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5330;[ 01a30/5 complete)HW Scone: 73.3396,3.67 0t5pe79.7Question Helpsuppord amelo 0l O-rings Wus colaited ard the wal tickrers $ (n inchos) Iich VLin rtd Ennunnren...

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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) $$\begin{array}{|l|l|l|l|c|l|}\hline \text { Chicago } & \text { Dallas } & \text { Miami } & \text { Denver } & \text { San Diego } & \text { Seattle } \\\hline 43,581 & 36,524 & 49,357 & 37,790 & 48,370 & 57,678 \\37,731 & 33,709 & 53,207 & 38,970 & 45,470 & 48,043 \\46,831 & 40,209 & 40,557 & 42,990 & 43,920 & 45,943 \\53,031 & 51,704 & 52,357 & 46,290 & 54,670 & 52,543 \\52,551 & 40,909 & 44,907 & 49,565 & 41,770 & 57,418 \\42,131 & 53,259 & 48,757 & 40,390 & & \\& 47,269 & 53,557 & & & \\\hline\end{array}$$

This is a problem. # 29 We are given a set of data regarding regarding the per capita disposable income for each of the 50 States and District Columbia. First we will construct a frequency distribution Given that the lower bound of our class is 20,000 and our class with our 2500. We will get the following for our income. Mhm. I'll just round both of these classes to make it easier to fit. So we have 20- 25 K. 22.5. Mhm. 22 5-25, 25-27 5 27.5-30 30 to 32.5 32 5-35, 35 to 37.5 37 5 to 40 and finally 40- 42.5. Now these are not inclusive of the bounds. This is obviously this will be like 22499, I just ordered like this to make it easier to ride out for our frequencies. If we count each of these data points, we get the following three, 10, 14, 12, seven, two, two, zero and one. And when we calculate our relative frequency we get 5.9%,, 19.6%,, 27.5%,, 23.5%,, 13.7%,, 3.9%,, 3.9%,, 0% And 2.0%. So as we can see most of our data is in the top part of our data set. And when you construct your hissed a gram, it does seem to follow this so it is very slightly right skewed. If we are then to change our class sizes to be 4000 wide you will find that the data is significantly more skewed left as our relative frequencies will drop to 19.6%,, 43 0.1% 27.5%,, 5.9 And 2% for the last two classes that we would have. So as you can see, this data is significantly more right. Skewed. This set of data with these smaller classes is a much more precise way of representing this data.

Following is the solution in number 14 at one way Innova test. Uh and this is about the mean sales prices for three cities. And the null hypothesis here is that the mean sale prices are the same for these certain houses. And then the alternative is that at least one of them is different. The second step is to find the critical value and you need three pieces of information to find the critical value. One is your alpha, Your significance level, that significance level. In this case, that's usually given to you is 10. They also needed the degrees of freedom for the numerator, which is the number of categories in this case, the number of cities minus one. So there were three cities that we looked at minus one is two, so degrees of freedom for the numerator is two degrees of freedom for the denominator is the total number of data values minus the number of categories. So in this case, if you counted up those data values, there were 31 -3 cities that we looked at. So 31 -3 is 28. So that's what we need. So from there you can use a table or you can use software. I'm gonna use software. So I wrote a program and I called it inverse. F. I'm not going to show you how to write this program. You can youtube it if you wish. But um it makes Makes it easier for me. So the area is the alpha value. So we'll put in .10 for that. Degrees of freedom from the numerator was too. And then degrees of freedom for the denominator was 28. And that's going to spit out my f. star my alpha value and my uh critical value which is about 2.503. Let's call it 2.503 is my F. stars 2.503. So anything greater than 2.503. We're going to reject h not anything less than 2.503. And we're gonna fail to reject the null. The next step is to find the f statistic and you can do that manually but especially with bigger data sets that can be really time consuming. So I went ahead and punch this into stat. Edit. And these are my data values. So these are the I think these are in thousands of dollars but these are the mean sale prices and if you go back to stat and then tests and the very last one in nova And you put in your columns just make sure you separate them. This is on the T. 84 by the way but make sure you separate them by commas otherwise it's not gonna read it right so nova for those three columns and that's gonna give us everything we need to. The f statistic is about 0.966 Let's go and write that down. So 0.9 66 which is somewhere over here. So that lands in the non rejection region. So that's actually gonna tell us why our fourth step which is the decision and we're going to fail to reject H not since the F statistic is less than the critical value. Now you can also use the P value method that's what this second piece of information is good for it. Now this other stuff doesn't really matter. Um You can just kind of ignore it because this is really what we need. We need the F statistic and we need the p value and the p values pretty large. It's about 0.39 And what you do is you explicitly compare the P value with your alpha value. So the p value in this case is greater than your alpha value. 0.39 is bigger than 0.10. And any time you're P values greater than alpha, you failed to reject H nine. If it's less than alpha then you you reject. And then the final step is to conclude this, you know, with actual words and bring it back to the question at hand. And so what we're going to say is that there is not enough evidence or there is not sufficient statistical evidence. So there's not sufficient evidence to suggest that the mean sales price prices of houses In the three cities are different. Okay that's the five step in Nova process.

For the match pairs given on the right. We want to conduct a sign tested matched pairs testing P does not equal 0.5 at alpha equals 0.5 significance. This question is testing our understanding of non parametric tests in particular how to conduct a scientist of matched pairs. We proceed there steps A through D below to solve. So first, in a we stayed alpha hypotheses, this gives alpha equals 0.5 H. And r P does not equal five. H. A P does not equal five. So the no hypothesis P is equal to have, the alternative is not A and B. We compute the test out. So are signs for these matched pairs are as follows and equals 12 is the total number of matched pairs. So X equals the number of plus or the total number plus and minus which is 4/12. Thus we have seen it equals x minus 5/45 over and equals negative 1.155 Thus the p value is for a normal distribution, P equals two PZ greater than zero equals 248 Thus we conclude that we fail to reject asian off because he is greater than alpha, which means that we lack evidence to support the claim P does not equal five.

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.


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