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Select one: wear gi 107 Question the 23 given that the standard sample size (n) must Not yetanswered 'Sevietion Gsigg9a} obea 1 confident that the sample . 3 q...

Question

Select one: wear gi 107 Question the 23 given that the standard sample size (n) must Not yetanswered 'Sevietion Gsigg9a} obea 1 confident that the sample . 3 qaatlon mean Milnin of the populaton0

Select one: wear gi 107 Question the 23 given that the standard sample size (n) must Not yetanswered 'Sevietion Gsigg9a} obea 1 confident that the sample . 3 qaatlon mean Milnin of the populaton 0



Answers

For Exercises 3 through $8,$ the null hypothesis was rejected. Use the Scheffé test when sample sizes are unequal or the Tukey test when sample sizes are equal, to test the differences between the pairs of means. Assume all variables are normally distributed, samples are independent, and the population variances are equal. Exercise 13 in Section $12-1$

What's up, stat Cots in this video, we're asked to perform either a shove or two key test after a one way Unova. So says the example we were given. And this is the average debt of public colleges graduation per state. So because the sample sizes are even, we are going to be doing a two key test. This is our test statistic, and we find our critical value with the end table using K. And okay, so let's go over to the Excel spreadsheet where I have this raw data and our Unova already performed and already paired them up. So this is New York and Virginia and Virginia and California, California, Pennsylvania, Virginia, Pennsylvania, California, New York and Pennsylvania in New York. So to calculate our numerator, we are going to take the differences in the means. So we're gonna take mean of New York minus the mean of Virginia. Then we're gonna take the mean of Virginia minus Orania for you. Pennsylvania, Virginia, Pennsylvania. Orania, you are, and Pennsylvania in New York. Okay. And to find our denominator, we're gonna take the square root of the main square within groups over our sample size. So it's gonna be five. So let's take our square root of our mean square within groups over and which is five for everything. So that number is just gonna coffee down. Oops. Are I'm just And do that use away. Okay, so we're gonna take our numerator over our denominator. And as you can see, some of them are negative. Um, but our test statistic is always gonna be positive. So we're actually taking the absolute value. So just ignore the negatives, and we're just gonna copy these values over as positive values. And last thing we're gonna do is calculate our critical value, which is using the end table. So K is equal to our number of means. So it's gonna be four, and V is equal to our total sample size of the population. So the five times for 20 minus our k so be 16. So let's pull up the end table for Alfa level 0.5 So, as we said K was for and B was 16 so are critical. Value is 4.5 All right now, I'm gonna copy all those numbers over just to give us some visual organization and again they're all gonna be positive. All right? And to decide if we have a significant pair or not, we're going to take our test statistic. And if it is greater than our critical value, then we can reject are null hypothesis. That means we're gonna have a significant pair. So is 0.33 larger than 4.5? No. Is 0.48? No. Is 4.9? Yes, it is. 4.42 Yes. Is 0.14. New is 4.75 Yes. OK, so we have three significant pairs. So let's summarize our results in this kind of small area we have just after a small. So there is sufficient evidence to conclude there is a mean difference between hairs three and four or into, and for one. So that would correspond to a California in Pennsylvania, Pennsylvania and Virginia and Pennsylvania and New York. So those colleges do you have a significant difference in the average debt of graduation? Alright, guys, that's it for this video. I hope you learned a lot. I'll see next time

