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#2polng MlniroSialg E032HowtAsk Yous TeacharComputers SOre uehice calculale Uannaut qudnttes reldted Detortanc One Inese efficiency; Msually expressem mles gallon (...

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#2polng MlniroSialg E032HowtAsk Yous TeacharComputers SOre uehice calculale Uannaut qudnttes reldted Detortanc One Inese efficiency; Msually expressem mles gallon (mpg). For one vehicle equipped this way; the mpg were recorded eac filled, and thc computer Nas (ncn rscu Hcre the mpg values tor random empic thetc recoros;nuleag8 , the gas tank48.3 50.4 46.8 44.7 43.0 45.0 39.2 44-1 41.4 37.5 37.5 43.! 47.9 474 43.3 46.2 344 42 3 43.1 36.6(a) Describe the distribution using graphical mcthods_ oppro

#2polng MlniroSialg E032 Howt Ask Yous Teachar Computers SOre uehice calculale Uannaut qudnttes reldted Detortanc One Inese efficiency; Msually expressem mles gallon (mpg). For one vehicle equipped this way; the mpg were recorded eac filled, and thc computer Nas (ncn rscu Hcre the mpg values tor random empic thetc recoros; nuleag8 , the gas tank 48.3 50.4 46.8 44.7 43.0 45.0 39.2 44-1 41.4 37.5 37.5 43.! 47.9 474 43.3 46.2 344 42 3 43.1 36.6 (a) Describe the distribution using graphical mcthods_ oppropriate neiyte thesc data usIng mcthods based on Normal distributions? Explain why Or why not: Yeta appropriate analyze these data using methods Dased an Norma dlstributlons There Jr0 outllers particular skewness appropriatc analyzc thcsc datu using mcthods bascd an Normal distnbutions Thcsc data JrC skcwco right: Yes, appropriate analyze these data Using methods Dased on Norma distributions: There ar the data ske wed to the right: outliers and appropriate analyze these data using methods based on Normal distnbutions. These data are skeweo Ihe lelt. not appronnate analyze these data using methods based on Normal distnbutions. There are extreme oueliccs . Find the mcon . (Round Your dnswici Mpq two declmal places ) Find tne standard deviation. (Round Vour answici four decimal places ) mpo Find the standard error: (Round vour anshier mpg four decimal places.) Find the marqin Emor for 9590 confidence (Round You 4nsaer didcim places. ) mpg (c) Rcport the 959 contidence interval four decimal places. ) moq; mpg | the mean mpg for this vchicle based on these data. (Round Your answers



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Preliminary data analyses and other information suggest that you can reasonably assume that the variables under consideration are normally distributed. In each case, use either the critical-value approach or the P-value approach to perform the required hypothesis test. Gas mileage estimates for cars and light-duty trucks are determined and published by the U.S. Environmental Protection Agency (EPA). According to the EPA, "... the mileages obtained by most drivers will be within plus or minus 15 percent of the [EPA] estimates...." The mileage estimate given for one model is $23 \mathrm{mpg}$ on the highway. If the EPA claim is true, the standard deviation of mileages should be about $0.15 \cdot 23 / 3=1.15 \mathrm{mpg} .$ A random sample of 12 cars of this model yields the following highway mileages. $$\begin{array}{llll} \hline 24.1 & 23.3 & 22.5 & 23.2 \\ 22.3 & 21.1 & 21.4 & 23.4 \\ 23.5 & 22.8 & 24.5 & 24.3 \\ \hline \end{array}$$ At the $5 \%$ significance level, do the data suggest that the standard deviation of highway mileages for all cars of this model is different from $1.15 \mathrm{mpg} ?$ (Note: $s=1.071 .$)

