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Tha P-value for a hypothosia jesl m shawn Uzn tho ftn 0.0481deoida #hathet rejct Ho whon the Iovel 0t sigriticarco (816=00' (b) 4 =0.05,and (c}a #0,t0_(3) Doy...

Question

Tha P-value for a hypothosia jesl m shawn Uzn tho ftn 0.0481deoida #hathet rejct Ho whon the Iovel 0t sigriticarco (816=00' (b) 4 =0.05,and (c}a #0,t0_(3) Doyou rejoct or fa | reject Ho at the 0.01 luvel of significanco?Fail = rejed Ho because ~vBlue 0461 than 020,01. Roject Hp becausa the P-va uo 0461. grualer Ihan 0.01 Rojoct Ho bacauso Ihd P ~luo ,0461, Josa Uun 0.01, relect Ha bacusa tha P-value, 0,0461 _ qieator Ihjn 0,01(0) Do yau reroctrojoct Ha 0t tno 05 lavel Hanlficanco?Rojort Ho

Tha P-value for a hypothosia jesl m shawn Uzn tho ftn 0.0481 deoida #hathet rejct Ho whon the Iovel 0t sigriticarco (816=00' (b) 4 =0.05,and (c}a #0,t0_ (3) Doyou rejoct or fa | reject Ho at the 0.01 luvel of significanco? Fail = rejed Ho because ~vBlue 0461 than 020,01. Roject Hp becausa the P-va uo 0461. grualer Ihan 0.01 Rojoct Ho bacauso Ihd P ~luo ,0461, Josa Uun 0.01, relect Ha bacusa tha P-value, 0,0461 _ qieator Ihjn 0,01 (0) Do yau reroct rojoct Ha 0t tno 05 lavel Hanlficanco? Rojort Ho bocaust Iha ~vnluo: 0461 Iass (han = 0 05 Rojuct Ho bucause Ihe IQvo 0461 Aryalot Wan Fnll = rujoc Ho bacdusa Iha Pvalua QO:6i Dfoutar Inan 0,05, Fail ! rojuc: Ho bocausu P-vlue 0461, Iona than 005. (c) Do you Iejuct = rojecl #o ot tho = lovol 0f elgiificance? ReJect Ha acau Ina P-valuo 0461 , Ihon 0.10 Foil rujeci Ho bocause Ihe ~value, 0.0461 greator Inan u = rejeci Ho because tha P-value, 0.0461_ I25s than 0.10, Fall E 0461, gruater tnan 0.10, Reject Ho becuuse the P-value



Answers

(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.

As we're looking at the mean price of agricultural books at the 10% significance level. The mean is supposed to be 8.45 So that's gonna be our hypothesis. $8.45 to be more specific. But that's it. And then we want to find out if there is a difference so that is a does not equal to tail test your hypothesis. And our green is gonna give us our calculated chi square statistics. So it's gonna be 28 minus one. So 27 divided by the deviation squared. So 8.45 mhm squared times 9.29 squared. And that is going to get us a value of put into my calculator because divided by 8.45 squared. And we get 32.63 Yeah. Yeah. And that's our chi square graph right? There does not equal means we chop it off here and here for a good dose. If we land in the shade of reach right here, we will rejecting all hypothesis at the 10% significance level. That means we're looking at half 5% on both ends. So we're gonna be looking at 5% for the right end and 95% for the left end at a degree of freedom of 27. So we're looking at 27 here and 5% is going to get us 40.113 Mhm. And for the left side, gonna get us at the 95% level. Yeah, and that's gonna be 29 degrees of freedom. So 17.7 all eight 32 is smack dab in the middle there. So we're good. We fail to reject the hypothesis and that is all that was required of us. Um mm The standard deviation is indeed 8045 cents.

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.


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