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Research shows that people from Asian descent are on average shorter and lighter than people from non-Asian ethnicity. Besides; people of Asian descent are also tho...

Question

Research shows that people from Asian descent are on average shorter and lighter than people from non-Asian ethnicity. Besides; people of Asian descent are also thought to weigh less for given height: To test this hypothesis yOu collect height and weight of 400 Asian students and 810 non-Asian students at your university. regression of the weight on constant; height; and binary variable, which takes value 0l one for Asian sludents and Is zero otherwise, yields the following result:Student weight

Research shows that people from Asian descent are on average shorter and lighter than people from non-Asian ethnicity. Besides; people of Asian descent are also thought to weigh less for given height: To test this hypothesis yOu collect height and weight of 400 Asian students and 810 non-Asian students at your university. regression of the weight on constant; height; and binary variable, which takes value 0l one for Asian sludents and Is zero otherwise, yields the following result: Student weight = 229.21 6.36 Asian 558 Height, R2 = 0.50 where student weight is measured in Kilograms and height is measured in inches L.A Interpret the results (6 coefficients and R2) Does make sense t0 have negative intercept? (5 Marks)



Answers

An investigator collected data on heights and weights of college students;
results can be summarized as follows.
$$
\begin{array}{ccc}{{ } & {\text {Average}} & {S D} \\ \hline & {\text { Men's height }} & {70 \text { inches }} & {3 \text { inches }} \\ {\text { Men's weight }} & {144 \text { pounds }} & {21 \text { pounds }} \\ {\text { Women's height }} & {64 \text { pounds }} & {21 \text { pounds }} \\ {\text { Women's height }} & {64 \text { pounds }} & {3 \text { inches }} \\ {\text { Women's weight }} & {120 \text { pounds }} & {21 \text { pounds }}\end{array}
$$
The correlation coefficient between height and weight for the men was about
0.60 ; for the women, it was about the same. If you take the men and women
together, the correlation between height and weight would be
just about 0.60 somewhat lower $\quad$ somewhat higher
Choose one option, and explain briefly.

Problem. 77. We have the mean of other deals. Heights equals 60 inches new equal 60 inches. Yeah, and it's normally distributed with a standard division of 2.5 inches. Sigma equals 2.5 inches. We will take one Asian adult by random, and we will measure inside. Random variable X will be defined as the height of the individual male. Then, for Bharti, a X follows a normal distribution because it's given in the problem or in the study. It follows a normal distribution with me in equal 66 and standard division 2.5 be. We want to find the probability that the person is between 65 60 nine inches. We can do so by calculating the score for each value. For X one equals 65 we can get that one, which equals X one minus mu. Divided by sigma equals 65 minus 66. Divided by 2.5, it equals one divided by 2.5. Then they called minus 4.4, and we can forex to calculate its that score. It's two equals 69 inches. Then the two equals x two minus mule divided by Sigma equals three, divided by 2.5 equals 1.2. Let's explain what this means for a normal distribution like this with a mean of 66 inches. We want to calculate the area under this curve for the in terrible between 65 and 69. You want to calculate this area. To do so, We have calculated the zero score that correspondence to the X values for 66 score zero for 65 as a score equals minus 4.4. And for X two, there's a score equals 1.2. Now we can enter the tables to get the area to the left of the score to get the area between them. If we get the area to the left of 65 or minus, opening four and the area, then if two 1.2. Then when we subtract them, we can get the area in between. Let's enter the table to get the area to the left of 65 left the 69. Or they left to minus open four and the left to 1.2, which corresponds to the probability for X to be greater than or equals 65 greater than 65 smaller than 69. It equals the probability for that at minus over in four. Yeah oh boy. Into three, 4458 and the other probability to the left of 1.2, it's 4.88493 Then it equals 4.88493 minus 4.34458 The final answer is Oh 0.554 Oh 35 which means the individual height as 54% chance to have height. The individual. Let's make it. The individual has a 54% a chance to have a height between 65 69 inches for birth. C. It's what is the expectation to meet any Asian overseas 72 inches, which means we want to get the probability for X to be greater than 72 inches as previous for the normal distribution table. We have here 66 we have here 72. We want to calculate the area to the right of 72. Let's get does that the school 4 72. There's a school which can be suppressed as the three equals 72. Minus mu, which is 66. Divided by Sigma Sigma, is given 2.5. Then it equals 72 9. 66 divided by 2.5 equals 2.4. Then the score she correspondence to 72 inches is 2.4. Let's enter the tables again to get the area to the left of the school, which makes X is greater. 72 will be equal one minus this area. We will get this area and subtracted from one to get the total Bravo. Let's enter by 2.4 Toby and forest here. Then the probability is Oh 0.9918 one minus 0.9918 which makes this area or the probability to meet an Asian over 72 inches equals four point Oh, it two. Then the probability and percentage almost equals 0.82 were sent. It's less than 1% then it's not likely, and it's not expected. Not expected to do so. To meet and other Asian and other Asian mill over 72 inches. The final berth of the Problem Board C. The middle 40% of whites we want to calculate its range again for this normal distribution with 66 as I mean, we want to calculate the middle 40% from the definition this area must be going for. And if we have here a value, and we have here a value for Does that the school we have here? Zero. We want to get the score here, which makes the area to the left of it equals 4.7 y 0.704 for example, because this area will be 4.3 and this area will be all by industry. Then we will enter the tables to get there's a score that corresponds to area of open seven or another that the score that corresponds to area equals over in three this body, let's enter the tables again. To get this score, there's a score which corresponds to a 0.7. It's almost this value at 4.5 or 123 it's at that equals 4.53 Then that four equals 0.5 city. We know that Z equals X minus mu divided by sigma. Then this X, you can say it's rubber X, our equals 4.53 multiplied by Sigma, which is 2.5 plus new 66. This gives the value 1.53 multiplied by 2.5 or 66 is 70 six inches. And by cemetery, of course, 67 point three and by Cemetery X, lower will be equals. It will be equal minus 0.53 multiplied by 2.5 plastic C six, which equals 64 0.7. Then we have 40 percent chance to select and Asian Asian model. I mean with I'll try it between 64.7 and 67.3 inches and this is the final answer. Our problem.