What's upstart cots In this video, we are given an example, and we are asked to perform either a chef or a two key test. So this is the example where provided. So considering we have a neat, uneven sample, sizes were gonna want to perform a chef test. So this is the formula to calculate our test statistic. So let's go ahead to the Excel spreadsheet where I have the raw data and heard. So I have entered the data and performed an Innova because through the example, were given, we are told that, um, the differences are there's already a significant difference. So I just wanna had ended theano of a test. So I think the first thing we should do is calculate our critical value. So our critical value is found by taking the critical value from our F test and multiplying it by the degrees of freedom between groups. So that is our critical value. And now we're going to calculate the test statistic. So what we're gonna do is I'm gonna go ahead and pair these categories up, so cars I'm just gonna use one SUV will be too, and trucks will be three. So I'm gonna pair cars and SUVs, SUVs and trucks and then trucks and cars. And the first thing I'm gonna do to calculate the test statistic is I'm gonna do the differences in means squared. So I'm gonna do equals parentheses. And I'm going to go to the mean of cars cause that's one and subtract the mean of SUV's gonna close the parentheses, and I mean square. So I'm going to do the same thing for the rest of these. All right, So that's the differences in our means squared. If we go back to our formula now, we have to do the variance within and multiply it by one divided by our sample sizes. Okay, so let's go back to our excel spread she So I'm gonna do okay first. Going to calculate that one over n nonsense. So let's do over here. It's too. He fools one over and then and will be the sample size for car. So 1/5 waas one over the sample size for SUVs. Well, I have us over a lot going on. Okay, So let me just type for than against, Okay, so I'm gonna do that for the rest of them. So one over. Sample size for SUV plus one over sample size for Chuck's man. Last one. Okay, that's just gonna make it easier for us when we do our numerator divided by our denominator. Okay, so next thing we're gonna do is we're gonna take our means square with n group, and we're gonna multiply it by that value we just calculated. And then now we can calculate our f our test statistic really easily. So we just take our numerator divided by our denominator, and I can just pull out formula down or not. Okay, so we've calculated our critical values and our test statistics, so let's go back to our whiteboard and fill them in. So are critical value. Waas e 0.52 on. Yeah, Okay. Just making sure 8.52 and our test statistic for cars and SUVs. Waas 2.1. And for SUVs and trucks, it was 17.64 And for cars and trucks, it was 27.9 to Okay, So in order to know which of these is significant, we want to ask which of these is larger than our critical value. So it was 2.1 greater than 8.52 No. Is 17 greater than a yes? And is 27 greater than a yes? So these so SUVs and trucks. So the differences between those means or different and the difference between the means of cars and trucks are different. So to summarize this, we're gonna say there is sufficient evidence to conclude a difference. And I mean cost the drive. These this between hybrid cars and trucks and between SUV's and checks. Alright, guys, that's it for this video. I hope you learned a lot else. You next time.

What's up? Stock cats. In this video, we're gonna be doing either a chef or a two key tuhs, depending on the example were given. So we're given this raw data we're told are out level 0.5 So because our sample sizes are uneven, we are going to be doing a chef test. So here is the formula for the test statistic and are critical value. So I gone ahead and entered this data in a in an Excel spreadsheet E and also performed the Unova because we're also told that we did have a significant and Nova. So I just went ahead and did that. So let's go ahead and calculate are critical value. So are critical. Value is found by taking the critical value from our Unova test and multiplying it by our degrees of freedom between groups. Okay, so now let's do our test statistic. So if dia A is one b is two and see us three, I'm gonna go ahead and pair them, So diet A and B, diet B and C and D I, c and a So let's go ahead and calculate our differences between means squared. So this would be for dia. A minus tie be. We're gonna square it. Gonna do the same for the rest of them. Okay. And next thing I'm gonna dio is I'm going to calculate part of the denominator. So I'm gonna calculate this portion right here so that it makes it easier for us to do the rest of the denominator. So let's go ahead and do take our one over our sample size for dia A and add that to one over a sample size for Diaby isn't. Then we're just gonna go ahead and do that first of them, Okay? And now we are going to take the our mean squares with in through and multiply it by that value we just calculated. So we're doing the denominator part of the test statistic. So actually, before we do that, I'm going to go ahead just to make the formula easier for us to do. I'm gonna take that means square within group number, and I'm gonna dio function core, and then I'm gonna multiply it by that value. And then in that way, I can just pull that down, see? Okay, so this these are our test statistic. I'm sorry This is the denominator. And this is the numerator. So in order to get our final test statistic, we're just going to take the numerator and divided by the denominator. I'm just gonna go ahead and pull that down. Okay, So these are our of justice is sick values for chef test. So let's go ahead and mark them over to just give us some visual organization. So are critical. Value was 7.96 and our test statistic between A and B Waas 9.81 and between diets, BNC is 11 point hey, and between A and C, it was 0.8 So we have a significant value if it so if our test statistic is larger than our critical value, then that means the means are significant between each other. So is nine larger than 7.96? Yes, it is 11.8 larger than some points 996 yes. Is 0.8 larger than 7.96 No. So these are our two significant results. So now we're going to summarize them in words. So we're gonna say there is sufficient evidence. You include. There is a difference. And I mean I mean, I think these were calories. Oh, no. They're mean. Pounds lost. Okay. And mean pounds lost. Sorry about that. So there is sufficient evidence to conclude that there is a difference in mean pounds lost between diets A and B and between diets, B and C. Alright, guys, that's it for this video. I hope you learned a lot and I see next time.

