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Suppose the mean daily peak power load (y) for power plant and the maximum outdoor temperature (x) for sample of days is as given in the ff. table214 1521561292541...

Question

Suppose the mean daily peak power load (y) for power plant and the maximum outdoor temperature (x) for sample of days is as given in the ff. table214 152156129254100266Find the correlation coefficient (2 decimal places)

Suppose the mean daily peak power load (y) for power plant and the maximum outdoor temperature (x) for sample of days is as given in the ff. table 214 152 156 129 254 100 266 Find the correlation coefficient (2 decimal places)



Answers

For the data in Exercise 21 of Section 10.2 "The Linear Correlation Coefficient" a. Construct the $95 \%$ confidence interval for the mean increase in strength at 28 days for each additional hundred psi increase in strength at 3 days. b. Test, at the $1 / 1$ 0th of $1 \%$ level of significance, whether the 3 -day strength is useful for predicting 28-day strength.

So for 21 the first thing we have to do is find the mean and standard deviation. So if you put him in your calculator, run won their stats. You will find that the mean is 53 point A with a standard deviation of 3.42 So we want to know how many of our items are within one standard deviation, two standard deviations and three standard deviations. So if we add the standard deviation to are mean and subtract the standard deviation, that gives us a range. A 50.38 2 57 22. There are seven numbers in that range. Two standard deviations. So subtract another 3.42 and add another three point for two gives us 46.96 to 60.64 Nine items are in that range and then going out one more standard deviation. 43.54 2 64.6 includes all Ted

Referring back to exercise 23 from this chapter. In part, they were asked to calculate the co variance of the random variables X and Y from this exercise. So using the values from the table in this exercise we have that expected value of X times. Y is, for example, we have next equals zero times y equals zero times the probability of X equals zero and y equals zero from the table is 00.71 So really summing over all possible values for X and Y kinds, the joint probability f of X y, and we can continue this some the whole table until we get to se X equals three y equals two, you get X equals three times to and then F of 32 according to the table is 0.1 And computing this find that, like the value of X Y, is 0.35 and therefore co variants of X and Y is going to be expected. Value of X Y minus expected value of X times expected value of why which from the problem we calculated affect the value of X was 0.35 expected value of why this 0.32 So we have 0.35 minus 0.35 times 0.32 in this equals 0.238 in part B were asked to calculate the correlation coefficient of X and y, so they calculate the correlation coefficient defined variance of x and Y. Now, using the marginal distribution which we found in that previous exercise way, we have that variance of X. This is the expected value X squared second moment of X minus the expected value of X weird and we have for the second moment of X is 67 and expected value of X once again is 0.35 This is 0.35 squared and so the variance of X is equal to 0.5475 Likewise, we have the variance of why again using the marginal distribution. Why is expected value of why squared second moment of why minus detective value of why squared. This comes out to be 0.3976 Therefore follows that the correlation coefficient of x and y, which we could also call row. This is going to be co variance of X and Y Just point we calculated before points to 38 Yeah, over the standard deviation of X, which is the squared of the variance of X squared of 0.5475 times the standard deviation of why? Which is the square root of the variance of why squared of 0.3976? Calculating this, we get approximately 0.51 for the correlation coefficient. Now we're asked to interpret what this means. Well, 0.51 is halfway between zero and one. So we would say that there is a moderate because of this size, the magnitude of the correlation coefficient. And then we would say direct because this is what the correlation coefficient measures measures when your relationships association between the random variable X, which recall that represents the number of syntax errors and the random variable. Why represents the number of logic errors in a program? And to me, at least, this makes sense. Figure that someone who's making errors and one way is probably more likely make errors in other ways because they're just often general Now. Directing association indicates that programs with a higher than average number of Syntex servers also tend to have higher than average number of logic areas and vice versa.

Welcome to new Madrid. In this problem we are given the average electrical usage is from around it to in the in the unit uh kilo or what hours. And we are also given that this is coming from 1000 hubs. So the data looks the given data which looks like uh tabular format that the first column will give us the usage and the second column would give us the number of homes. That is the frequencies. So we can mention this as excise and this as events. So we have to find the problem excited into the Florence and then find the mean. So this kind of a situation is known as we did frequency distribution or weighted data. And the formula for mean would be expert. Me too one day and will get replaced by one by submission. F. I. One runs from right to end. Want to end into submission X. I If I I run from one to him, no one is submission. If I submission. If I use this, if we do, we will find 1000. As the look now. They have to find exciting. We will write these values 500 into 32 will be 11 council. So we're going to be 680 will be for the example. Yes. Mm 42 City City 2200 next week. Baby one lack 48,000. Next up we will have 342 000 one totals triple zero. Yeah, 99 triple zero. And finally. And the Yeah. Culture. So this would give us the sound. 86 photo title. So beautiful. This especially solution of exciting to F. I. And this expression is submission. F. So we have to divide this expression. This number with 1000. So therefore we have eight books, 86200 Divided by 1000. Therefore the answer will be 8 62.2 kilo. What in? Can I walk in hours, jake a small bit as his answer. So keep. So this is that average or the mean of the given usages from 1000 houses where the data set is given innovated frequency table format. Make sure we use this particular formula in such scenarios to obtain the cult answer. I hope you could understand the problem. Let me know if you have any questions, but by


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