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The data from exercise I follow.$$\frac{x_{i}|1 \quad 2 \quad 3 \quad 4 \quad 5}{y_{i}|3 \quad 7 \quad 5 \quad 11 \quad 14}$$$$\begin{array}{l}{\text { a. Compute the mean square error using equation }(12.15) .} \\ {\text { b. Compute the standard error of the estimate using equation }(12.16) .} \\ {\text { c. Compute the estimated standard deviation of } b_{1} \text { using equation }(12.18) .} \\ {\text { d. Use the } t \text { test to test the following hypotheses }(\alpha=.05) :}\end{array}$$$$\begin{array}{c}{H_{0} : \beta_{1}=0} \\ {H_{\mathrm{a}} : \beta_{1} \neq 0}\end{array}$$$$\begin{array}{l}{\text { e. Use the } F \text { test to test the hypotheses in part (d) at a } .05 \text { level of significance. Present }} \\ {\text { the results in the analysis of variance table format. }}\end{array}$$

a. 4.133b. 2.033c. 0.643d. 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

Temple University

University of North Carolina at Chapel Hill

Cairn University

Boston College

Lectures

0:00

11:56

The data from exercise 2 f…

11:04

The data from exercise I f…

Here's a solution for number 33 in the book. So we have h not equals uh the p. Is not equal to 0.5 wherein equals 10 and x equals nine and then not equal 2.4 and then greater than 0.5. Whenever n equals 1009 And x equals 545. So we're doing three things here, we're doing the normal approximation and then we get the exact with the binomial distribution and then we get the exact with the simple continuity correction. So there are three different items to do now. The first one the normal approximations. Whenever its normal approximation we just do a normal one sample or one proportion I should say one proportion Z. Test. And just to save some time. I'm gonna do this in the calculator. So you just follow the formula as closely as you want. But if you just go to stab and then arrow over to test and it's that fifth option down the one property test And the peanuts going 2.5 The X. is going to be nine and then the end is 10 so you're just filling that in and the alternative is not equal to. And then whenever you calculate that You get this p value of .0114. So let's go and write that point 0114. And that's basically what we're gonna be doing for this is whenever it's approximately normal, I should say it's normal. So that's what we're gonna do for the rest of them. So essentially the same thing go to stat and then tests And then one prop z test And that's .4 And then everything else can stay the same. We calculate that gives us .0012 For this next 1.0012. Right? And then the last one slightly different here we have stat tests fifth option And this time it's .5 But this time the x value is 545 And it's 1009 as the sample size. And then we need a greater than symbol there. So the alternative is greater than and then we calculate That gives us zero. I'm just looking at this p value here .0054. So .0054. So that's the normal approximation. That's kind of our baseline. That's what this whole entire section is about. Now to get the exact remember this is a binomial distribution, it's either X or it's not. And um with the binomial distribution uh now you can do the formula, it might take you a while especially whenever it gets bigger. But um all we're gonna do, well I should get out of this For this 1st 1. It's a it's a two tail test. So what I'm gonna do is find the probability of it being greater than nine I should say greater than equal to nine. And then I'm going to double it because it's a two tail test. So to do that I do one and then I go to second distribution and then it's binomial CDF. The trials is 10. The p the probability of success is 50% and then the X value is actually gonna be eight Because I want the probability of nine and 10. Okay? And go ahead and click enter And then we're going to multiply that by two Because it's a two tail test. So .0215 will be That's a zero will change colors here. So point zero 215. Sorry? So whenever it's exact we use the binomial CDF. Okay, So the same thing for the next one slightly different changed. We're gonna go one and then distribution will go binomial CDF Trials is 10. This time the P is .4 And then we'll go ahead and paste that in there. And then to get the other tail, we multiply by two so we get .0034, 10034. And then for here it's actually just a one tailed test. So this one's a little bit sneaky, We'll do 1- still and we're going to go to second distribution. And then The trials this time is 1009. The P values back to or she said the probability of success is back to .5 And then the X value what I'm going to use is 544 not 545 because it's strictly greater than. So I want 545 or more. So I want to not include 545 that I'm subtracting. So 544 and then we paste that in there and we get about .0059. Now we do exact with the simple continuity correction so exact with the correction and I should probably change College for this one. Okay so the formula book says that we find the probability that it's um unless they are less than or greater than and then minus One half of the probability of exactly that value. And then we're going to multiply by two for the two tail. So this one's a little bit more involved. We're gonna go one and then go back to binomial CDF trials is 10 .5 and then we're back at eight. Okay so we'll figure out what that is and then we're gonna subtract 1/2 2.5 times the binomial pdf. So the exact probability of 10 trials probabilities five and exactly nine. Okay so we get that and then we're going to multiply that by two for the two tails. We get zero 117. Okay so I'm just kind of following the formula there. So point 0117. So technology can be really really nice um cuts down sometimes. So the second part it's almost exactly the same things. That same process running a one minus and then it's the CDF. So second distribution CDF trial is 10 It's time the P is a .4 and it's eight. Yes we do that. And then we're going to subtract one half times Binomial pdf. Okay. And then we're gonna we'll try that bite you. So we have a point 0018 whenever we round 0.0 018. Okay. And then the last one is it's a one tailed test so it's actually not as bad. We're gonna go one minus and then we'll do the CDF of 5 44 again. So binomial CDF 1009 Trials Probabilities .5 and then 544. Okay, figure out what that is. And then we'll subtract 1/2 times PDF of 545 of the actual x value In 1009 five and 545 And we get about .0054. Okay so you can kind of compare that now and you can see that as the N as in increases the normal approximation is a better fit. Okay so one of those problems that definitely the technology can be your friend there if you know what's going on

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