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Describes the following study: Arecent article titled "Action Movies Make You Eat More students at Cornell University to watch either Rescarchers randomly assi...

Question

Describes the following study: Arecent article titled "Action Movies Make You Eat More students at Cornell University to watch either Rescarchers randomly assigned 64 undergraduate Bay movie "The Island' clip showing an action sequence from the 2005 Michael 20 minute participants were provided with MCMs, cookies minute clip from the Charlie Rose talk show: The the snacks snack on while watching the shows The researchers weighed carrots and grapes to how much each person had eaten

describes the following study: Arecent article titled "Action Movies Make You Eat More students at Cornell University to watch either Rescarchers randomly assigned 64 undergraduate Bay movie "The Island' clip showing an action sequence from the 2005 Michael 20 minute participants were provided with MCMs, cookies minute clip from the Charlie Rose talk show: The the snacks snack on while watching the shows The researchers weighed carrots and grapes to how much each person had eaten before and after the students watched the clips and determined The following table summarizes the results measured in grams; The Island Charlie Roset 206.5 Sample mean 104.3 45.2 Sample SD Sample size You can assume the distribution of the amounts eaten for each group is roughly symmetric with no apparent outllers You will first compute the p-value below: Then; for each of the following issues, write to 3 clearly worded sentences summarizing; in the context of this problem; the appropriate conclusion regarding the issue; and why you can make that conclusion based on the information pro vided: Be sure you address each issue separately and only discuss the information /concepts/numerical values that are most pertinent to each individual issue When necessary; justify your answers with appropriate numbers and show any work in your calculations: If you cannot compute appropriate values,make up reasonable values to use in your explanation:



Answers

Filling Bottles A certain brand of apple juice is supposed "to have 64 ounces of juice. Because the punishment for underfilling bottles is severe, the target mean amount of juice is 64.05 ounces. However, the filling machine is not precise, and the exact amount of juice varies from bottle to bottle. The quality-control manager wishes to verify that the mean amount of juice in each bottle is 64.05 ounces so that she can be sure that the machine is not over- or underfilling. She randomly samples 22 bottles of juice and measures the content and obtains the following data: $$\begin{array}{llllll} 64.05 & 64.05 & 64.03 & 63.97 & 63.95 & 64.02 \\ \hline 64.01 & 63.99 & 64.00 & 64.01 & 64.06 & 63.94 \\ \hline 63.98 & 64.05 & 63.95 & 64.01 & 64.08 & 64.01 \\ \hline 63.95 & 63.97 & 64.10 & 63.98 & & \end{array}$$ (a) Because the sample size is small, she must verify that the amount of juice is normally distributed and the sample does not contain any outliers. The normal probability plot and boxplot are shown. Are the conditions for testing the hypothesis satisfied? (b) Should the assembly line be shut down so that the machine can be recalibrated? Assume $\sigma=0.06$ ounces and use a 0.01 level of significance. (c) Explain why a level of significance of $\alpha=0.01$ might be more reasonable than $\alpha=0.1 .$ [Hint: Consider the consequences of incorrectly rejecting the null hypothesis.] FIGURE CANT COPY

All right. So in this question, we want to compare the empirical formula to actual data and then we're going to look at the natural history am and decide whether the empirical rule is actually useful in this case. But for part a right you the empirical rule to estimate the percentage is Well, the empirical rule tells us that within 12 and three standard deviations of the main, respectively, You are going to have a party, you're going to have 68%,, 95 in 99.7 of the observations. And this is respective 2, 1, 2, three standard deviations from the mean. All right. So that is our empirical formula estimation. And now what we're going to do, we're going to take this data and we're going to see what percentage of the actual data fall within 12 and three standard deviations of the mean. So what I've got here, I've got a number line which is centered at 45.3, so it's centered at our mean, and we're just going to determine essentially these buckets. Okay, we're going to determine these boundaries for 12 and three standard deviations. So in order to do that, all we have to do is add Well. So, for example, for one standard deviation above the mean, all we're going to do is add 45.3, we're going to add one standard deviation above the mean. So that's 4.16. Now, when we do that, we're going to get 49 0.46 So this is 49.46 and we can keep right, we can keep doing this. This one is going to be 49.46 plus 4.16 49.46 plus 4.16 Right, And that's going to give us 53.62, 53.62. And if we do that once more, that's going to give us a third boundary. Just 57.78 57.78. And likewise, on the left hand side, we could do this by subtracting the standard deviations and right. Why am I using, Why am I using the standard deviations as these buckets? Well, we want to compare it to the empirical rule which uses the standard deviations as buckets as bench works essentially. All right. So 45.3 45 .3 -4.16 gives us 41.14. And then if you subtract 4.16 again, um That gives a 36.98 and once more To get the three standard deviations below the mean gives us 32.2. All right. Now, our data here are organized from flow, like the biggest risk is quite nice, um which gives us the ability to uh drop them in various buckets. So 36.3 is going to be in this bucket, 37.7 in this bucket, 38 in this bucket. Uh 38.8 38.9, 39 39.3. Right, everything up to um 41.1 39.3. So we have 40.9, we have 41.1 Then 41.3 is in this bucket. 41.5 41.8, 40 to 40 to 42.1 42.5. Uh 42.5 again, 42.8. Just to be clear, you could also um simply, you know, mark them off as you go. So we've done all of this first row, we've done all of the second row now. 42.9 43.3 43.4 43,45, 44 44.4. There's a lot of here and there at 44.7 44.8, 45.2 45.2 45.2. And then, right, We have done all of those, okay? And then the rest of them. So 45.4, what I'm actually gonna do now, I'm just going to count. I'm literally going to count in our table. Um up to 49.6 so 49.6 happens right over here. So all of these guys here, 1 2 3 4567, uh plus nine, seven plus nine. So at 16 plus another nine, That's 25. Fall within here. Um and then between 49.46 and 53.62, 53.62, that's going to be here. So then there's four and then to Sorry about that, then there's four and then to and I suppose we could have also done this for these guys. So this guy would have been one um 12345678 This guy would have been eight and 123456789 10 11 12 13 14 15 16 17 18 1920. So then you have 20 here. All right, Yeah. Um so let's compare. All right, so how many are within one standard deviation? While 20 plus 25 That's that's equal to 45. And how is that as a percentage? Well, how many data do we have in total? 123 So this is nine rows 1234569 times six is 54 plus 2, +3456 So they're 60 in total. Okay, so 45 out of 60 so 45 out of 60 Gives us our uh percentage or proportion rather. So 45 out of 60 is 75%. The 75 are within one standard deviation. Alright, Now, let's look at within two standard deviations. So within two standard deviations, you're gonna have eight plus 20 plus 25 plus four, so eight plus um 20 plus 25 plus four. So we're just adding essentially 12 to the previous ones. That's gonna be 57. So then 57 out of the 60 total observations are going to be within Um two standard deviations, which is 95 Actually matches quite nicely, so this is 95%. Within two standard deviations, and finally, how many are within three standard deviations? Well, in this case, actually, all of the data notice all of the data fall within three standard deviations of The mean, so 100 of data of observations, all within three standard deviations. Alright. Um so actually when we when we put these together, so 75 95 100 that's not too far off from the empirical world. Alright, And the empirical rule, of course, is using a perfectly smooth mathematical model, and if we were to look at the actual data in a history RAM and you can do that by flipping to page 76 looking at figure 2.100 um it's not perfectly smooth, right? It's it's a bit jagged, but it is pretty symmetrical, it's pretty bell shaped and for this reason um in conjunction with the data that we've discovered, I would argue that the empirical rule is quite useful here because the data is pretty bell shape.

