Data Analysis: Wildlife A wildlife management team studied the reproductive rates of deer in thr

Data Analysis: Wildlife A wildlife management team studied...

Key Concepts

Least Squares Regression

This is a statistical method used to determine the best-fitting model to a set of data by minimizing the sum of the squares of the offsets (the residuals) between the observed values and the values predicted by the model. It is a fundamental concept in both simple and multiple regression analysis and provides a viable way to estimate model parameters by reducing errors overall.

System of Linear Equations

When fitting a model such as a quadratic curve to data, a system of linear equations is generated from the normal equations, which are derived from setting the partial derivatives of the sum of squared errors to zero. These equations are solved simultaneously to obtain the coefficients of the regression model, illustrating the relationship between independent and dependent variables.

Quadratic (Parabolic) Regression

This specific type of regression involves fitting a quadratic function (a parabola) to the data, which is useful when the underlying trend in the data is non-linear. It extends linear regression by allowing for curvature in the data while still using the least squares criterion to find the best-fitting curve.

Data Modeling and Prediction

Data modeling involves using statistical models to represent and explain the relationships between variables within a dataset. Once a model is built, it can be used to predict future values or understand underlying patterns, which is crucial for planning and decision-making in various fields.

Graphical Data Analysis

Graphing both the raw data and the fitted model is a key technique in assessing how well the model represents the data. It provides a visual representation of the relationship between variables, allows for checking the goodness-of-fit, and helps to identify any potential discrepancies or trends that may not be immediately evident from numerical summaries alone.

Transcript

00:01 All right, okay, so here we have a table, right, that gives us data on the percent of female deer who actually gives birth, right, given the number of female deer present at a particular tract of land.

00:26 Right, okay, so let's say there are, there's at least three tracts of land.

00:34 Equal size but of course of differing female deer population okay so the first track of land would have 100 dose female deer my only 75 percent of them all right has an offspring on the second tract of land right we have 120 females 68 percent of them have an offspring and finally on the third track of land, right? we have 140 females, but only 55 % of them, all right, have an offspring.

01:17 All right.

01:18 So the objective is to figure out a quadratic model to represent, right, to represent this data so that we can use that to actually predict, all right, predict further values, all right depending on the number of deer present or maybe the percent of female deer that we would like to have an offspring all right so since the number is represents x values and the percent represents the y values we can actually put this values in into a into this particular quadratic equation right so equation one all right if we substitute 100 right for the x we expect y to be 75 so that would be 75 is equal to a times 100 all right okay so this one eventually becomes 75 is equal to 10 ,000 a plus 100 b plus c right equation two on the other hand of course uses the second data point wherein it's 68 percent if there will be 120 females.

02:55 So this one should give us 68 is equal to 14 ,400a plus 120b and then plus c.

03:09 All right.

03:10 So now we have equation two.

03:12 All right.

03:13 Then the third data point for equation three, we have 55 is equal to a times 140 square plus plus c is equal to 5519600 a plus 140 b plus c all right now we have our three equations so what we're going to do right is we're going to try to eliminate the variables one by one right so that we'll be able to identify their values okay so the first thing we're going to eliminate would be c so c is a good choice because if you look at the values per c, they have the smallest coefficients and lucky for us, their coefficients are all one.

04:09 They all have the same coefficient of one.

04:13 So eliminating them would not be too difficult.

04:16 All right.

04:16 So we'll start off combining equations 1 and 2.

04:30 Okay, that's equation 1.

04:32 We now go to equation 2.

04:40 All right.

04:41 So to eliminate c for equations 1 and 2, we'll go into actually.

04:47 Subtract the two equations from each other.

04:50 75 minus 60, this is 7, 10 ,000 minus 14 ,400, this is negative 4 ,4008.

05:02 All right.

05:04 100 minus 120, this is negative 20b.

05:07 And then of course, c minus c, that of course is offset.

05:12 All right.

05:13 So now we have a new equation here, which we will call equation four.

05:16 Equation 4 is the result of combining equations 1 and 2.

05:21 This time equation 4 does not have any c with it.

05:28 All right.

05:30 We're going to repeat the process.

05:31 This time choosing a different pair.

05:33 So we'll still use equation 1, right, as one of the two equations we're going to use.

05:43 All right.

05:44 But this time, we're going to pair that with equation 3.

05:56 Again, we're going to eliminate c.

05:59 By simply subtracting.

06:05 All right.

06:06 So 75 minus 55, this is 20.

06:11 10 ,000 minus 19600.

06:14 This is negative.

06:15 9 ,600.

06:18 Oops, i think this one should also be minus.

06:24 All right.

06:24 100 minus 140 is negative 4 to p.

06:29 And then of course, c minus c is offset.

06:32 All right.

06:33 So now we actually have equation 5.

06:35 But if we look carefully notice, we notice that all of them actually has a gcf of 20.

06:41 The coefficients have a gcf of 20.

