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(a) identify the claim and state $H_{0}$ and $H_{a},(b)$ find the critical value and identify the rejection region, $(c)$ find the test statistic $F,(d)$ decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, the populations are normally distributed, and the population variances are equal. If convenient, use technology. The table shows the salaries of a sample of individuals from six large metropolitan areas. At $\alpha=0.05,$ can you conclude that the mean salary is different in at least one of the areas? (Adapted from U.S. Bureau of Economic Analysis) $$\begin{array}{|l|l|l|l|c|l|}\hline \text { Chicago } & \text { Dallas } & \text { Miami } & \text { Denver } & \text { San Diego } & \text { Seattle } \\\hline 43,581 & 36,524 & 49,357 & 37,790 & 48,370 & 57,678 \\37,731 & 33,709 & 53,207 & 38,970 & 45,470 & 48,043 \\46,831 & 40,209 & 40,557 & 42,990 & 43,920 & 45,943 \\53,031 & 51,704 & 52,357 & 46,290 & 54,670 & 52,543 \\52,551 & 40,909 & 44,907 & 49,565 & 41,770 & 57,418 \\42,131 & 53,259 & 48,757 & 40,390 & & \\& 47,269 & 53,557 & & & \\\hline\end{array}$$

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.

Okay, so we are conducting a one tale test at the 1% significance level Arnold hypothesis is that they're the same. And the alternative is that the first is less than the second the computer have statistic is going to be s one squared divided by S two squared. So this turns out to be approximately 0.17 Our degrees of freedom are one less than each of the sample size is making 18 the numerator and 50 the denominator, We can go to the back of the book in table eight to figure out what are critical value is. Since this is a left tail test, our f statistic is going to be f of one minus point to one, which is going to be equal to F of 099 which has a statistic of 2.78 Of course we take the reciprocal of that. This turns out to be approximately 0.36 So for illustration purposes, we have our F curve, whatever it looks like, and 0.36 is going to be somewhere right here, and 0.17 is going to be clearly on this side in the rejection region. Therefore, we reject the null hypothesis at the 1% significance level.

In question six were given information about oxides of nitrogen and then we're given information about one light truck model in the, uh the amount of emissions given off tells us that the mean is unknown and the centre deviation is 0.4 gram per mile. We test a simple random sample of 50 trucks, and we get a mean knocks level that estimate siendo mu. And then we get different X bars for each repeated sampling and a it says describe the shape, center and spread of the sampling distribution of X bar. On this case, um, we know that X bar is approximately normal. This is true for the sampling distribution because we have a large enough sample size with a sample size of 50. The center, which we're gonna right as mu of X bar, is going to be the exact same thing as mu. We don't really know what that is, but we do know that the center of the central distribution will match the true population mean and then the spread is gonna be the centre deviation here. We know that the sample I'm sorry, the sampling distribution is going to be represented by standard deviation of export, which is gonna be the population of center deviation divided by route in. And in our case, we do know the population standard deviation is 0.4. We have a in A 50 and this gives us a standard deviation of 0.566 and be were asked to sketch the sampling distribution of export. And we're gonna mark the mean and the values 12 and three standard deviations on either side of the mean. So in this case, we don't actually know what I mean is, But we do know that it's the same as the population. So we're gonna market as you and then on each side we're gonna add one center deviation than two Standard deviations three. So we're going to do this as mu plus one centered aviation, which was from above +1057 Approximately the next one will be mu plus. Two of those were gonna just multiply that, get two of those values that's 20.114 and then for the third value mu plus point 171 which is three center deviations. We're also going to the to the left of the mean. So we have u minus one senate deviation. With those points you rough up. Seven U minus two Senate deviations, which was 20.11 four and then lastly, mu minus three. Center deviations, which was 30.171 in SI. It says, According to the 68 95 99 7 rule, about 95% of values of X bar lie within the distance film of the mean of the sampling distribution. Well, the 95% refers to that second center deviation on either side. Mark this in our picture above, we have to senator deviations above the mean here to center deviations below the mean here. And we have this area that represents approximately 95% of the curve. So that value of em that it's looking for is that to center deviation mark on either side. So our M value in this case is approximately 2.114 and then we had that value shaded above. Then lastly, in D, it says whenever x bar falls in the region, you shaded the unknown population. Mean mu lies in the confidence interval. X bar plus reminded him for what percent of all possible samples. Does the interval capture you? Because our confidence interval was created using are 95% critical value. What we're gonna find here is that about 95% of all the intervals created will capture the truth. Um, parameter, that we're looking for in this case.