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Date: 05/23/19 Time: 15.35 Sample: 1980M01 2019M02 Included observations: 470AutocorrelationPartial CorrelationACPAC 0-Stat Prob0.956 0.956 431.92 0.000 0.908 -0.05...

Question

Date: 05/23/19 Time: 15.35 Sample: 1980M01 2019M02 Included observations: 470AutocorrelationPartial CorrelationACPAC 0-Stat Prob0.956 0.956 431.92 0.000 0.908 -0.055 823.05 0.000 3 0.862 -0.012 1176.2 0.000 0.810 -0.099 1488.3 0.0o0 5 0.750 -0.111 1756.5 0.000 6 0.691 -0.015 1985.0 0.000 7 0.638 0.031 2180.3 0.000 0.589 0.018 2346.8 0.000 9 0.535 -0.078 2484.7 0.000 0.479 -0.075 2595.3 0.000 0.430 0.035 2684.7 0.000 0.387 0.031 2757.2 0.000 0.341 ~0.042 2813.8 0.000 0.305 0.069 2858.9 0.000 0.2

Date: 05/23/19 Time: 15.35 Sample: 1980M01 2019M02 Included observations: 470 Autocorrelation Partial Correlation AC PAC 0-Stat Prob 0.956 0.956 431.92 0.000 0.908 -0.055 823.05 0.000 3 0.862 -0.012 1176.2 0.000 0.810 -0.099 1488.3 0.0o0 5 0.750 -0.111 1756.5 0.000 6 0.691 -0.015 1985.0 0.000 7 0.638 0.031 2180.3 0.000 0.589 0.018 2346.8 0.000 9 0.535 -0.078 2484.7 0.000 0.479 -0.075 2595.3 0.000 0.430 0.035 2684.7 0.000 0.387 0.031 2757.2 0.000 0.341 ~0.042 2813.8 0.000 0.305 0.069 2858.9 0.000 0.276 0.030 2896.0 0.000 0.250 0.007 2926.6 0.000 0.226 -0.017 2951.7 0.000 18 0.201 ~0.049 2971_ 0.000 19 0.175 -0.043 2986.4 0.000 20 0.151 -0.005 2997.7 0.000



Answers

Stock Prices A random sample of stock prices per share (in dollars) is shown. Find the 90 $\%$ confidence interval for the variance and standard deviation for the prices. Assume the variable is normally distributed. $$ \begin{array}{cccc}{26.69} & {13.88} & {28.37} & {12.00} \\ {75.37} & {7.50} & {47.50} & {43.00} \\ {3.81} & {53.81} & {13.62} & {45.12} \\ {6.94} & {28.25} & {28.00} & {60.50} \\ {40.25} & {10.87} & {46.12} & {14.75}\end{array} $$

Hello. Hi. Here In this question, we have been given a question from Atkins weight loss program. Right. The question is given like there is a weight loss program. Is there on the number of people use these 40 people has been used on after toll month. They're mean weight loss is found 2.1 l b. On question is asking whether this conference in trouble given in the question information, it will be effective for the date. That is the question. Okay, so basically, we need to find the confidence in trouble. Estimate for standard deviation. So to do that, we know how to find confidence. Interval estimate first and activation. Right. So we're going to find what are the points given here? First we know number off samples given it is 40 Onda. We can find the degrees of freedom for that we can use and my instrument. So we get 39 on when it check the table. We know there is no 39 in the chi square table, so we know 39 is greater than 30. So in the table, using the road 30. Okay. In order to get the value of K square and to find ice core value, we need to know, See value. See values Given in the question. 90% Asian 90% means that you can write a 0.9 on alpha value. We can find out one minus. C one minus t will be equal to one minus point night you get 0.10 Okay, now first guy square value that is left tail critical. Value will be given by Chi Square one minus sigma by two That is equal. Do Chi Square one minus Seema Baidu Alfa by Toby Help Founder That is equal to 0.1. So one minus signal by two you'll be getting one minus 10.5. You'll be getting 0.95 years. Europe lined 95 alphabet. Right? So from the table we get this values ical do 18 point 493 Similarly right tailed critical value Chi square Alfa by two that is equal Do chi square 0.5 From the table. We get that values equal to 43.773 So once you get guys core values next step we can determine the conference in trouble. Estimate right. We know the formula is given by wrote off and minus one by chi Square stigma by two times standard deviation less than Sigma Less than against Kyle Rudolph and minus one bye bye square one minus Alfa by two one minus Alfa by two times standard deviation. So we get we have to substitute all these values here we know and minus one is 39 so 39 bite 43.773 We found out into standard deviation value. We know that is equal toe 4.8 that is given in the question is the sequel to 4.8 less than Sigma Less than here also 39 by keister one minus alphabet to be found out 18.493 times standard deviation. So we just started toe Simplify this value standard deviation value we know which is 4.8 er So when you take a multiply here and simplify you get This value is equal to 4.53 08 less than Sigma and less than 6.9706 This big conference in trouble estimate four standard deviation we got. Now the question is asking us whether this confidence in trouble used the information about affect, you know, subject it will give affecting us out diet because the value we're getting is 4.53 to 6.9. Is there the required value? The given value is what? Standard divisions? 4.8. So this conference interval estimate is enough for affecting us off. That that will be the correct answer. I hope this answer your question. Thank you.

Part one. The regression for a are one serial correlation years row hat of minus 0.110 and the T statistic is minus 0.63 The estimate of row therefore, is small and statistically insignificant. So a are one serial correlation does not appear to be a problem. Mhm part two Theo O L S N P W Estimates of beta one are both 10.152 rounded to three digits. Given that part, one indicates little problem of serial correlation. We expect that they would be very similar.

All right. So we're looking at the normal probability plot of the big executives and $350,000 a year, which is a little bit more than I make and probably what my boss makes. But maybe my boss's boss, but this is in Arizona and they make a lot of money. We're looking for a nice linear line and we see it. My goodness. That is that is a prime candidate for the T. Procedure. We got also about like what? Four times five we got 20 such executives. Um I think that's pretty reasonable sample size. Especially considering that there won't be so many executives to sample from. Like you could probably take a look at every public company in Arizona. But for this one, you're just looking at the one company. So 28 executives is a perfectly fine sample. Our line is nice and straight. So good. We have the go ahead for this.

Mm. Just look at that line. Nice and straight. There's no outliers. No funny business going on here. We got all of our data and none nice straight linear line right here. No outliers. We've got a decent sample size of 20 executives. Which is reasonable because there's not gonna be 200 of them to choose from. There's just 20, maybe 30 if you can't like the ceos and all the other guys up top. I don't know how business works but 20 it's pretty reasonable seeing movies. There's like a table right here. This you know the guy that sits there and then a couple of guys there there there there there there so Perfectly reasonable to have a seat of 20. And right here, nice straight line. Once again, beautiful, definitely a green go ahead for this T procedure. Prime candidate.


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