Given the data points X. Y. At the top of this document. We want to answer the following questions. A through F. In order to start off with in part A on the left, we want to produce a scatter plot of our sample data points, X. Y. I've already done so. And as you can see, I've included the scatter plot right below part A The data points are marked with exit or crosses next in part B. We want to compute the sums listed below as well as the Pearson correlation coefficient. R. I've included the sums already. Since they are simply computed by following, the formula is given where some access to some of the X values, someone in some of the Y values and so on. The correlation coefficient R. Is given by the formula. Then it follows, it takes us input the sample size N. And the sounds listed, plugging them in. We obtain articles negative 0.9463 next in part C. We want to find the line of best fit and to do so we find our mean for our X and Y values as well as A. And B. Values explore and wine bar are given by these formulas. Excess bar is 37.4 wine buyers. 21 next we identified being a. We remember is a very similar definition to our correlation coefficient. Are plugging an end and the relevant sums gives B equals negative 0.601 As our slope plugging in Y bar, B annex bar into a formula gives 43.3 to 6 for intercept giving us line of best fit. White hat equals 43.3 to 6 minus 60.601 X. Next let's plot Y. Have a scatter plot. We make sure to mark our mean X bar. And I mean Y bar on that line max let's calculate R squared and interpret it. R squared is simply the square, a correlation coefficient 0.8955 From this, we can say that 89.55% of variations in the data can be explained to the data itself presented here, and roughly 10% cannot be. Finally, we predict Y for X equals 38. Using our formula, we obtain why equals 20.488

So here we have an auto manufacturer that's claiming that the cars it sells have an average fuel efficiency of 35 MPG. But a consumer group thinks that the true figure is lower. They obtained five of the company's cars and determines an average of 32 MPG is is significant. The company runs a simulation by randomly picking five cars 160 times from a fleet of cars with no 1 35 MPG efficiency and calculating the resulting averages. To show that 32 is possible, the company makes the following dot plot, which was given and as the company's arguments seem reasonable, given the stop plot. So, um, just judging by the answers of E is the correct answer. And this is no because the possibility of picking five cars with an average of 32 MPG or lower from the fleet of cars no. 1 35 MPG efficacy is very small. Only four out of 160 or 1600.25 um, so the consumer group randomly picked five of the cars and found an average average of only 32 MPG and the chance of picking five cars with an average of 32 or lower from a fleet of cars with real 35 MPG. Efficacy is so small that it seems reasonable to conclude that the consumer groups set of five carbs cars was not picked from a fleet of cars with 35 mile per pound embassy and that the company's claim is wrong. So the chances it mhm having yeah okay, yeah, yeah, yeah.

All right guys. So for this question, were given a data set that contains Uh, information on uh, the fuel economy of different cars. From 2012 were asked to create a cross tabulation report for all the cars with the size of zero highway mpg as the column, and then group them by highway mpg, An interval starting up five, Beginning at 15 and ending at 14. Well then comment on the relationship between size and highway mpg. We'll create another cross tabulation report this time with dr as the row and highway mpg at the column grouping again by the same intervals as we did in question one then comment on the relationship between the drive and the city npg. And finally, we'll look at the relationship of fuel type and city mpg by creating another cross tabulation report. So let's create our cross tabulation reports on a different tab here to do so we'll go to another tab and insert a pivot table, collect our data from our data set here. Yeah. All right. And then we need to fill in the rows and columns. So this first one we were asked for uh size and highway mpg so size as the rose highway mpg as columns. Mhm values will be the account of highway mpg s. And then we were asked to look at only car. So we need to filter this. Um, these rose to just the values that our cars so large car. Midsize car. Small car. Yeah, there we have it. So looking at our relationship, what do we see? Let's movies around a little bit. So large car in the bottom, Small car up top. Yeah. So what we see here is that these small cars, a lot of them are In the fuel range of this 22 35 We see as we get a larger car, um We see also a smaller amount of cars but also that the fuel economy seems to go down. So mid sized cars, we see similar trends with most of the Um car. Fuel ranges being from 25 to about 35. But as we get a larger car we see the fuel economy go down so we can answer this one. I think it's Yeah. Mhm. That fuel economy trends downward as vehicle size increases. Next. We want to create a cross tabulation for drive and highway mpg. So we'll do the same thing here. So insert a pivot table again, Select our data set. Yeah. Okay. Yeah. And then this one we're going to put dr on the rose and city mpg on the columns and here. So we'll also have a filter by um uh huh. By size or only cars. So when you go back here, yeah, we want to filter this down to just the cars again. So, large cars, mid sized small cars and then group Our Fuel Economy Guide five. You took Up to 45 In groups of five. So what can we say about the relationship of this? Uh We're really, what we see is that Uh two wheel drives tend to get better gas mileage. We can see that here by these um columns out here where we have a lot of cars are getting really good gas mileage and your are no four wheel drive cars that are even getting that kind of gas mileage. So For this question # four, we can put two wheel drive cars tend to get better mpg. Last question, we're looking at fuel type in city mpg for our cross tabulation reports. So we'll do the same thing and we can actually copy this pivot table and just cut it so that it's easy to update. So instead of drive right, we're not looking at dr we want to look at a fuel type, I believe. Yes, your tape. So here, what do we see? We see most most are gasoline. Um, we see most of the gasoline cars are getting anywhere between 15- 30 for most of them. We see electric cars are getting really good gas mileage and your diesel cars are getting uh, slightly better than your gasoline cars. So You can put that for question six years. So which yeah, electric cars get the best mpg followed by you saw and then gasoline. So that's it. You just created a couple cross tabulation reports to look at the data set by size of car and gas mileage, the drive and the type of fuel and we concluded, um, fuel economy turns downward as vehicle size increases. That too, will drive. Cars tend to get better gas mileage than four wheel drive, um, and finally, based on field type electric cars get the best image E, followed by diesel and then data.

