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The data from exercise 2 follow.$$\frac{x_{i}|3 \quad 12 \quad 6 \quad 20 \quad 14}{y_{i}|55 \quad 40 \quad 55 \quad 10 \quad 15}$$$$\begin{array}{l}{\text { a. Compute the mean square error using equation }(12.15) .} \\ {\text { b. Compute the standard error of the estimate using equation }(12.16) .} \\ {\text { c. Compute the estimated standard deviation of } b_{1} \text { using equation }(12.18).} \\ {\text { d. Use the } t \text { test to test the following hypotheses }(\alpha=.05) :}\end{array}$$$$\begin{array}{c}{H_{0} : \beta_{1}=0} \\ {H_{\mathrm{a}} : \beta_{1} \neq 0}\end{array}$$$$\begin{array}{l}{\text { e. Use the } F \text { test to test the hypotheses in part (d) at a } .05 \text { level of significance. Present }} \\ {\text { the results in the analysis of variance table format. }}\end{array}$$

a. 76.6667b. 8.7560c. 0.6526d. Reject the null hypothesis $H_{0}$e. Reject the null hypothesis $H_{0}$

Intro Stats / AP Statistics

Chapter 12

Simple Linear Regression

Linear Regression and Correlation

Missouri State University

University of North Carolina at Chapel Hill

Piedmont College

Lectures

0:00

08:57

The data from exercise I f…

So as we go through this problem, we're going to find out the different parts that we need to complete our and nova table. So I just put that over here so that as we go throughout, we can just fill it in. And I also, um, it specifies in the problem. That's it. This is from exercise number two. The data is from exercise number two and in exercise number two, we got the expected regression equation. Where? Why? Hat is equal to 68 minus three x. So check that video out if you don't know where that came from. Problem number two of the same Ah chapter. But we're just going to use this information to come up with some the sum of squares of the air in the main square of the air and other factors involved with this another test in general. So let's start toe first. We need to find a means square of the error in part a not part B, in part, a part a. And what we need is to come up with the sum of squares be error, which is, um, the following formula. And what we need is the number of elements minus two. So first, let's come up with the sum of squares of the air and some scores of the error is simply equal to each individual wide data point, minus the predicted value at that X value squared. So, for example, if we we know that our why hat is equal to 68 minus three X if we go thio um, X equals three. Why had at X equals three is equal to 68 minus three times three witches, and I so 68 59 is equal to negative four. So what we do is we would square negative four, and we would get the first value of this, some of that 16. And now we would do the same thing with the second, um, X value we have, which is equal to 12. So I had at X equals 12 is equal to, um 68 minus three times 12 which is 36 68 minus 36 is 32 and we would square 32. Um, or yet we would have 30 two and then, um, 68 minus sorry. Ah, and the Y value at that point is 40. So 40 minus 32 is equal to eight. All right. Ah, so now we would square eight to get 64 we would keep doing the same thing for all our different Why variables over here until we get ah to the end. And then we would take the sum of all those variables. So eventually we get a sum of squares of the error equally, um, 230 230. And now we have to come up with and minus to re see that we have, ah, five values, right? 123455 Values s or and is equal to five with that and minus two is equal to three. So this would be equal to 230 over three, which is approximately equal to 76.67 So this is the main square of the air and the sum of squares of the air. We got those. So for the air, this is 230 degrees of freedom is three in the main square of the error is 76.67 And now, using this information, we can come up with how We're, um our estimate of the ah standard error. So Ah, what we're interested in is the estimate of the standard deviation for our beta sub are the standard error of the estimate. So we can use this equation. We would just take the square root of the main square of the air, which is just equal to the square root of 76.67 which is equal to approximately 8.7 six. All right, and now we have to come up with Ah, the estimated standard deviation of beta someone. So the estimated standard deviation of beta sub one. This is part See, um, Speed one is equal to the S that we just discovered center. That s that. We just discovered over the sum. The square root of this. Some of the differences between each individual X value and the X for the expert is just the mean of the exes. So the mean of the exes if we add all these values together and divide by five is equal to 11. So for this we found our s value to be equal to 8.76 8.76 divided by the square root of the some of the differences. So our first X value is three. So three minus 11 squared. Plus, next value is 12 12 minus 11 squared plus, and then you keep going until you get to the end of the data set. And from this we get a, um, standard deviation of beta sub one to be 10.65 to 6. So this is the answer to part C. And now we have to ah, test the hypotheses, um, at Alpha equals 0.5 and, um, with the following hypotheses. So what this means is the no hypothesis is that there is no ah, when you're really or there is no relationship between the X and Y variables, whereas the alternate hypothesis is that there is a relation. Get it. So in order to test this, we have Thio come up with a test statistic and this is the formula for our tea test statistic which is equal to ah the beta value. Our Betas have one divided by the standard bearer for the standard deviation of beta someone. So we got this to be equal to negative three and that is from over here. The beta suborn values equal or beast of one is equal to negative three negative three divided by the standard deviation of our beast of one is equal to 10.65 to 6. So we get a ah test statistic of negative 4.5968 And now we have to find the degrees of freedom. Degrees of freedom is equal to and minus two. And we already discovered that to be five minus three, which is equal to three. So our p value, given that our T te statistic is equal negative 4.5968 and our degrees of freedom is equal to three is, um it corresponds to a ah value between 0.1 more. So this would be a member of the set 0.10 point 005 to 0.1 But because we this is a two tailed test, we would have to multiply this by to to get our multiply this by two to get to our p value. So our P value is in between 0.1 and 0.2 This is where a P value lies. And now you have to test this against an Alfa 0.5 So against this 0.5 because our P value is less than 0.2 just less than zero lessons Your 00.5 um, we can reject it all. Okay. And now we have to do the same thing, but with a f test statistic. So we found out the sum of squares due to the air already. We just need to find some squares due to the total. And then ah, this column here. That was a terrible rectangle. Okay, s. So the sum of squares for the total is equal to the following. It is the sum of each individual, each individual. Why value minus the mean of the wise square? So we have to find our why bar, which is equal to defy plus 40 plus 55% plus 15 all over five, and we get a ah value of 35. So our why? Why bars equal of 35 now we're just going to take the ah difference squared between each of our individual. Why values and our Y bar. So this is going to be equal to 55 minus 35 squared plus 40 minus 35 squared plus 55 minus 35 squared and so on and so forth until we get a sum of squares. Of the total equaling 2 18 50 this is equal to 1850 and oh, sum of squares total is equal to 18. Oh, goody, not the regression. And now, to find the sum of squares to you too, regression, we're just going to take the difference between the sum of squares of total and the sum of squares you air, which is equal to you 1850 minus 230. It is equal to 16 20. And now we just need to find the number of independent variables that we have, which is one we have one independent variable, and the means square of the regression is equal to the sum of squares over of the regression divided by the number of independent variables just equal to 16 20 over one, which is 16 20. And now all we need to do is come up with an F test statistic. So the F statistic is equal to the main square of the regression over the means square of the error, which is equal to 16. 20 over 76.67 which is equal to 21.13 approximately. And now with this, we have to come up with a P value. So we have, um we have three degrees of freedom. There are degrees of freedom equal to three. Because it is, um um, the number of elements that we have minus two. And now we have to come up with the F statistic. So, um, or a P value so r p value, given that our f statistic 21.13 and our degrees of freedom is three is equal to 0.19 which is less than our Alfa 0.5 Therefore, we reject the no

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