Finding the Best Model
In Exercises 5–16, construct a sc...

Question

Finding the Best Model

In Exercises 5–16, construct a scatterplot and identify the mathematical model that best fits the given data. Assume that the model is to be used only for the scope of the given data, and consider only linear, quadratic, logarithmic, exponential, and power models.

Deaths from Motor Vehicle Crashes Listed below are the numbers of deaths in the United States resulting from motor vehicle crashes. Use the best model to find the projected number of such deaths for the year 2025.

Accepted Answer

Welcome back, everyone. The following data lists the number of electric vehicles sold in our region over several years. Construct a scatter plot and determine the best mathematical model, whether linear, quadratic, logarithmic, exponential, or power for the data. Use the chosen model to predict the number of electric vehicles sold in 2026. Here we have a table of the years and electric vehicles sold and A says 70,000 electric vehicles will be sold in 2026. B 40,000, C 68,186, and D 52,000. Now let's start with the first part of this problem. Let's construct a scatter plot, OK, which will help us to determine the best mathematical model for the data. Now, for our plot, OK, here we have our years on electric vehicles sold, but in this case, we're going to list the years or the number of years starting from 2013, OK? So let's let X be equal to the number of years. Starting. Let me move this over a bit. From 2013. OK. And let's let Y be equal to the number of electric vehicles sold. OK. So now, in this case, for example, for the years 2013 would be zero or for 2013 X would be zero for 2015 X would be 2, 2017 it'd be 4, 2019, it would be 6, 2021, 8, and 202,310. So we're gonna plot these values against the number of electric vehicles sold. Now, I'll leave that to you to do, but I already have a scatter plot here handy. And when you do, you should find that uh you get a graph that looks like this, OK? Over here on the right. And know for these values we can try to figure out the best model to use. What do you notice by looking at the scatter plot? Well, you probably realize that there is a rapid increase, OK? And because of this rapid increase, that means an exponential model would be the best one to use for this data. Now what do we know about exponential models? We recall that an exponential model is usually written in the form Y equals a multiplied by E to the pore of B X. So now, if we take the natural log of Y, then we can linearize, OK. Linearize our equation, OK. To get the natural log of Y. To be equal to the natural log of a. Plus B X. And know that it's in a linear form, OK? Then we can perform linear regression on X and LNY basically, OK? And by that linear regression, we should be able to come up with a regression equation and thus use that to help us predict the number of electric vehicles that will be sold for any given year, OK? So what do we know about this equation that can or rather how can we perform linear regression on this equation? What do we need? Well, it would help for us to identify uh the our our values here, the natural log of a, the intercept, and the B or slope. B, OK, is going to be equal to N multiplied by the sum of X multiplied by LNY values, OK, minus the sum of X values multiplied by the sum of LNY values. OK. And all of that will be divided. By n multiplied by the sum of X squared values minus the sum of X values squared, so that would be B. And that means then that A would be or sorry, the natural log of A. Is going to be equal to the sum of LN Y values minus B multiplied by the sum of X values all divided by n. So, if you look at our equations here to find A and B for our regression equation, you notice there are a few things we need. We'll need the sum of Xs, sorry, the sum of X multiplied by LNY values. We'll need the sum of X values as well, and we'll need the sum of LNY values. And we'll also need the sum of X grade values. So let's create a table to help us find those values, plug them into our equations for LN A and B, and thus come up with our regression equation. So we can create a table, OK, of X values. Uh, we, we also need our Y values to use them to find the LN Y values. Then X values and X multiplied by LNY values, OK? So let me just make a table here. OK. And now, we can write down our X and Y values. So remember for X values, it would be 02468, and 10. For corresponding Y values, that's 2000. 3500, 6000. 10,500, 18,000, and 30,000. Now we can use that to solve or figure out the rest of values. For the natural log of y values, we just take the natural log of the values we have. So for example, when Y equals 2000, the natural log of 2000 is 7.6009. When it's 3500, you get 8.1605. For 6000, it's 8.6, sorry, 8.6995. For 10,500, it's 9.2591, 18,000, it's 9.7981, and for 30,000, it's 10.3089, OK? Next, for X values, it would be 04, 1636, 64, and 100. And finally, for X multiplied by N and Y values, for example, it'd be 0 multiplied by 7.6009, that's 02 multiplied by 8.1605, that's 16.3210. And for the rest, we'd get uh 34.7980. 55.5546 78 03848 and 103.0890. Know that we have these values, we can find their sums, OK. Because no, for their sums for the X values, the sum would be 30, for the LNY values it's 53.8269. For the X quad values, it's 220, and for the X multiplied by LNY values, it's 288.1474. Now that we have the sums we can solve for B and A. Starting with B, then we know that N or sample size is 6. So this is going to be 6 multiplied by 288.1474 minus 30 multiplied by 53.8269. All divided by 6 multiplied by 220 minus 30 squared. Now when we go ahead and solve here, we should get B to be approximately equal to 0.27. If that's the value for B, that means the natural log of A is going to be equal to 53.8269 minus 0.27 multiplied by 30. OK. Multiplied by 30. All divided by 6, which would be approximately equal to 7.62, OK. No, that's the case, OK. Then this means, so now we have a value for B, we have LNA. This means that A, let me write it below it, actually. A itself would be approximately equal to E to the power of 7.62, which would be approximately 2,038.56. So now that we have a. OK, have a value for A and we have a value for B. We can substitute it into our linear, or, sorry, or exponential model. Because no, then this means. That our model will be Y is approximately equal to 2038.56 multiplied by E to the power of 0.27 X. OK? And now, in 2026. OK. In 2026, where X would be equal to 13, because that's 13 years after 202013, then, Why, that is, the number of electric vehicles sold would be approximately equal to 2,038.56 multiplied by E to the port of 0.27 multiplied by 13. And now when we go ahead and solve here to the nearest whole number, this would be approximately 68,186. Therefore, we predict that in 2026 there will be 68,186 vehicles sold. If we look back at our answer choices, then that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Scatterplot

Mathematical Models

Extrapolation

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