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QUESTION 12HolenncaneTat Tured nunior Lnani Icccm Innnsis 0l dati for the 50 U5 staies 0n sate Imcz at or bolow th pover curl) vlelded the {ezrestlon & ualion I...

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QUESTION 12HolenncaneTat Tured nunior Lnani Icccm Innnsis 0l dati for the 50 U5 staies 0n sate Imcz at or bolow th pover curl) vlelded the {ezrestlon & ualion Icercenl olpconic inunteDuloc haly anok nove ny(aleJ = 209.9+25.5xInterrcltne soptIncrease In Ine numbe? of fcople Lizat Dr below theavctace 235 + more violent crittes per IOO,0IO peoplc in Ihe sntetor @veni0a There would / povetty [Evel Wneanumoc peoplc Fing &1 bertue crimes per 100, O0 people in Lhe state tor very 1t increase The

QUESTION 12 HolenncaneTat Tured nunior Lnani Icccm Innnsis 0l dati for the 50 U5 staies 0n sate Imcz at or bolow th pover curl) vlelded the {ezrestlon & ualion Icercenl olpconic inunte Dulo c haly anok nove ny(ale J = 209.9+25.5x Interrcltne sopt Increase In Ine numbe? of fcople Lizat Dr below the avctace 235 + more violent crittes per IOO,0IO peoplc in Ihe sntetor @veni 0a There would / povetty [Evel Wneanumoc peoplc Fing &1 bertue crimes per 100, O0 people in Lhe state tor very 1t increase There would De Jverage 25 5 more povcrty Iev2 Erery additiarial I030 people Inving Or bacn tra poverty levrl mora Vlolent crinesrr Jon ponceopl Rhc €al Tnemekomd average 709 Increuse Kthenumbtrol numnnei vioient cnunes Dcr I00 WO0 people Intnsale Ior @ttny There would 0.255 percent increase In the wverage peop # Iiving at or be Dxthe poverty level for exth addtonal 1c4,00D Ecale livng Jt or beluw tne povetty more wolent crnes [hcatle Voud be On average 25



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Please do the following. (a) Draw a scatter diagram displaying the data. (b) Verify the given sums $\Sigma x, \Sigma y, \Sigma x^{2}, \Sigma y^{2},$ and $\Sigma x y$ and the value of the sample correlation coefficient $r$. (c) Find $\bar{x}, \bar{y}, a,$ and $b .$ Then find the equation of the least-squares line $\hat{y}=a+b x$. (d) Graph the least-squares line on your scatter diagram. Be sure to use the point $(\bar{x}, \bar{y})$ as one of the points on the line. (e) Interpretation Find the value of the coefficient of determination $r^{2} .$ What percentage of the variation in $y$ can be explained by the corresponding variation in $x$ and the least-squares line? What percentage is unexplained? Answers may vary slightly due to rounding. Does prison really deter violent crime? Let $x$ represent percent change in the rate of violent crime and $y$ represent percent change in the rate of imprisonment in the general U.S. population. For 7 recent years, the following data have been obtained (Source: The Crime Drop in America, edited by Blumstein and Wallman, Cambridge University Press). $$\begin{array}{c|ccccccc}\hline x & 6.1 & 5.7 & 3.9 & 5.2 & 6.2 & 6.5 & 11.1 \\\hline y & -1.4 & -4.1 & -7.0 & -4.0 & 3.6 & -0.1 & -4.4 \\\hline\end{array}$$ Complete parts (a) through (e), given $\Sigma x=44.7, \Sigma y=-17.4, \Sigma x^{2}=315.85, \Sigma y^{2}=116.1$ $\Sigma x y=-107.18,$ and $r \approx 0.084$. (f) Critical Thinking Considering the values of $r$ and $r^{2}$, does it make sense to use the least-squares line for prediction? Explain.

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.

We are given a sample of data points listed at the top of this document and using that data, we want to answer the following six questions they threw. Steph, we'll proceed together one by one as follows. First, in part A We want to produce a scatter plot of our data points. I've already included the scatter plot right below part AM left with the points X Y are marked with black Xs or crosses next directly to the right. We want to compute for part B. The sum is relevant to this data as well as the Pearson correlation coefficient. R I've already included the values of the sums since they are found simply by taking the exact formulas for those sums of some X is some of the X values and so on, correlation coefficient. R is given by the following formula which takes as input our sample size and and the someone who just calculated Putting those in, we get our equals .084 next in parts. Do you want to find the line of best fit for this? Do that To do so? We have to identify the following parameters. 1st. First are excelling and winning X bar and wine bar which are simply with some of our X and Y values about it. I am. We can use these now to find the slope, be an intercept A. Of the line of best fit first and left. The slope B is given by the formula here, which takes us input and and the sums again. This is very similar to the correlation coefficient. R formula From this. We get b equals .1-9. Next for a wee plug in ry bar expire and be to get intercept equals negative 1.66 or equation for line of best fit, Y equals negative. 1.66 plus 1.29 x. Next report. Do you want to return to the scatter plot and plot? Ry by that we just produced or rather our white hat doing so. And making sure we include our X. M and Y me produces the following plot. Finally, party we want to calculate a correlation or rather a coefficient of determination R squared and interpret the coefficient of determination. R squared is simply the square of our correlation coefficient. So .007. We take this to mean that .7 of our data or rather our variation in our data can be explained by the corresponding variation and x and the least squares line. However, that means that 99.3 of our variation cannot be explained by this. Finally important if we want to answer considering our value for R and R squared, should we trust why hot making predictions for why? Based on X? No, we should not, because R and R squared are so small that we aren't really seeing a very strong relationship between these variables.

