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Average of 3YrReturn% Column Label Row Labels Avjerage Hlghe ILow 'Large 20.40 036,54 17.428 Growth 21.70 1B6.54 17.94 Value 17.95 16.22 Mb- Cap 22.69 126 30 2...

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Average of 3YrReturn% Column Label Row Labels Avjerage Hlghe ILow 'Large 20.40 036,54 17.428 Growth 21.70 1B6.54 17.94 Value 17.95 16.22 Mb- Cap 22.69 126 30 20.28 Growth 12281 26 46 19.65 Walue 21.58 26.11 21,801 ISmall 24.82 25.36 1241501 Growth 25.11 24.39 26 85 Ialue 21.64 2706 22 15 Grnd Total 21.90 26181 j18.091 Copy the pivot table for the standard deviations into Box 2.Grand Total 19.68 21.10 17.01 23101 122179 2378 25.14 24.78 126.08 I2186StdDav @f 3YrRetum% Column Label Roww Label

Average of 3YrReturn% Column Label Row Labels Avjerage Hlghe ILow 'Large 20.40 036,54 17.428 Growth 21.70 1B6.54 17.94 Value 17.95 16.22 Mb- Cap 22.69 126 30 20.28 Growth 12281 26 46 19.65 Walue 21.58 26.11 21,801 ISmall 24.82 25.36 1241501 Growth 25.11 24.39 26 85 Ialue 21.64 2706 22 15 Grnd Total 21.90 26181 j18.091 Copy the pivot table for the standard deviations into Box 2. Grand Total 19.68 21.10 17.01 23101 122179 2378 25.14 24.78 126.08 I2186 StdDav @f 3YrRetum% Column Label Roww Labels Average High Lowi Grand Total Large 4,79 118.181 3.72 003 Growth 4,81 118 18 3 90 I7 93 Value 13.72 3119 3,52 Mid Cap 4.92 4.89 4159 15420 Gmowwth 5417 474 5185 5.43 Ia luel 1149 5.34 1128 432 Hhismmall 3i719 51.09 348 4.60 Growth 8.79 4,79 13,72 433 Malue 2,38 5631 lo.59 5 22 Grand Total 4.94 111 Ha2t 6.47 Answer these 2 questions (to Bet; all points; YQL must awver both 07these correctly}: 1Which category (just 3 words: cap; tvpe; risk) had Ite highest average 3-Year % return? shallep 2 wnich categoly Fad ee towest vomtiy



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Average Income The following data represent the per capita (average) disposable income (income after taxes) for the 50 states and the District of Columbia in 2003 With the first class having a lower class limit of 20,000 and a class width of 2500 (a) Construct a frequency distribution. (b) Construct a relative frequency distribution. (c) Construct a frequency histogram of the data. (d) Construct a relative frequency histogram of the data. (e) Describe the shape of the distribution. (f) Repeat parts (a)-(e) using a class width of 4000 . Which frequency distribution seems to provide a better summary of the data? (g) The highest per capita disposable income exists in the District of Columbia, yet the District of Columbia has one of the highest unemployment rates (7\% unemployed). Is this surprising to you? Why? $$\begin{array}{llllll} 24,028 & 30,641 & 24,293 & 22,123 & 29,798 & 30,507 \\ \hline 36,726 & 28,960 & 42,345 & 27,610 & 26,356 & 27,837 \\ \hline 23,584 & 30,063 & 25,929 & 26,409 & 27,033 & 23,567 \\ \hline 23,889 & 25,900 & 32,637 & 34,570 & 27,275 & 30,397 \\ \hline 21,677 & 26,317 & 23,528 & 27,865 & 28,188 & 31,251 \\ \hline 35,411 & 23,301 & 31,527 & 25,307 & 26,902 & 26,684 \\ 24,169 & 26,102 & 28,557 & 28,365 & 23,753 & 27,149 \\ \hline 26,314 & 26,922 & 22,581 & 27,750 & 29,683 & 30,288 \\ \hline 22,252 & 27,508 & 29,600 & & & \\ \hline \end{array}$$

So here we have created, um, cross tabulation, a table for types, off fun and average annual return. Over five years, period Onda, we have calculated the row totals and column totals. Once we have done, we are done with cross tabulation. We can easily we can easily make Quincy distribution out off the cross tabulation. For example, we have created this frequency distribution out of this cross revelation table. And this frequency distribution consists off data on five years average return. So here are the classes. We have just destroyed them in rows. Here in the cross revolution, they are in column. But here, in frequency distribution, we have rested them in rows. And these are the sub particular frequencies. You can see that. And this is the somewhat product. So in the end, we have created African see distribution on, uh, on the data front type, So it is very simple. This is the first column off the cross regulation, and this reconstitution will be the same. And here we will pick this last column, and that's appropriate hair to the So there appears to be a relationship between fun type and average return. Over the past five years because the frequency is not off. Frequency is not roughly the same in each row and each column of the cross tabulation. So there has to be some relationship between the front type and average rate of return over the five years period.