It's a kind of an interesting question if we we would be assuming that the anxiety level if we are going from easy to difficult is equal to the anxiety level. If the questions were difficult to easy and alternately, we just want to know if there's a difference and easy to difficult and not equal to difficult to easy. And so we're going to assume that that difference is actually zero so that the, I'll just write E G minus D. E. That that difference is zero. And when I put this into my calculator, um we have the lists have different sample sizes. We have that first group of the E. T. D. I have, there are 25 numbers versus the D two E. There are 16 numbers and we could use the degrees of freedom of 15, but I'm going to use the degrees of freedom from the formula. And that degrees of freedom ends up being uh 30 almost 39. So 38 point well, I'm just going to call it 39 it's approximately 39 degrees of freedom. And that test statistic that we're going to get, we need to take the mean, which was 27.1152 minus the mean of the other group, which was 31.7 to 8125 And then divided by the square root of. And the first standard deviation all around that a bit is 6.857 one square, divided by the sample size, which was 25. And then the second standard deviation was 4.26 square divided by the sample size of 16. And when I got that test statistic, the test statistic came out to be, I'll just read it up here negative 2.6 566 And so it came out down here. And since we're doing a two tailed test, we also use this one. And so what's the likelihood of getting a test statistic in this distribution that is less than or equal to that negative 2.6566 That gives us this tale and then we want the other tales to double it. And that p value comes out to be a 0.114 So at a 5% significant level, this is smaller than 5%. So we would have sufficient evidence to reject now. Yeah. And say that the mean anxieties are different so that they're different. The means are different. However, at a 1% significance level we would fail to reject the novel. Mhm. And we'd have to say here that they're actually not. They appeared to be the same. So it does depend on our significance level. How did pick you want to be. But it is an interesting idea in all my years of teaching, I never thought about the arrangement of having them all go easy, too difficult or difficult to easy. So what interesting concept.

Soon. Number 28. What's your name? The level of significance is mentioned is all for equal 0.1 on our hypothesis, which state that the population mean new is equal to the value mentioned in the claim, so it would two new equal 4.55 g. The alternative hypothesis stated the opposite off the novel hypothesis according to the clean, thus using lesson. So each one to mu is less than 4.55. Dreams is alternative hype with his old listen. Then the test is left. 15 is the alternative. Hypotheses uses Biggers in, so the desk is right field. The alternative hypothesis uses non equal, so the test is toe field, so the answer will be left field question. Number be given export equal 3.75 being equal 4.55 Zita equal 0.7 and unequal six. The sampling distribution off the sample mean export is normal because the population distribution X is assumed to be normal. The sampling distribution off the sample mean has mean new and stirred and standard deviation. Zita over square root and does it? Value is a sample mean decreased boy. The population mean divided by the standard division so that equal explore minus new over Zita over square root in equal 3.75 minus 4.55. Mhm 0.7 over square root six equal. Minus 2.8. Question number C Riddle Sport A and B. It snowed too. Mu equal 4.55 g each one to mu is less than 4.55 g and did equal minus 2.8. The B value is the population off obtaining a value more extreme or equal to the standard test static Zet that remind the probability using tepidly. So, probability equal probability off export is less than 3.75 Equal probability off. That is less than minus 2.8. Equal 0.26 Question number We? Yeah, given Alpha giving Goma equal 0.1 Result port. See, P equal points here is you 26 If the value is smaller than significance level Alfa, then then I'll Hypothesis is rejected. Mm. Is listening 0.1 So I reject h No. If we rejected them on hypothesis the data is the data easy, statically significant at level Alfa Question number e result, 40 each node to new equal, 4.55 g and each one to mu is listening four point 55 graham result for the project. Each node. There is sufficient evidence to support the claim that the mean, the mean weight off these periods in three sports off the Grand Canyon is less than 4.55 g.


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