13:52 Okay, all right.

13:54 So i'm not sure whether, um, this one right here negative 96b minus 770 is equal to 645 would still have a common factor i'm going to assume there's none because it might be very tedious to determine whether they still have common factors or not so we'll just call this equation 5 all right so notice in equation 4 we don't have any c there question 5 doesn't have any c as well so what we're going to do is we're going to combine this two and then eliminate b, because b would have the smaller coefficients.

14:36 All right.

14:37 So, equation 4 is 20b plus 1 ,600a is equal to 1333.

14:46 This is equation 4.

14:48 Equation 5, on the other hand, is negative 96b minus 7 ,720a is equal to negative 645.

14:58 All right.

14:59 So to eliminate b, we need to multiply something such that the coefficients become the lcm of each other.

15:08 20 and 96, its lcm is actually 480...