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Is there a relationship between area somewhere lives in Spainand who they think is the best footballer? (I made up the data)alpha = .01Use the File Chi Square Socce...

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Is there a relationship between area somewhere lives in Spainand who they think is the best footballer? (I made up the data)alpha = .01Use the File Chi Square SoccerHint: Create a Pivot Table, player on one axis city on theotherAlpha = .011) What is the Chi Square test statistic? (answer must be 2 decimalplaces)2) What is the Chi Square Critical Value? (answer must be 2decimal places)3) What is the PV? (answer must be 2 decimal places)4) What is the decision? (choose a letter)A. Do Not Reject Ho

Is there a relationship between area somewhere lives in Spain and who they think is the best footballer? (I made up the data) alpha = .01 Use the File Chi Square Soccer Hint: Create a Pivot Table, player on one axis city on the other Alpha = .01 1) What is the Chi Square test statistic? (answer must be 2 decimal places) 2) What is the Chi Square Critical Value? (answer must be 2 decimal places) 3) What is the PV? (answer must be 2 decimal places) 4) What is the decision? (choose a letter) A. Do Not Reject Ho B. Reject Ho and Accept Ha 5) This means: (choose a letter) A. There is a relationship between player and city B. There is not a relationship between player and city What is the relationship? (Choose a letter) A. Barcelona leans towards Rinaldo and Madrid towards Messi B. Barcelona leans towards Messi and Madrid towards Rinaldo C. Palma prefers Barcelona D. There are no preferences based on city and player Row Labels Count of Madrid M 810 Other 390 R 649 Grand Total 1849 Row Labels Count of Barcelona M 809 Other 389 R 648 Grand Total 1846 Row Labels Count of Palma M 808 Other 389 R 648 Grand Total 1845



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(a) identify the claim and state $H_{0}$ and $H_{a},(b)$ find the critical value and identify the rejection region, $(c)$ find the test statistic $F,(d)$ decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, the populations are normally distributed, and the population variances are equal. If convenient, use technology. The table shows the salaries of a sample of individuals from six large metropolitan areas. At $\alpha=0.05,$ can you conclude that the mean salary is different in at least one of the areas? (Adapted from U.S. Bureau of Economic Analysis) $$\begin{array}{|l|l|l|l|c|l|}\hline \text { Chicago } & \text { Dallas } & \text { Miami } & \text { Denver } & \text { San Diego } & \text { Seattle } \\\hline 43,581 & 36,524 & 49,357 & 37,790 & 48,370 & 57,678 \\37,731 & 33,709 & 53,207 & 38,970 & 45,470 & 48,043 \\46,831 & 40,209 & 40,557 & 42,990 & 43,920 & 45,943 \\53,031 & 51,704 & 52,357 & 46,290 & 54,670 & 52,543 \\52,551 & 40,909 & 44,907 & 49,565 & 41,770 & 57,418 \\42,131 & 53,259 & 48,757 & 40,390 & & \\& 47,269 & 53,557 & & & \\\hline\end{array}$$