Problem. 20 Which an eight. Using the scatter plot, we know that the scatter plus slope is upward, and thus the correlation should be buzzed. So the point does not seem to lie on the same line and that the correlation should be close to not not close toe person or negative one. Thio. To calculate that our confession, we had to get the X and the Y corresponding to X and X times. Why and x Times X, which is X squared and why times Why was she is the widespread And then we get the signals for this valuable, which is 396 case off fact. 41 14 case wide and 27339 In case off, next time is why 26158 In the case of X Squared 28 598 in case of Y squared so our is equal to end times 2733 No, I'm chasing Max. Why minus C. Max Times Sitting A line over square roots off end times segments squared minus Sigma X squared times six times two 85 98 minus 414 square, which approximately ableto open 5653

In this problem. We are comparing the Asian population in the Lake Tahoe area to the Asian population in the Manhattan area, and you were given some data about their race and their frequencies. So in the Tahoe region versus the Manhattan region and this was self reported data. And there were 1, 419 Asian, um individuals in the South Lake Tahoe region and off those 131 reported as Asian Indian 118 reported as Chinese 1045 reported as Filipino, 80 were reported as being Japanese, 12 Korean, nine Vietnamese and 24 were other. And then in Manhattan, we had 174 Asian Indian, 557 Chinese, 518 Filipina, 54 Japanese, 29 Korean, 21 Vietnamese and 66 other. And in both cases they add up to a some of 1 419 individuals, and we're trying to conduct a goodness of fit test to determine if the self reported sub groups of Asians in the Manhattan area fit the Tahoe area. So we're going to have to write up a null hypothesis and are null hypothesis is going to read the self reported sub groups of Asians. Yes, in the Manhattan New York area fits that of the Lake Tahoe area. Actually, it was the South Lake Lake Tahoe area, as described in the table. And if we're going to test that hypothesis, I try again. Hypothesis. We need an alternative to fall back on if we reject it. So our alternative hypothesis is going to be the self reported subgroups of Asians in the Manhattan New York area do not fit that of the South Lake Tahoe area as described in the table. And again, we want to run a goodness of fit test. And to start the goodness of fit test, where you're going to have to calculate a Chi Square test statistic. And to calculate that test statistic, you will apply the formula some of observed, minus expected quantity squared, divided by expected. So we're going to go back up and label the Manhattan as what we have observed, and we're hoping it matches the Tahoe region. So we're going to refer to this as being the expected. We're going to need to add on another column, and we're gonna call it observed minus expected a quantity squared all over expected. And the fastest way to get that set of numbers would be to utilize our graphing calculator. So you pull in the graphing calculator and we're going to go to the stat feature and edit, and we're going to clear out any lists better there. And we're going to put the observed values, which are the Manhattan values. Enlist one. And we're going to put the corresponding Tahoe values enlist to. And after you have those in, we're going to sit on top of list three, and we're going to tell it to find the Chi Square test statistic by doing the quantity of the observed values. Enlist one minus the expected values. Enlist to that quantity is squared before it's divided by the expected values enlist to, and we get these values and we're gonna round those to two decimal places as we write them in our table. So our values for the Asian Indian is going to be 14.11 for Chinese will be 1633.23 For the Filipino, it will be 265 77 Japanese 8.45 Korean 24 0.8 Vietnamese 16 and other 73.5. And to find our Chi Square test statistic, we need to total these up or add thes and again, there's a fast way of adding them up. Since they're already in our graphing calculator, we'll bring in our calculator, were going to quit and go back to the home screen hit second step. Scoot over to the math operations and we want to sum up list three and we get approximately 2035.146 So that is our chi Square test statistic 2035 point 15 We now need to find a P value. And when we find our P value, what you're finding is the probability that Chi Square is greater than that test statistic. So in this case, it's greater than 2035.15 and our best bet is to draw a graph to represent the situation which ties all of this together. So we're dealing with a chi square distribution, so it will be a skewed right distribution, and our distribution is going to be determined. The shape of it is determined by the degrees of freedom and degrees of freedom are found by doing K minus one. And the K represents