We're gonna be using Chevy Chefs rule from the Z score, we calculate uh that's the score is going to be Our X Value 205,550 -220258. So some big numbers with a pretty small difference, relatively A standard deviation of 5037 for the population gets us a Z score of negative 2.82 So we're gonna be plugging that into the cake for Chevy chefs of inequality. Here it is. Mhm. Well, I don't know why bother writing the negative. It's gonna be squared so positive by the way, And this is approximately equal to 87.4, Pretty high percentage. Uh we can therefore conclude that the other houses are above $205,000 and that this home is a good deal. And obviously we can use tv shows because it's um relatively high Z score with a good percentage

Okay, so I actually learned something new today. Apparently, there's a formula that kind of tells you when a fledgling bird is going to be able to fly on its own. And it's the ratio of two functions of time, one of which kind of gives an indication for how long their wings are and the other one is their body mass. Okay, so whenever these ratios kind of approach one, whenever FFT approaches one thing, the fledging fledgling is able to fly on its own. You didn't know all this question is asking us to do is to interpret the physical meaning behind those and described the units associated with them. Okay, so the 1st 2 shouldn't be too difficult, right? Because M prime of tear is the time derivative. It's the time derivative of em of tea. It's a rate of change. It's how fast this function changes with respect to time. And since we're given that the average body mass is measured in grams, this is gonna be essentially grams per per unit time and that the time is in weeks, by the way, since grams per weeks and just like I said, ah, the interpretation of physical meaning is how fast the body mass is changing with respect to time. It's the rate of change of body mats. Okay, for w prime of tea, that's again the time derivative of W T. Which is the length of the wings. And since they said that wing length was gonna be measured in millimeters, this rate of change is millimeters per week. And again, it's just how fast the length of the ones you're changing F prime of tea has to have special analysis because, um, f prime of TIA is the time derivative of f of tear. But f of tia is the ratio. Oh, the length of the wings and the average body mass. These air two functions that change. So we actually do have to use the quotient rule, which is lo de I minus high. Do you low swearing the bottom and we're gonna determine the units of this function As far as a physical meeting goes out, I said half of Tia as FFT approaches one, then the fledgling is gonna be more able to fly. This is a rate of change of that. So if this is a really, really positive number, then it's going to rapidly go towards, and it's gonna take a lot less time for to be able to fly. Maybe it's some sort of growth spurt. So all that remains to do is to find the units of this I'm of Tia. Ah was measured in grams w prime of tea we said was millimeters per week. Okay, W of tea waas um millimeters and M prime of tea is grams per week, divided by AM of T squared M of T is measured in grants. This is just grams squared. Okay, so we have a grams times of millimeters minus of grams, times of millimeters, all divided by a week. Okay, so this is gonna be some sort of of grams times millimeters. We don't know. Of course you know how exactly these variables are changing, so they're obviously not gonna be the same necessarily. So when you subtract two things that are like each other, you're gonna get something else that's like each but that's like those two things. So this is gonna be some grams millimeters divided by weeks, all divided by Graham Sward, which is gonna be grams times, millimeters times, weeks times one over. Graham squared. One of these grams cancels with one of those, and then we're left with millimeters Her grands week. Here we go.

In this exercise were given a sample of size 20 and we would ask We were asked if we would use a one sample T test to determine whether there is sufficient evidence to conclude that the population mean is less than 100% or in other words, is less than one. And since the sample size is only 20 before using a T test, we'd want to know that the population is approximately normally distributed and one way to investigate this is to do a normal probability plot with the sample data. So we can do this in our. So in this first step, we can enter the sample information with this command and then make a normal probability plot using the QQ Norm Command. And this is what the pot looks like now in the middle sections, it's fairly linear. But in the tales the lower tail, you can see it drops down, the upper tail rises up. This inverted s pattern is representative of a distribution that has a slightly heavy tail on the right side and a slightly light lower tail, which implies that it is probably right, skewed distribution. And if we want to be cautious, then this would suggest that we do not use a T test to test our hypotheses because we're not convinced that the population is approximately normally distributed. Now that's for part a. So for Part B, we're told that a normal probability plot of the natural log of the sample values shows a pronounced linear pattern, which suggests that the distribution is log normal. Now we can test this in our if we call an object log, underscore sample and simply take the natural log of the sample and then make a normal probability plot of log underscore example. And we can see that indeed, this is a much more linear pattern than the normal probability plot for the UN transformed sample data. So this suggests that the population is approximately log normally distributed with parameters, mu and sigma, and were instructed in the question that this also means that e to them you is the median of the UN transformed distribution. So if we want to test to see if there's evidence based on the sample that the UN transformed population is less than 100 or less than one, we can make an alternative hypothesis that E. To them, you is less than one, which means that mu is less than zero. If we take the lawn of when we get zero, the no hypothesis can then be eating them. You is equal to one. So let's call the transformed variable y, and it's equal to the line of X. And since why is normally distributed with parameters, mu and sigma, we can say that our test statistic is given by the sample average of the transform data minus mu of the log. Normal distribution over the sample standard deviation for the transform data over the square root of the sample size. And so this mu is for our no hypothesis. So let's calculate some of these values and are we need the transformed sample average and the transformed sample standard deviation so we can use the mean of the log of the sample and the standard deviation for the log of the sample. And we get these values now. Mu is zero for the sample standard deviation. We got 652 and divide that by the square root of 20. This comes out to minus 0.181 Now our P value will be the probability of getting the test statistic smaller than negative 0.181 This comes out to about 0.4 to 8, which is a very big P value. So therefore we would fail to reject the null hypothesis for any reasonable significance. Level 0.4 to 8 would be bigger than the significance level, and we can conclude that there is insufficient evidence to suggest that the population means is less than one or 100%.


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