So for this problem, what we're given is, and as our sample size, that's 3 11. We're going to let x one br displacement X two will be class. Mid size X three will be class Large x four will be fuel premium x five will be. Actually, it's not x five. Why is going to be fuel efficiency? And that's going to be calculated by a highway mpg. So now we have our regression equation, and we first have to determine the necessary sons. So you look at the sum of X I and that's 10 38. Then you look at the sum of X, I squared, and that's 3833.68 Then we look at the sum of why I and that's 80 36. And the sum of why I squared is 212,638 than the sum of X. I Y II gives us 2 25 432 7. So now we want to determine the slope of B um, and that's accomplished by B equals and times the sum of X y, minus the sum of X times the sum of why all over end times the sum of X squared minus the sum of acts square. This is going to end up giving us approximately negative 28825 So the mean is the sum of all the values divided by the number of values. So we're going to get that X mean is 3.3145 and the mean of why is 25 points 8392 The estimate of a of the intercept Alpha is going to be the average of wide decreased by the product of the estimate of the slope. So this is going to look like a is equal to the mean of Roy minus B times the medevacs, which is gonna give us 35.3933 Um, so we can replace Alpha or a with Alpha and be with beta. So now that's going to give us this value right here, which will now be 35 933 minus 2.8825 x one and then for part B, we will use. We can use excel to generate the multiple in your aggression model, and it will give us an output. Since this can't be done on Excel, Uh, we get a bunch of values, but if we put in the correct values will end up getting the correct output. So now we can move on to purchase E um, or actually, with part B, we can do part of us. So once you get the numbers in Excel, we see why hat is equal to be not. Plus B one x one plus b two, x two plus B three x three. Um, so we see that be not is going to equal 29 0074 b one is negative. 16581 B two is 4.4860 b three is 1.8190 and with that we end up getting as a result that are y is equal to 29. 0074 minus 1.6581 x one plus four point 44860 x two Move this for our last term, which is B three x three. That's gonna be plus one point 8190 x three. That's our final answer for beam and then foresee. We have our significance level Alpha. That's gonna be a 0.5 And the given claim is that b I or beta? I rather equals zero. So the null hypothesis states that even claim that the slope is zero. Um, So what we have now is that a church not equals. Uh, actually, h hot is given as b I equals zero and h A is given as b I. There's not equal zero. So the p values in this case we would end up getting abated. To corresponds to P equals zero, and beta three will correspond to p equaling zero point. Um, it's 1234 12345678 zeros 704 It's a very close to zero, but not quite. So if the P value is less than the significance than we will reject the null hypothesis hypothesis. So in this case, he is, um, less than zero and b two case, it's less than zero. So we reject the null hypothesis, which is h not. And in this case, um, it's less than zero is that we reject the significance is too Well, so you reject the null hypothesis again. Then, for part D, we do another excel generation of multiple linear regression model. Um, so we can't do anything with the sell side of things, but we have another one of these equations where it's be not, plus B one x one plus B two, x two plus B three, x three plus B four x four and then with all of our values what we end up getting as our estimated regression equation. It's going to be 29 7123 minus 1.6383 x one plus 3.9984 x two +16700 x three minus 1.585 x four In this case, the significance level. This is the last part. E. Alpha is going to equal 0.5 again. Um, and we and then all hypothesis we'll have data. One is equal to beta two, which equals beta three jiggles beta four, which is zero. And our H sub A is going to, um, ST that at least one of the FBI's is non zero. Besides, for some hi then the P values, uh, corresponding we want to look at, so we C p equals zero and P is less than zero. So we reject we reject h not because the P value is less than the significance. So I reject each nut are null hypothesis.


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