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Taxi tires A taxi company monitoring the safety of its cabs kept track of the number of miles tires had been driven (in thousands) and the depth of the tread remain...

Question

Taxi tires A taxi company monitoring the safety of its cabs kept track of the number of miles tires had been driven (in thousands) and the depth of the tread remaining (in mm) Their data are displayed in the scatterplot: They found the equation of the least squares regression line to be tread -36 0.6miles , with R2 = 0.74 and T=- 86. (5 Points Each)a) . Explain (in context) what the slope of the line means_b) Explain (in context) what they-intercept of the line means_c) Explain (in context) what

Taxi tires A taxi company monitoring the safety of its cabs kept track of the number of miles tires had been driven (in thousands) and the depth of the tread remaining (in mm) Their data are displayed in the scatterplot: They found the equation of the least squares regression line to be tread -36 0.6miles , with R2 = 0.74 and T=- 86. (5 Points Each) a) . Explain (in context) what the slope of the line means_ b) Explain (in context) what they-intercept of the line means_ c) Explain (in context) what R2 means. Describe the correlation. Predict the tread remaining on a tire with 35,000 miles_ f) Dave has 35,000 miles on his tires, and his tread remaiing is 15 mm_ Find and interpret his residual.



Answers

An automobile tire manufacturer collected the data in the table relating tire pressure $x$ (in pounds per square inch) and mileage (in thousands of miles): $$ \begin{array}{cc} \boldsymbol{x} & \text { Mileage } \\ \hline 28 & 45 \\ 30 & 52 \\ \hline 32 & 55 \\ \hline 34 & 51 \\ \hline 36 & 47 \\ \hline \end{array} $$ A mathematical model for the data is given by $$ f(x)=-0.518 x^{2}+33.3 x-481 $$ (A) Complete the following table. Round values of $f(x)$ to one decimal place. $$ \begin{array}{lcc} \boldsymbol{x} & \text { Mileage } & \boldsymbol{f} \boldsymbol{x} \boldsymbol{)} \\ \hline 28 & 45 & \\ 30 & 52 & \\ 32 & 55 & \\ 34 & 51 & \\ 36 & 47 & \end{array} $$ (B) Sketch the graph of $f$ and the mileage data in the same coordinate system. (C) Use values of the modeling function rounded to two decimal places to estimate the mileage for a tire pressure of $31 \mathrm{lbs} / \mathrm{sq}$ in. and for $35 \mathrm{lbs} / \mathrm{sq}$ in. (D) Write a brief description of the relationship between tire pressure and mileage.

And this problem were given the graph of the sales of a certain company in billions of dollars from the year 1995 which is five to the year 2004 which is 14. First question is, find the year of Greatest increase, sort of find it. You have to look for the line that has the greatest slope or the line. That's steepest when you look from left to right. So I'd be from here Thio here and that is your 13. So 2003 to 2004. Now for the your greatest decrease, you're looking for the line that goes to doubt. Downhill the quickest, The most steepest line downhill. So from here to here, Yeah, 1997 Tonight what to 90 1997 to 1998. Can I wants me to write the equation of the line from this point to this point. So this point is five 13.2, and this point is 14 18.4. So first find the slope M equals 18.4, minus 13.2 over 14 minus five. 5.2 over nine. 50 to over 90 which is 26/45 26/45 equals. Why? Minus 13.2 over X minus five. Cross multiply 26 X minus 1 30 equals 45 Y minus 5 94. Simplify and you get 26 X plus 4 64 over. Whoops over 45. Because why, what's the meaning of the slope in this equation? Well, is the change and sales from one year to the next? So the change in dollars per year. Okay, let's use this equation to predict the cells in 2010 with 1995 is five. 2010 is 15 years later, so that will be your 20. So you're 20 so 26 times 20 plus 4 64/40 over 45 equals about 21.87 Okay, Is this a good prediction? Well, what you have here is the equation of the line joining this point to this point personally as a line. Okay. And first of all, it could Onley predict things between 1995 and 2014. 2nd of all, it can't predict anything for this because the graph is not a line. The graph is a a group of lines put together. So no, this is not a good prediction.