The following is a solution to number 15 goodness of fit test. And this is about Benford's law. So the first the distribution of first non zero digits in accounting files apparently follows a certain distribution. I didn't know that, but It follows this distribution here. And it's Benford's distribution, I guess. So here are the percent, so 30.1 Have, you know, a certain number and then 17.6%, there's a certain number and so on. So here the observed data values for this certain accounting file, their 275 total. And then here the expected. So I just took the Binford numbers here and then multiplied by the sample size to get the expected numbers. And if you look at these, they look pretty darn close. So I think this is going to be a pretty good match. But let's go and answer this with some technology. And first off, we need to find well, the significance level. That's point no one. All right. So they tell you the 1% significant level. So Alpha's 10.1 And then we need to write the alternative the Nolan alternative hypothesis. The null is always that these two distributions would match or they're the same distribution. The alternative is that they're not the same distribution. So in this case, I'm gonna change colours to right and up a little bit the distribution of first non zero digits. And this accounting file follows Ben for God's law. Okay. And then the alternative is that it does not follow Bedford's law. Okay, The chi square value, we'll get to that in a second. We're gonna use technology to find it. Especially with that many data values. I don't really want to use the formula that many times. Uh We need to make sure that the expected values are greater than five and they are. So if you go back here, you know, the smallest expected value here is 12.65 So we definitely have conditions of inference have been met, so we can use the chi square distribution. This is the next part of the question with eight degrees of freedom. So the reason why it's eight degrees of freedom is you take the number of categories minus one, So non zero means one through nine, there are nine here, 9 -1 is eight, so there are 8° of freedom. Okay, so that's the setup. Now, we can turn to our calculator. Now I use T I. T four, I like the T I T four for smaller data sets, but um if you want to use Excel or are or mini tab s. P s s ass, you know, whichever you prefer, you're certainly welcome to do that. So and edit if you go to stat, edit here. I've already done the work but L one is where you're gonna put the observed values and then L two is we're gonna put the expected now you don't have to put them in those but you'll need to change it where it is later on. But um I just put it now one and L two I think it's easier this way and if you like to stat and air over to test and it's the chi square G O. F. Test that stands for goodness of fit. So here is where I put the observed was the one and expected was L. Two. But again if you have a different option in L two L three for example you just change that to L. Two and L three degrees of freedom, remember was eight. And then when we calculate that that gives us everything we need. So the chi square value is about 3.56 let's say. So the Chi Square is 3.56 which is pretty small, especially if that many degrees of freedom. And that's verified with that. A. P. Value. It's a pretty big P. Value of 89 I mean that's that's a pretty darn good P. Value. So 89 is the P. Value which is definitely greater than our alpha of one. Which means we fail to reject. H not meaning that this accounting file does in fact follow Benford's law. So that's how we're going to kind of summarize this. In the last part. We can say there is not sufficient evidence To suggest that the first non-0 digits in this accounting file does not follow Ben for God's law. So if you double negatives in there, I'm not a big fan. So you could actually just say that The first non dear zero digits in this accounting file does in fact follow Benford's law.

This is problem number 30. We are given a set of data for the poverty level in each state and the District of Columbia. First, we will construct a frequency distribution given that are lower bound is five and we have a class width of one, so we have our percent and frequency. So we will have 18 classes or 13 classes 56, 7, 8, 9, 10, 11, 12 and 13 for the integers. And they go to 5 to 5.96 to 6.9. And so so for our frequencies we get the following 11 three seven nine six. An attorney. Mhm Five three five. And then for over here, we'll have our second set to just make sure we can fairly again. Mhm. 14 15, 16, 17 and 18. And we'll have 412 three one. So now finding our relative frequency, we have 2.0, 2.0 5.9, 13.7 17 6, 11.8, 9.8 5.9 9.8 7.8 2.0 3.9 Who, What is that? 3.9 5.9 And 2.0. So if you turn this into a frequency distribution, you will find that the shape of the graph is something similar to this. They small hump over here. So we can say that the graph is skewed right, as the data is shifted over to the left. If we are to change our class size to two, we would find that the data becomes more normally distributed. They're still slightly skewed to the left and we'll take this sort of shape with a bit more of a tail on this side compared to a normal distribution, there's still fairly centered.


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