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Assuming the esiduals are normally distributed_ consinct 95% confidence interval about the slope of the true least-squares regression line_Determine the lower and u...

Question

Assuming the esiduals are normally distributed_ consinct 95% confidence interval about the slope of the true least-squares regression line_Determine the lower and upper bound for the confidence interval using statistical software , rounding to three decimal places_

Assuming the esiduals are normally distributed_ consinct 95% confidence interval about the slope of the true least-squares regression line_ Determine the lower and upper bound for the confidence interval using statistical software , rounding to three decimal places_



Answers

Use a computer or calculator to find the $95 \%$ confidence interval for estimating $\mu_{d}$ based on these paired data and assuming normality:

And problem number eight were asked to construct a 95% confidence interval using data from section 10.2. Number eight. Now this actually isn't data per se. You're actually giving summary statistics. So we actually have to use the formula here. And um, you know, basically here's the formula. So it's beta one hats, that's the point estimate plus or minus the margin of error in the margin of air is the critical value T alpha over two. And we'll find that here in a second times S. E and S. E. Is the um the sum of the squared residuals divided by n minus two, so degrees of freedom and square rooted. And then that's divided by the square root of S. S. X. X. Now, fortunately for us, a lot of the stuff we already found, so Beta one hat we found in a previous problem as 10.599 or 0.6 maybe. And then the S. S. X. X. We found to be 60.8. So I went ahead and wrote those down already because we already did them. And now we just need to find S. E. And to find sc we need to find the sse first, fortunately for us, the SSC is just S S Y Y minus beta one half times Ss X. X. And like I said, we've already found all this stuff, so I'm going to change colors here. The S. S. Y. Y. Again from a previous problem is 25 to beta one hat. So minus beta one hand is that 10.599. And then S. S. X. Y. Is again found from a previous problem is 36.4. So whenever you plug all that stuff in, you should get 3.3964. So that's the S. S. E. And now we can find the S. E. We just take 3.3964 divided by the degrees of freedom. So in minus two is the degrees of freedom and in equals five in this case. So in five minus two is three. So it's a square to 3.3964 divided by three. So now we're pretty much ready to roll except for that critical value. And I'm gonna use you can use a table if you want. I like to use a calculator though. So if you go to second on a. T. I. T. For second bars and go down to inverse T. The area here is going to be it's one minus alpha over two or just alpha. Alfa over to it. It doesn't really matter how you do it but um 95%. So one minus the 95% is 5% and um you cut that in half and you get point oh 25. So the area is is half of one minus the percent confident you want to be. So point oh 25 is the area. Or you can do the complement of that 250.975 degrees of freedom is going to be three in minus two. And then whenever we paste we should get about three point and I'm gonna ignoring the negative here because the negative positive doesn't really matter. 3.182. That's our critical value. So I think we're pretty much ready to roll. So beta one hat is .599 Plus or minus. And then that critical value was 3.182. So like I said the negative doesn't really matter because it's plus and minus. And then the S. E. was this big old square root three 3964 divided by three. And then that's divided by the square root of SSX. X. Which is the square root of 60.8. You really just use a big old skirt if you want And that's a bit of a mess. You may need to go in pieces, you may need to solve this first and then solve this and then divide them and then multiply by 3182. But you should get 0599 plus or minus 0.434. So you can write it like that where you have the point estimate and then plus or minus the margin of error. I like to write an interval notation though. So I actually solved out .599 -134. And I got about .165 And then .599 plus .434. You should get about 1.033. So the interpretation here would be we can be or we are 95% confident that the true population um Slope for this data is between .165 and 1.03.

The following is a solution number one. From section 10.5 it says constructed 95% confidence interval for the slope. So beta one of the population regression line from section 10.2. Number one. So I'm gonna use a t i a t four. If you go to stat and edit you can see here's where I put my ex values in L 10135 and eight. And then L two is where I put my Y values 2465 and nine. Then if you go back to stat and then air over two tests and it's almost the very last one. It says lin Wreg T. Ent that stands for linear regression T interval. Go and click enter the excellence should be a one. The wireless should be L2 frequency can be one and the sea level. We're asked to find a 95% confidence intervals. 1.95. Then if we calculate this top band here, that's our Our confidence intervals .16502- 1.3- 04. Okay, so let's go and write that down 0.16 502 All the way up 1.3-0 4. So it doesn't ask us to interpret it, but we would just say we are 95 or we can be 95% confident that the true population slope for the regression line of this data is between .16502 and 1.3-04.


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