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Hypothesie Testing and CI for / from two or more independent samples Question A study Ws conducted t9 explore the offects & ethanol on skcep time. Fifteen rando...

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Hypothesie Testing and CI for / from two or more independent samples Question A study Ws conducted t9 explore the offects & ethanol on skcep time. Fifteen randomked t0 one o three treatments. Treatmnent rats were got only wator (cntrol). Treatmont per kg of body welght, and treatmen 2 got Ig of athanol got 2g/kg: Tuc amount of REM skp In in minutes. 24hr period was recorded, Data are below: Trentment I: 63,54, 69 50, 72 Treatment 2: 45 , 60, 40 , 56 Treatment 3: 31, 40 , 45, 25 , 23 28Make p

Hypothesie Testing and CI for / from two or more independent samples Question A study Ws conducted t9 explore the offects & ethanol on skcep time. Fifteen randomked t0 one o three treatments. Treatmnent rats were got only wator (cntrol). Treatmont per kg of body welght, and treatmen 2 got Ig of athanol got 2g/kg: Tuc amount of REM skp In in minutes. 24hr period was recorded, Data are below: Trentment I: 63,54, 69 50, 72 Treatment 2: 45 , 60, 40 , 56 Treatment 3: 31, 40 , 45, 25 , 23 28 Make preliminary graph o the data_ Why did YOU choose the graph that You did and what does it tell jou? Calculate relevant summary statistics that will be useful in an ANOVA analysa. Create an ANOVA table for the data using the originul or AATI Ilues ahove: (It is usc RIs relevant functions: ) good idea t Evaluate the ANOVA assumptions mmerically and graphically: Was ANOVA appropriate for this data? Based on the ANOVA table mak colclusion the context of the problem Use R to obtain the relovant multiplier and then create 95% CIs for nll pairwise using tbc Tukey method , Do this bv comparisons f meAIL hud ad show Jour #ork: You W usO R to check your antuurs. Summarize your results using letter cdes What do Son conelude? Compar the values of tha Tukey CT uutiplier And that usex] with Bonftroni mljustutent. Explaiu theimplications of choosing Ot' multiple comprison method Owr the other for tho o thia data sL . MuIWi cOmpisons How dors your conclusion change If at AlL; i instead YOu chose to use the Kruskal- Wallts folloued Piirwie comprisons with bonferroni Adjustment? Wlnt test and coucluslors woukl JOu recomend the sclentist use based evnluation of thia Oll Vuutr findings Anl yaur Msumptians & the tests"



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This question asks you to study the so-called Beveridge Curve from the perspective of cointegration analysis. The U.S. monthly data from December 2000 through February 2012 are in BEVERIDGE.RAW.
(i) Test for a unit root in urate using the usual Dickey-Fuller test (with a constant) and the augmented DF with two lags of curate. What do you conclude? Are the lags of curate in the augmented DF test statistically significant? Does it matter to the outcome of the unit root test?
(ii) Repeat part (i) but with the vacancy rate, vrate.
(iii) Assuming that urate and vrate are both I(1), the Beveridge curve,
$$u r a t e_{t}=\alpha+\beta vrate +u_{t}$$
only makes sense if urate and vrate are cointegrated (with cointegrating parameter $\beta<0 )$ . Test for cointegration using the Engle-Granger test with no lags. Are urate and vrate cointegrated at
the 10$\%$ significance level? What about at the 5$\%$ level?
(iv) Obtain the leads and lags estimator with $cvrate_{t}$, $cvrate_{t-1}$ and $cvrate_{t+1}$ as the I(O) explanatory variables added to the equation in part (ii). Obtain the Newey- West standard error for $\hat{\beta}$ using four lags $(\mathrm{so} g=4$ in the notation of Section 12.5$) .$ What is the resulting 95$\%$ confidence interval for $\beta$ How does it compare with the confidence interval that is not robust to serial correlation (or heteroskedasticity)?
(v) Redo the Engle-Granger test but with two lags in the augmented DF regression. What happens? What do you conclude about the robustness of the claim that urate and vrate are cointegrated?

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.

First one. There are 248 families do not want the apples at any price Or two. So distribution is not continuous, There is focal points and rounding for example. Many people report on powder and either 2/3 of a pound or one and a third pounds. This the fact that the distribution of quantity demanded is not continuous violates the underlying assumption of the topic model. Yeah. Which is the Layton error has normal distribution, but we will still explore the Tobin approach in this context. Yeah. It may work better than the linear model for estimating the expected demand function, do you? And Along with Part eight. The estimates from their topic and L. S.. Models are reported in the same table or the tablet model. The price variables ali them are Statistically significant at the one level. The sign over these prize coefficients are in accordance with the demand theory, the own price effect is negative and across price effect is positive. Cross price is the price of this substitute good, which is regular apples part Let's do part 6. 1st part six we will obtain their fitted values and we find that the ranged from 27 8 798 to you. 3.33 at five. The null hypothesis is later one plus beta two equals zero. This is something you can easily test regardless of your statistical package. Yeah. Yeah, you should get a small T statistic about minus point you and a P value of buying eight. So we are unable to reject the North part seven. The squirt correlation between it is E call B. S and it's fitted value is about .04 and that is there are square hard eight. Given the linear model estimates, even this result, we find that the old LS estimate are smaller than the top bit estimate. And in terms of our square oops, you compare the goodness of fit between the two models we look at. There are square mhm. There are squared of the topic. Model is still smaller, slightly smaller than the old LS model and serve. We can conclude that the topic is yeah, no better than the old LS. It doesnt suite the data better. The Last Part, Part nine. The statement is simply incorrect so you could run into a uh counter example. We have valid price effects, but we cannot explain much of the variation in the dependent variable. It's simply difficult to estimate the demand for a fictitious product.

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.


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