What's up? Stock cats? In this video, we are asked to perform either a chef or two key test, and those are post talk tests for a one way and over. So this is the example were given, and these are. These are the costs of what costs oven oven her watts of the oven. So because the sample sizes are uneven, we are going to perform a chef test. So this is the test statistic. This is the formula for the test statistic. This is our CV says are critical value. And we are told that the Alfa level for the one way Unova tests was performed at a 10.1 Alfa level. So let's go to the Excel she that I have our raw data on. So I've already performed the one way and Nova because that's how they have the example set up. So now we are going to pair up our oven watts because a chef test is a pair wise test. So we're gonna pair up 1000 watts and 900 watts, 900 watts and 100 watts and 100 watts and 1000 watts. So to calculate the numerator for our test statistic. We're gonna want to take the difference of the means squared. So let's go ahead and do that. So we're gonna take the mean for the 1000 watt of in. We're going to subtract the mean of the 900 watt of in, and we're going to square that value, and we're gonna do that for the rest of the pair. Wise pair wise pairs. No cheese. Hey, off to a rocky start. But we can recover. So the this is our numerator, So are enumerators done. And now we want to calculate our denominator. So the first thing I'm gonna do is I'm actually gonna calculate, um, this portion here and then because we already have our difference within groups. We already have our means squared within groups already calculated. So we need to calculate the other part, which I'm gonna do right here. So one over sample size of 1000 want plus one over sample size of 900 watt. All right, so now too calculate our denominator. We are going to take our means square within groups, and we're going Teoh, multiply it by the other value we just calculated I'm gonna go ahead and pulls. Actually, I can't pull it down. Okay? Finally, to calculate our test statistic, we're going to take our numerator over our denominator, and I can pull the this value, these values down. So for 1000 and 900 watt oven, we have two point 91 for a test statistic. And for the 908 100 we have eight point 40 And finally, for the 800 the 1000 we have 19 0.28 Okay, so last thing we have to do is calculate are critical value, because that's how we're going to decide which pairs air. Significant. So we're just going to take our critical value from the Unova test and multiply it by our degrees of freedom between groups. So are critical. Value is 5.21 Okay, so let's go ahead and go back to our white boards so you can copy these numbers down, make it easy for us. Teoh reject or failed to reject, so to decide whether we have a significant para not We're gonna take the critical. I'm sorry. We're going to take the test statistic of that pair, and if it is larger than our critical value. Then we can reject our no hypothesis. So that means the pair will be significant. So is 2.91 larger than 5.21 No. Is 8.40? Yes. Is 19.28? Yes. So we have to significant pairs at the 0.1 council level. So let's go and summarize our results. So there is fish in evidence. Uh, the 0.1 for love will to conclude there is a difference in me and cost between between 908 100 watt of ends and 801,000 walk ovens. All right, that's it for this video. I hope you guys learned a lot Chelsea next time.


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