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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

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To determine how much buoyant force is needed to make the to float. We first and all of the forces acting on it. So we need to determine the boy and force. In this case, if we draw all the forces until on this cylindrical tube we have that only it volume force. And wait are up in on these cylindrical tube.

Then apply Newton's second law. We assume all of these forces. And we obtained that the fully enforced minus the weight is equal to in this case because the tube is in equilibrium. There is no net force update on it. And then we obtain That the Volume four must be equal to the weight of any liquid.

It is of any liquid. It is in this tube. No. To obtain the bowl liam, we must remember that the boy import is defined as D the city of the liquid times the volume displaced by this liquid, by the math and the consideration of gravity. Using the previous resolved we know that the boy is ultimately went and and finally stopping for the volume we obtained that the volume is equal to they weighed.

Mhm density times the acceleration of gravity. A boolean is blazed by acid. We use the way given the information given for the wade which is to your own when Gade Gade I was sent to the minors to mutants and the acid density, which is in this case. And the value of the consideration of gravity. We don't obtain that the volume displaced by as did it 4.7 I stand to the minus sets nature, cute.

No, in the case, uh the undo freezer we have that the same weight because it is the same cylindrical tune bad. The density changes it is this the density of the under freezer and the separation of gravity from these. We obtain a value of fine poin 59. I'm stan to the minus its return as quickly. Then from these we know that the volume is placed by the anti freezer is greater than the volume displaced by acid.

Finally to obtained. They hate they hate the I hate that the cylinder called tube reaches when it is when there's is filled with a deep freezer or acid, calculate this. We must remember that the volume is variant times. Hey, in this case the area of the crow sexual audio of the cylindrical to an age. The hate of that sooner part of it then from our equation of it, the buoyant force is equal to the weight we can obtain.

We know that wrong. We're not that uh, the volume force is equal to density times B. It's the acceleration of gravity is equal to that weighed now we substitute these into these and we obtained and and if we solve for behave, we obtain God Hate is able to the weight both murdered density, conceptual barium and this generation of property. Then we know all these values and or acid. We have the weight which is the same over in the city of acid.

The gross optional area is also given for this problem. And the acceleration of gravity. Then for these, we obtain A Volley of Fine 97 times 10 Today -2 m. On the other hand, or the anti freezer, we have the same weight. And the only thing that changes is the density of the antifreeze ERT, which is The cross sectional area is the same because cylinder Co two is the same and the acceleration of gravity.

These are the best About the of 7.12 I was turned to the -2 features. We can see from here that the herd mark and the tube corresponds to the anti freezer and the lower one corresponds to the mark of the acid..

First one. The test for a unit root in series you rate unemployment rate using the usual dickey fuller test with a constant yeah. And the augmented dickey fuller with two legs of change of unemployment rate. I find that seven both times we are unable to reject the now hypothesis that unemployment rate series is a unit fruit. The legs are not significant.

However, the significance of the legs matters. So the outcome of the unit root test, we will repeat what we have done in part one two series vacancy rate and report the result in part two. I guess similar result. So the rate is a unit root. Well part one and two.

I use package the R. Package A. T. S. A.

And the function is a D. F. Dot test. R. Three.

We assuming that unemployment rate and vacation re rate are both integrated of level one. We test for co integration using the angle grandeur test with no legs. So the step the steps are as follow. We first regress, you read on the rate then we yet the residual and we run the key fuller has on the residual to see whether the residuals our unit root. I find that you're right and we rate Arco integrated at the 5% level.

Yeah Heart Forest. I get the leads and lacks estimator of the change in vacancy rate and I did note that uh CB rates up minus one. This is for the lack and plus one is for the lead. This is a regression result. So the usual centered errors are in green and in round brackets, the robots that Iran's are in blue and in square brackets you can see that the main estimate on vacancy rate is highly significant.

This one is not correct. So the centered errol the usual one for the estimate of the first lack of change in vacancy rate is 164 In all cases except for the estimate of the lead of C. V. Right. The robust standard Iran's are larger than the usual standard errors.

This is usually the case it happens but rare that the robot standard errors are smaller than the usual standard errors. The r square of this regression is 0.77 So for the rate, because the robot standard error is larger than the usual standard error. So we will get a wider confidence interval if we use a robot standard error and for confidence interval you will run this function in our count in and you impose the name of the regression. It was spits all the 95% confidence intervals for all explanatory variables. The default version is the 95% interval.

But because the standard barrel of this estimate is are very close, two versions are very close to each other so the confidence intervals should be roughly equal. Yeah. Last part. What you could say about real business of the claim that you rate and the rate are co integrated. Yeah.

When I run the test and good grandeur, the results are not consistent across alternative types of process. In one case I can reject the notion that the residuals are united and for all the cases I cannot reject. So I conclude that the claim that you rate and be rate our co integrated is not robust..

Following is a solution video to number 24 and this is where we compare to means uh for the soil, water content for field a compared to feel b. And the first part is just to verify that the mean and the standard deviations are in fact these numbers, So the mean for field day is 12.53 and the standard deviation is 2.39 And then the mean for field B was 10.77 with the standard deviation of 2.4, and it says use a calculator something use this T I 84 I went ahead and typed in the means, or sorry, the data values L one and L two, so L one represents fuel day and then L two represents, you'll be and if you go to stat falcon and it's one of our stats and we're going to change that to L one and calculate and that gives us everything we need. So the X bar, is that 12.53 So that's verified. And then we're looking at the s the standard of the sample standard deviation is about 2.39 So for the field A. That is correct.

And then let's just go and double check field be. So we're gonna change it to L. Two now because that's where field B. Is, and then that verifies its 10.77 for the mean, then about 2.40 for the standard deviation. So the first part is done, that's verified.

And then the second part it says conduct not formally, but we're basically just going to conduct a uh uh two sample T. Test with an alpha value of point oh five. And we're gonna go back to the calculator because doing it by hand can be pretty cumbersome. So if you go to stat and then test now, since we already have, we're gonna go to two sample t. Test since we already have the data in there.

I'm just gonna use the data instead of the summary stats. So list one is L. One that's the field A. List to is L. Two.

That's field B. And then these frequencies just keep them the same and then we have the the alternative hypothesis and you kinda have to use your context clues here. But this one actually explicitly states is field A. Is the soil content or water content higher or greater than. So I'm going to change this to greater than So μ one is greater than you to warm you.

A. is greater than YouTube pulled is usually zero or no unless they tell you otherwise. So we can go ahead and calculate and that's gonna give us everything we need. So the T. Value if you want you can put it down there.

But really it's just this P. Value that I want to see. So 00.27 is the p value. So P value equals 0.27 And then we compare that with the alpha value and it's less than the alpha value. So any time it's less than the alpha value.

That means we reject the no hypothesis. So we're rejecting the statement that these means are the same. And we are I guess you could say accepting the fact that these two means are in fact the water content and field A. Is greater than the water content and field be on average. So we can type this out as there is enough evidence to suggest that field A.

Has on average a higher soil content soil. Sorry, water content. Banfield beat. Okay so this field A. Has on average a higher soil water content on field because since we are rejecting that null hypothesis and accepting the alternative hypothesis..

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