14.69 -297 1037 ~0.63 0.15 0.27 4,59 6.38 024 40.32 10.294.16 1107 9.24 267 8.97 187 3 6.87 4698.31206 7.66 10.83 14.488.6313 42 6.546.95Find tne Mutual Fund Return...

(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 sale prices (in thousands of dollars) of a sample of one-family houses in three cities. At $\alpha=0.10$ can you conclude that at least one mean sale price is different from the others? (Adapted from National Association of Realtors) $$\begin{array}{|c|c|c|}\hline \text { Gainesville } & \text { Orlando } & \text { Tampa } \\\hline 139.0 & 169.9 & 184.7 \\111.5 & 127.1 & 69.7 \\156.6 & 111.3 & 165.0 \\152.3 & 113.5 & 157.5 \\214.7 & 133.9 & 103.9 \\172.4 & 160.8 & 120.8 \\52.8 & 179.2 & 88.1 \\170.6 & 70.7 & 168.2 \\140.5 & 89.9 & 59.5 \\186.0 & 99.3 & 170.2 \\139.0 & & \\ \hline\end{array}$$

The following is a nova test based on the mean salaries for different metropolitan areas. So the alternative or the null hypothesis is that all the means are the same. So there are six metropolitan areas, I think it goes Chicago, Dallas Miami, Denver san Diego and Seattle. Uh So the null hypothesis is that all the means are the same. And then the alternative is that at least one of them is different. The second step is to find the critical value and you can do that using either software or a table, But they're essentially three things you need. The first thing is your alpha value, your significance level and that's usually given to you the problem and that's .05. Then you need the degrees of freedom for the numerator and the degrees of freedom for the denominator. And the way you find that Is the degrees of freedom for the numerator is the number of categories -1. So there were six cities that we looked at our metropolitan areas, so 6 -1° of freedom would be five for the numerator. And then for the denominators, the total number of data values minus the number of categories. So there were 36 data values minus the six metropolitan areas. So 30 is your degrees of freedom for the denominator. So that should be enough to use a table. But I use a calculator and I wrote a program in here called inverse. F. I'm not going to show you how to how to write the program. You can youtube it if you wish. Um But this is what I do. So um I put in my area which is my alpha value, my degrees of freedom is five and then my degrees of freedom for the denominator is 30 and That gives me my critical value. About 2.534 2534 is my critical value. I call f. star. So 2.534. Okay so anything greater than 2.534. We reject the annual hypothesis that all the means are the same And anything less than 2.534. We failed to reject meaning the h not is true. Okay so the second step is to find the F statistic and there's a formula but it's a bit of a mess. I always use software you know technology is a great thing. So if you go to stat and you can type in your data values. So these are the mean salaries um So again L1 I think was Chicago and then this is the mean salary for Dallas Miami Denver San Diego and Seattle. So there are six categories. And if you go to stat tests and then we're gonna go to the Unova test and then you just type in your columns separated by commas remember there were six columns, six data columns that we used and we need to make sure that all of them are in there and last one and then also you know make sure you separate those by commons, otherwise it's going to read it wrong. So then um that gives us everything we need. So the F. Is the F statistic, that's the third step. So we're looking at this it's about 2.281 as our F. Value. So two point 281 is our f statistic Which is actually barely in the non rejection region 2.281. So that means we fail to reject. Okay and also we can verify that with this p value here. So the p values 0.7 which is a pretty small p value, but it's still in this case greater than the alpha value. So the alpha value remembers point oh five, so it's barely greater than the alpha value. And whenever it's greater than the alpha value, uh we failed to reject, I should probably put H not there, so we failed to reject H not whenever the P values greater than the alpha. Okay. So then the last step is to summarize everything with actual words. So what does this all mean? It just means that there is not sufficient evidence, there is not sufficient. I guess you could say statistical evidence to suggest that the mean salaries from the different metropolitan areas are different. Okay. And that's the five step process for an Innova one way and over test

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.

Part one. We use the ADF regression with one leg and a time trend. We get the coefficient on the locker wage in area 2 32. A Tam T -1 to be -1056 and the T statistic is minus one point 39 The estimated route is Very close to one .9944. The coefficient on the lack of employment in the same area and same period in the A. D. Every question is slightly positive .0008. The route is estimated to be one and the team study is also very small. We don't care about that. We have a little evidence against a unit root. So we should treat these time series as having unitards are too without a time trend. The regression of delta U. T. Hat on U t minus one, had Delta UT -1 Hat and delta U. T monastery hat. Mhm yield A coefficient on UT -1 hat Of .00008. And so the estimated route is one beauty. Here are the residuals from the co integrating regression. When time is added to the equation, they need show regression. The coefficient on U. T minus one hat Is 0- two. And again there is no evidence for co integration or three when we use the real wish and a time trin the coefficient on new t minus one hat is minus point 044 with a T. Statistic of -3.09. The 10 critical value for the test with the time trend is -3.5. And so we cannot recheck the no hypothesis that the two series lock of wage and lack of employment are not into co integrated lot of wish and lack of employment and lack of real witch. Sure. A factor that is submitted from the co integrating relationship could be labor productivity. Mhm. We could also include other sources of income, say non wage income. This factor can affect the supply of labour.