Hey guys, my name is Colin and let's go ahead and jump right into this problem that deals with two different tests for tire wear and whether or not they may or may not be similar in result. So let's go ahead and first tackle part. They were asked to look at whether or not there is a that confidence interval that were given there is correct on DSO to do that. Of course, we're gonna start by finding our Alfa value. Since this is a 99% confidence interval, our confidence level is 99%. So our Alfa ends up being 0.1 and that means that are critical probability, which we calculate is one minus Alfa over to is just going to be one minus 10.0 1/2 or 0.995 and then the last kind of piece of information you know that we need for a T test is our degrees of freedom. We always calculate that is end the number of trials minus two since in this case, they mentioned that they ran that test on a random sample of 16 different tires. We know that our end is 16. So our degrees of freedom equals 16 minutes two or 14. So now we can go ahead and calculate our T statistic using Ah, those numbers that we just calculated. So we're going to be using a T statistic with an Alfa value of 0.1 and 14 degrees of freedom. And from the tea table chart, you can get that that is 2.977 So from this, we need to just go ahead now and calculate our margin of error, which we use our t statistic that we just found. And we multiply it by the standard error of the slope of the regression line, which in this case is 0.7104 When we multiply those two numbers together, we get a margin of error for our confidence interval as 0.211 for eight eso. Lastly, that construct that confidence interval. We're going to take thes slope of the regression line with 79021 from our many tab I'll put there and we're going to add and subtract points to one one for eight. And when we do that, we get a 90 99% confidence interval that matches the one. Not that we were getting a problem. 0.57 85 and the upper boundary is 1.0 one. And so now we're asked to look at different tests. So Part B s whether or not researchers that use that test there where we've got the null hypothesis as beta equals one and alternative hypothesis is beta is not equal to one whether or not that's an acceptable test to run. And, uh, I would say that this isn't appropriate pair of hypotheses, because if the tests yielded the same output, the two different tests showed the same output for each value or for each tire. Then that slope would be one. Um and so this is a good test to see if there is a difference between the two or not. Because if there is a difference that will be able to reject that null hypothesis, and if we can't of them will have to say that there is not sufficient evidence. So basically, we're looking to see whether or not beta equals one is statistically significant. And to do that, uh, that's kind of the lead in here to part C. We're asked to look at our confidence interval, which I'll remind you from. We calculated that down in part a that conference animals 0.5785 to 1.1 I mean, this is important because you'll see that this actually contains zero. And so because our 99% confidence interval contains zero, we cannot reject the null hypothesis. So we do not reject, which means that we cannot say that there is sufficient evidence to support the claim of a difference between the two tests we cannot reject. H not and cannot from this test anyway say that there is a statistically significant difference between the two tests.

All right, we are going back to a little bit of review whether you're doing regression equations. I've used my online graphing calculator. Does most feel free to check it out. But I basically plugged in X values. Why values? And it can spit out of linear regression equation for you. And so that equation is and you can kind of see the n value there. Why equals negative 0.16 if I run to the nearest 100th X cause 24,700 22 point and actually looked up the bba here because around it to the nearest 100 that's gonna be 1.26 There's a regression equation. Find the our value, Will. They already did that for us. It is. You can see it there. Let me just highlight it. R is negative 0.970 Run it to three decimal places. This is a strong correlation answering part C like Is this accord good predictor Bad court predictor. It is a strong predictor. So it is a good predictor. Um, we say anything where the our value or I should say the absolute value of the our value is greater than 0.75 It is a good predictor. You can also see it's really close to all the points, all right? And so Part D 19,000 is that good for 60,000 bucks? Well, it is definitely definitely knocked. And here's why. If our X values air all over my alleges here, if you plug in 60,000 into this equation, I actually went ahead and found what output you get. And so let me just copy down our equation. And if you plug it in, you actually get the output of I haven't written down here 14,000 $972 and 63 cents. So according to the equation, it should be just below $15,000. Another way to say this is Hey, this car that has 30,000 miles is priced at 19,000. So why is one with twice as many mileage twice as many e miles? I should say, price, that the same thing. This is overpriced for this reason here and that reason there

But we're given this equation here, and we need to explain the Y intercept and the slope. So this equation is telling us that given the age of the car, we confined the value of the car. So the slope, the slow tells us that every time X goes up by wine, why goes up by the slow? So for the age of the car is in years. So every year, the value of the car goes down by free 0.4 and this whole famous times $1000. So $3000 it goes down because of this minus sign right here. And the Y intercept gives us what happens at X equals zero. So in the car zero years old, we should get $34,000.


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