The following is a solution video to number 11 and it compares the well being index for different parts of the country, Northeast, ah Midwest, South and West. So the first step is to state the hypothesis. The null hypothesis is that the mean the means are all equals the mean for the Northeastern people is the same as the Southern people and Western people and so on. So the four means are the same. And then the alternative is that there they're not the same or at least one is different. The second step is that we need to find the critical value, critical value, I called it F Star. And there are three pieces of information we need to know and I wrote it, I wrote it down right here, We need the alpha and that's usually given to you at the .10 in this case. And then we need the degrees of freedom for the numerator and the denominator. So for the numerator, it's the number of categories minus once. There are four categories. The four regions of the country, Northeast, Midwest, South and West. So four minus one is three. So that's the degrees of freedom for the numerator. And then the degrees of freedom for the denominator is the total number of data values in this case. There were 34 of them minus the number of categories. 34 minus four is 30. So those three pieces of information is what you need. Now you could go and look at a chart. Usually most books have them uh chart or uh you can search online or you can program something in your calculus. That's what I did. So I made a program in the calculator. I called it in verse norm. I'm not going to show you how to write this program. You can google it or youtube but if you want. But um it's pretty nice and it's easy to see. So I call it in verse norman now it's gonna ask me for the area, right? And That's 10%. That's your alpha value. And then d. one. I just wrote that as for the numerator. So the degrees of freedom for the numerator remember was was three. And then for the Denominator D. two was 30. And then I press enter again and that's going to give me my Critical value of about 2.276. So f. star equals Uh 2.276. So that's going to go right here. So anything greater than if our test value of R. F. statistic is greater than 2.276. Then we're gonna reject. If it's less than 2.276 then we're gonna fail to reject. Okay so anything to the right? That's where it's in the rejection region. That's that .10. This represents 10% of the distribution. Then we reject. So now we're going to turn to our calculator again or at least I am because doing this manually can be a pain. Um If you had a Staten edit here your data value. So I think this was north Northeast. The L one, L two, I'm pretty sure was midwest L three I think was south and then L four was west. So these are the well being index means and you punch those in. Then you go back to Staten test. It's the very last one where it says innova and we're going to go to a nova and then let's just go ahead and type in all of our columns, all of our categories. We want to test all of them and make sure you're separating those with comma So it's all one comma L two comma. L three, comma. L four. And then whenever you press enter that gives you you know pretty much everything you need. So F equals 7.49 Okay so our f statistic is about 7.5 which is definitely in the rejection zone and that tells us on number four what we do we're going to reject are null hypothesis. So any time it's greater than the F. Star it's reject. Another way to double check. That is you can look at this p value. So this is another good thing about this. One way innova test um This p value here tells you whether to reject or not as well and this is about seven times 10 to the negative fourth. So the p value is not seven or it's not six point 95 It's 6.95 times 10 to the negative fourth. So this is actually a very very small number, basically zero. Any time something gets put in scientific notation it's basically zero and any time the p value is less than alpha in this case it's definitely less than it's less than 00.10. We reject. H not. So that's just a nice way to double check. And then the 5th and final step is to summarize this, you know, an actual plain words. So we're gonna write the sentence. There is sufficient evidence, evidence um to suggest that at least one mean score of well being index is different than the other regions. And we'll see in a later problem. I think it's south, that's quite a bit different than the other regions. So that's the five step process for one way an over test.

Let's use what we know about linear regression and correlation coefficients to determine whether or not there's any significant linear correlation that exists between the diameter and circumference here of different sports balls. So let's go ahead and start by inputting these data points into our stat list on our graphing calculators right now. Just keep in mind we're only working with diameter and circumference, ignoring volume for now doing that, let's produce a scatter plot. You'll see it looks something like this, which does indeed look fairly linear right in here. So I'm gonna guess that we have a pretty high correlation coefficient. But to test that, let's go ahead and run a linear regression T test doing that. You'll see that we have a correlation coefficient, which is equal to 0.999 actually, so almost one. But let's see how significant that is by looking at our P value, which is given to us as 3.167 That's given to us in scientific notation times 10 to the negative 19. So this is pretty much zero, so we can definitely conclude that this is a significant correlation coefficient value where we have a positive linear relationship here of 00.99 Uh, and again, if we wanted to just confirm this by looking at scatter plot, the reason we do this scatter plot first is just to give ourselves that visual representation in case the calculate calculations suggest something other than what it really is. Maybe it's suggested linear if it really isn't. That's why we're looking at the scatter plot, and you can see that it's definitely linear, so we can certainly count on this correlation coefficient to be significant.