the number of categories that your data has been separated into. So if we go back to our original table, we have broken our data up by race into 1234567 different categories. So therefore, R k value it's seven, and our degrees of freedom will be six. And the degrees of freedom are also telling us what our average is and are mean in this case is six. So on our picture, just to the right of the peak is where you'll find your average, so we could put six on our chi square horizontal axis right around there. Now we want to also plot this test statistic, and that's gonna be way, way, way, way out in the tail. So if I were to keep this going and this curve got closer and closer and closer, we're going to be putting that 2000 35.15 way out here, and we're trying to figure out the probability that were greater than that. Well, there's not much area between the horizontal axis and the curve at that point. But that's where you're gonna find your p value. And when it comes time to calculate that P value, our best bet is to use our chi squared cumulative density function in our calculator. And when you do that, it looks for a lower boundary on upper boundary and the degrees of freedom. And in our picture, the lower boundary of that shaded region is going to be 2035 15 Now the upper boundary. Keep in mind, that does keep on going. So therefore, we're going to pick a very large number. We're gonna say 10 to the 99th power and then our degrees of freedom was six. So we're going to bring in the calculator again, and we're going to quit and clear, and you're gonna hit the second button and the variables button, and it's number eight in the menu, you see, right now. So we're gonna type in our low boundary, followed by a comma or upper boundary, followed by a comma, and then our degrees of freedom, which was six, and the probability or the area to the right of 2035.15 is zero. So our P value is zero. Now that we have our test statistic and we have our P value, we are ready to make our decisions. And our decision is based on what we call our level of significance. And our level of significance usually is 0.5 Every once in a while you'll see a level of significance of 0.1 And based on that level of significance, if the level of significance is greater than your P value, then your decision is going to be to reject the null hypothesis. So if we were testing at the significance level of 0.5 0.5 is certainly greater than zero. So we would reject the null hypothesis if we were running this test at a 0.1 Significance level 0.1 is also greater than zero. So either way, our decision is to reject the null hypothesis. So we're gonna go back to that null hypothesis, and we're going to reject it. So if we're rejecting it, weaken, basically cross it off and say That's not true. So therefore, this is what is true. So we are ready to draw our conclusion. And our conclusion would be that there is sufficient evidence that the self reporting subgroups of Asians in the Manhattan New York area do not fit that of the South Lake Tahoe area, and that concludes your chi square goodness of fit hypothesis test.

So we're gonna be examining the freshman 15 with a couple of different cases, we're gonna look at the range variance and standard deviation of this phenomenon And look at a sample of 18 students. Yeah, So our high minus started low is going to get us a range and their high value is 11 And our low value is -5. So that's going to give us a range of 16. Next we can get the variants out of this, but first we'll have to get the average, so the average is the mean And the mean is calculated by summing up each of our turns and dividing by the number of observations, which is 18. And this is going to get us 189 mm. So now we can go back up with our 1.89 for our mean, like that in right here. Okay, so what we're going to get is our term x, I -1.89. And then we're gonna square that And do that over and over for all 17 remaining terms. And then we're going to do that for 18 -1, divide that. So it's going to get a 17 on the bottom. And once we complete this calculation, we find that there is a variance of 16.46 pounds squared, Oh well my dad, it's supposed to be kilos, I'm just using pounds, so it's gonna be kilograms squared and to get that into a standard deviation, all we have to do is take the square root of that. So we'll take these square roots Of 16.46 Kelly Graham Squared. Mhm. Mhm. This will get our units into kilograms, which is nice And this is going to be approximately form. So the exact 4.0 zero six kg, Let's not worry about the extra four digits that come after. So this is our standard deviation of 4.06 and if we see that there is a weight gain of More than 6.8 kg, the freshman 15, Uh this is greater than R 4.6 kg standard deviation. So we would consider this unusual because it's uh more than one standard deviation. Mhm.


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