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In exercise 9 , an estimated regression equation was developed relating the top speed for aboat to the boat's beam and horsepower rating.a. Compute and interpret and horsepower rating.b. Does the estimated regression equation provide a good fit to the data? Explain.

a) $R_{a}^{2}=55 \%$b) good fit

Intro Stats / AP Statistics

Chapter 13

Multiple Regression

Descriptive Statistics

Linear Regression and Correlation

Cairn University

Oregon State University

Boston College

Lectures

0:00

03:18

(a) Compute the power regr…

07:01

01:11

Driving speed and fuel con…

04:24

In Exercises 43–46, tell w…

02:03

04:37

02:54

06:41

An issue of Popular Mechan…

06:14

Let's try the re-expr…

03:50

(a) Use the method describ…

02:41

More horsepower In Exercis…

08:28

Recall from Exercise 55 of…

03:28

In $9-13 :$ a. Create a sc…

02:24

As the example in the chap…

05:31

In exercise loan estimated…

03:46

06:43

The following data were us…

02:19

03:23

02:11

So in this question, were given some data that's relating the top speed off the boat too. The beam and the horsepower ready. And we're asked first to compute R squared and adjusted r squared and interpret it. So let's go ahead and do that. So we're gonna put in our input wide range, which is right here. So cause from he won two the 21 and our input X range goes from we want to see 21 we set our confidence level. We're going to put our output over here. So let's put our output in the 24 for this multiple regression equation with all the summary statistics. So we have here our summary output, and we can see that they are squared is point 597 and the adjusted R squared this 5970.55 So let's first write that down. So our our squared iss iss and are adjusted R squared, please. 55 We're going to put that down and now to interpret R squared and adjusted R squared is basically saying that, but the adjusted our script 55% off the theory ability off the data can be explained by the multiple regression creation. That's our interpretation off we r squared values. The second part of the question, we're asked if the regression equation provides a good fit to the data. So we see that since are adjusted, R squared equals to about 0.55 only 55% off the variability. Let's explain by the equation. So no, it does not provide a good fit. We want something to provide a good figure, probably want t r squared or they just are square to be more than 1.7 toe. Provide a reasonable thing. But this is less than 0.7. So they said that it does not provide a good fit on This is our answer toe the end.

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