Following is the solution in number 14 at one way Innova test. Uh and this is about the mean sales prices for three cities. And the null hypothesis here is that the mean sale prices are the same for these certain houses. And then the alternative is that at least one of them is different. The second step is to find the critical value and you need three pieces of information to find the critical value. One is your alpha, Your significance level, that significance level. In this case, that's usually given to you is 10. They also needed the degrees of freedom for the numerator, which is the number of categories in this case, the number of cities minus one. So there were three cities that we looked at minus one is two, so degrees of freedom for the numerator is two degrees of freedom for the denominator is the total number of data values minus the number of categories. So in this case, if you counted up those data values, there were 31 -3 cities that we looked at. So 31 -3 is 28. So that's what we need. So from there you can use a table or you can use software. I'm gonna use software. So I wrote a program and I called it inverse. F. I'm not going to show you how to write this program. You can youtube it if you wish. But um it makes Makes it easier for me. So the area is the alpha value. So we'll put in .10 for that. Degrees of freedom from the numerator was too. And then degrees of freedom for the denominator was 28. And that's going to spit out my f. star my alpha value and my uh critical value which is about 2.503. Let's call it 2.503 is my F. stars 2.503. So anything greater than 2.503. We're going to reject h not anything less than 2.503. And we're gonna fail to reject the null. The next step is to find the f statistic and you can do that manually but especially with bigger data sets that can be really time consuming. So I went ahead and punch this into stat. Edit. And these are my data values. So these are the I think these are in thousands of dollars but these are the mean sale prices and if you go back to stat and then tests and the very last one in nova And you put in your columns just make sure you separate them. This is on the T. 84 by the way but make sure you separate them by commas otherwise it's not gonna read it right so nova for those three columns and that's gonna give us everything we need to. The f statistic is about 0.966 Let's go and write that down. So 0.9 66 which is somewhere over here. So that lands in the non rejection region. So that's actually gonna tell us why our fourth step which is the decision and we're going to fail to reject H not since the F statistic is less than the critical value. Now you can also use the P value method that's what this second piece of information is good for it. Now this other stuff doesn't really matter. Um You can just kind of ignore it because this is really what we need. We need the F statistic and we need the p value and the p values pretty large. It's about 0.39 And what you do is you explicitly compare the P value with your alpha value. So the p value in this case is greater than your alpha value. 0.39 is bigger than 0.10. And any time you're P values greater than alpha, you failed to reject H nine. If it's less than alpha then you you reject. And then the final step is to conclude this, you know, with actual words and bring it back to the question at hand. And so what we're going to say is that there is not enough evidence or there is not sufficient statistical evidence. So there's not sufficient evidence to suggest that the mean sales price prices of houses In the three cities are different. Okay that's the five step in Nova process.

So I'm gonna let d stand for just for convenience. The beginning time minus what the end time is. And we can subtract the opposite way as well. But it talks about is there a difference? So we would assume that that difference Is non existent, that there is no difference between the beginning and the end so that the main difference is zero and alternately that it's different. So they didn't say that they wanted to test to see if the time had gotten smaller. They just asked if it's different. So We have a total of eight people And we're doing a two tailed test and we want our significance level to be 1%. So we need to split that 1% between the two tails. So this needs to be 20.5 and this needs to be 0.5 And we'll find that T. Value with eight degrees excuse me, seven degrees of freedom. That T. Star value with 5% of 50.5% in that upper tail and that corresponds to with 7° of freedom. Looking like that is 3.499 and this one is negative 3.499. So we would reject the null Any time we have our test statistic which is a t. value with 7° of freedom being less than negative 3.499. Mhm. Or if that test statistic is greater than 3.499. So we can actually plan all this out before you actually look at the data and that was basically what we're doing. So part C. I subtracted all those values and hopefully I didn't mess up, I just did them in my head And I got the difference to be 8.2375 and the standard deviation of those values It was 5.2- two and it would round off to five. Now we want to find the test statistic and I have these values started my calculator And so our test statistic was 7° of freedom is going to be that mean that we got minus the mean we're assuming which is zero and then divided by the standard deviation Over the square root of and make sure you use eight and not seven. And when I did that I got my test statistic to be 4.461. Now look where that lies, that's going to lie up here and I can see that is in the rejection zone. Therefore part E we have evidence to reject the null. Mhm. Yeah. And that means that we would conclude that there is a difference, There is one is a difference in the amount of time. It doesn't ask us to say what direction, but it surely looks like the before is higher than the after.


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