06:44 So what we're going to do is we're going to divide everything by 20 in order to reduce the magnitude of the coefficients.

06:54 All right.

06:55 Reducing the magnitude of the coefficient is always beneficial because having to deal with small numbers makes determining lcms much faster.

07:07 20 divided by 20 is 1.

07:10 Negative 9600 divided by 20 is negative 480.

07:16 All right.

07:16 And negative 20 divided by negative 40 divided by 20 is of course negative 2.

07:24 All right.

07:25 So now we have our equation 5.

07:31 Okay.

07:32 Now that we have equation 4 and 5, notice that both of them do not have any c with them.

07:38 We're going to combine them again.

07:39 And this time, the objective is to eliminate a second variable.

07:43 But we're going to eliminate, all right, would, of course, do b, because b would have, you know, smaller coefficients in it.

07:52 All right.

07:53 So we'll start off with equation four, which is 7 is equal to negative 4 ,400a minus 20b.

08:02 Right.

08:03 And then we have equation 5, which is 1 is equal to negative 480a.

08:13 Minus 2d.

08:14 All right.

08:15 To eliminate b, we need to multiply something, all right, so that the coefficients become the lcm.

08:24 The lcm of 20 and 2 is 20.

08:27 So equation number 4 is already set.

08:29 We don't need to do anything because it's already 20.

08:32 For equation 5, we need to multiply it by 10.

08:36 All right.

08:37 So besides multiplying it by 10, we're going to, of course, negate the value so that instead of subtracting later we can just add because adding you know i think might be less i think it's a you we can be more careful with adding as opposed to subtraction right especially with integers right subtraction has more complex rules than addition of integer so we'll go with making sure that they have different signs so we can just add them direct all right equation four does not change stays the same equation 5 on the other hand changes right so negative 10 times 1 negative 10 times negative 480 this is 4 ,800 negative 10 times negative this is plus 20 all right so as you can see now all right as you can see now the bs when added together when the two equations are added together the bs of course would cancel each other out so 7 plus negative 10 this is negative 3, negative 4 ,400 plus 4 ,800, this is 400a.

10:04 Notice the resulting equation combined, the combination of equation 4 and 5 actually leads to a simple equation with only one unknown, which is a.

10:14 So from here on out, we can actually solve for a by dividing both sides by 400.

10:22 So dividing negative 3 by 400, you should get negative 3, divided by 400, we have negative 0 .0075.

10:41 All right.

10:43 Now that we have the value of a, all we need to do is now to solve for the values of b and c.

10:48 All right.

10:49 So what we're going to do to solve for b.

10:51 We're going to go back to equations 4 and 5 and choose which of that we would actually use to back substitute the value of a.

11:00 5 would be a better option because it has a better option.

11:03 Smaller coefficients.

11:08 All right.

11:08 So we're going to use equation 5 and back substitute a into it.

11:16 So equation 5 is 1 is equal to negative 40080a.

11:22 Now a is already known as negative 0 .0075.

11:28 B is still not yet known.

11:31 All right.

11:32 So all we have to do is now try and simplify this one.

11:39 We have, okay, we have this one times negative 480 and we have 3 .6.

11:53 All right.

11:54 Okay.

11:55 So we have to subtract 3 .6 both sides.

12:01 We get negative 2 .6 here.

12:04 So, equal to negative 2b, dividing both sides by negative 2.

12:11 All right.

12:12 We now have b is equal to 1 .3.

12:17 All right.

12:19 Okay.

12:20 Now we have the value for b.

12:22 All right.

12:23 Okay.

12:24 Now then we have the value for a and b.

12:26 All right.

12:26 The next thing we're going to do is to figure out the value of c.

12:30 To do that, we're going to go back to equations 1, 2, and 3 and select which one of those three would have the smaller or we have the smallest coefficients.

12:40 In this case, the smallest coefficients would be for equation 1.

12:46 All right.

12:47 So therefore, we're going to use equation 1 and solve for c.

12:50 All right.

12:51 So, equation 1 is 75 is equal to 10 ,000 a, which is negative 0 .0075 plus 100b, which is 1 .3.

13:09 And finally, we have c.

13:11 All right, so we just need to solve this on.

13:14 75 is equal to, so 10 ,000 times this one should be negative 75, 100 times 1 .1 .1 .000 times 1 .1 .1 .1.

13:23 3 is 130.

13:27 All right.

13:28 Combining 75 and negative, negative 75 and 130, we get positive 55.

13:36 So to solve for c, all we have to do is to subtract 55 onto both sides.

13:48 All right.

13:48 We now have the value for c.

13:51 So therefore, the quadratic, right, the quadratic function that could model this data here, is y is equal to negative 0 .0075 x squared plus 1 .3x plus 20.

14:16 All right.

14:19 Okay.

14:22 All right.

14:22 So here we go.

14:23 So now that we have the function.

14:27 We're going to check whether this one is graphically true...

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