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Srikar K.
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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 58 Problem 59 Problem 60

Problem 26 Hard Difficulty

In exercise 18 the data on grade point average and monthly salary were as follows.
$$\begin{array}{l}{\text { a. Does the } t \text { test indicate a significant relationship between grade point average and }} \\ {\text { monthly salary? What is your conclusion? Use } \alpha=.05 \text { . }} \\ {\text { b. Test for a significant relationship using the } F \text { test. What is your conclusion? Use }} \\ {\alpha=.05\text {.}} \\ {\text { c. Show the ANOVA table. }}\end{array}$$

Answer

a. Reject the null hypothesis $H_{0}$ or significant
b. Reject the null hypothesis $H_{0}$ or significant
c. See table

Related Courses

Intro Stats / AP Statistics

Essentials of Modern Business Statistics

Chapter 12

Simple Linear Regression

Related Topics

Linear Regression and Correlation

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Watch More Solved Questions in Chapter 12

Problem 1
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Problem 10
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Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
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Problem 24
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Problem 35
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Video Transcript

and this problem. We want to know whether a T test indicates significant relationship between grade point average and monthly. Sarah salary. So here we have grade point average. Here we have monthly salary. Let's first come up with our hypotheses. And these were gonna be our hypotheses that there there are no hypothesis will be that there is no relationship, whereas our alternative hypothesis well be that there is a relationship of some sort and we don't know what that is. And we're going to be testing this against Alphas. You're a 0.5 So in order to come up with a T test statistic, we need to find, um, the detail statistic we need to find our s and our standard deviation for our Ah, be said one value and ah, this is from and this data's from problem 18. And we already calculated, uh, the, um, regression estimated regression equation. So if you are wondering about that, go check that video out. Um, And with this now we can find out Ah, what we need. And in order to come up with this value for our S R. S is simply going to be the square root of the mean square of our error. So this is the formula for the main square of our error. But for a mean square of error, we need a sum of squares of the air. The sum of squares of the error follows this following formula. It is the difference between each individual value and the predicted value for that same X squared. So, for example, let's how are have our first live value of 3300 at X equals 2.6. So 3300 minus. Ah, Why? Hat of 2.6, which is equal 23301 0.36 squared. Plus our next. Why value of 3600 minus. Um, ready Down here. Why had actually you just need this to anyone? Plus ah, 3600 minus. Why? Hat at 3.4, which is equal to 3766 0.24 squared. And you do this for each of the values down here? Um, and you would eventually get a sum of squares due to the error equaling 85,000 135.1378 So in the end, we're going to come up with a nova table, so I thought we would just fill that out as we go along. So, uh, sum of squares do The air would be eight 5135 point 14 I'm just going to truncate it to fit it on the table. So this would be our sum of squares due to air? No, we have to come up with Ah, that sum of squares due to her minus, um, And over to. And our end is equal to six because it wanted to you 3456 and and minus 26 minutes to is equal to four. So this is equal to 85,135 0.14 over four, and we gotta meanest square means square Q t error equaling 21,000 too. I'm introducing different color 21,000 283 points of it eight. And now we're going to take the square root of that starts getting clinic quartered, but, um, yes. Oh, this is gonna be the square of 21,283 00.78 So that's gonna be 1 45 0.8896 So now we found our s value. So this is equal to 14.8896 But now we need to find this value down here, and that is equal to our s. Over the, uh, sample standard deviation really is what this bottom formula means. So our ex bar is going to be equal. Our expert is going to be equal to the mean of our X values. So 2.6 plus 3.4 plus three points expose 3.2 is 3.5 plus 2.9 divided by six. And eventually we get an X bar equaling 3.2. And now this is going to be equal. Thio Um, yes, value rediscovered of 1 45.8896 divided by the square root of the sum of all these differences. So our first X value is 2.6, 2.6 minus 3.2 squared plus 3.4 minus 3.2 squared and so on and so forth. Until we finally get a value, Miss finally equals 16 are this bottom part. This over here equals 0.74 So then we get a s, um, or estimated standard deviation of beasts of one equaling 169 169.59 So this is going to be equal to one 1 45.8896 over 169.59 And that is equal 23 Uh, I made a mistake. Somewhere out, this over here should not be s. It should be bait a subway or be sub one. It is our estimate, um, divided by, um our standard deviation. So this up top should be equal to Okay, hum. So we got our estimated standard deviation to be equal. Thio 169 0.59 And from the problem, we got our be value to be 581.1. So from this we get a t te statistic of 3.4 to 6. And now we can test this hypothesis, um, and R p value even that we have a, um t te statistic of 3.4 to 6. Ah, and with a degrees of freedom of four. Because we're going to take a number of elements that we haven't subtracted two from it. So and minus two would be four is a member of this set 0.1 are 0.0 to 5 and 0.1 the other way around Sierra 0.1 and zero point certified. But this is not good enough because, um, we have a two tailed Tess statistics. We're gonna take this and multiplied by two. So we get that r P value is in between 0.2 and 0.5 Who and which this is equal to or Alfa. But because we know that RP is less than 0.5 we can reject a No. And we have this for our t te statistic. But now we have to come up with our F test statistic. So if we go back here, we discovered our mean square of the error to equal 21,283 0.78 So I'm just gonna update this table. Um, what I said about you was 21,000. Yes, the mean square of the air, equal to 21,283 0.78 All right. And our degrees of freedom is for Okay, so now we have to come up with some of squares of the total. Okay. And the sum of squares of the total is simply equal to the difference between, um each Why value and the y mean are the mean of the Y. So be found. Are y bar to equal? Um 36. 50. Now we're going to take the some of the differences. So our first y values 3300 minus 36 50 squared plus 3600 minus 36 50 squared. And we're gonna do this for the entire data set until we get a sum of squares of the total equaling 335,000. And now we have to come up with some of squares due to the regression in sum of squares of their aggression is equal to the sum of squares of the total minus sum of squares of the air is equal. Do 335,000 minus the sum of squares of the error, which we got down here 85,000 135 0.1378 and we go to sum of squares of the regression equaling 249,000 864 0.86 and I'm rounding off. And now we have to come up with the mean square. Do you two regression, which is equal to the sum of squares due to depression over the number of, um, element groups that we have the number of independent variables and we have one independent variable. So the means square due to regression is just going to be equal to the sum of squares Future aggression, this 249,864 0.86 Now we have to come up with an F statistic, and our statistic is going to be equal to the main square of the regression over the main square of the total And these bodies or 249,864 0.86 divided by 21,283. 21,283 0.78 Oh, sorry. This should be mean square here. Mean square of the year. Um, so we're going to take this quotient, which is equal to 11.7397 and now we have to come up with a degrees of freedom in order to find our P value. Our P value is equal to 11.7397 There are degrees of freedom is six minus two, which is four, which is equal to, um something which is equal to not something 0.26625 which is less than 0.5 Therefore, you can reject, you know, and let me update my statistic in this table. 11.7397

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David R. Anderson, Dennis J. Sweeney,Thomas A. Williams

Essentials of Modern Business Statistics

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