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An increase of 1.0 in grey matter density corresponds to alannumber of Facebook friends:Subject has recorded grey matter density of-0.6_ Use the regression cquation...

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An increase of 1.0 in grey matter density corresponds to alannumber of Facebook friends:Subject has recorded grey matter density of-0.6_ Use the regression cquationestimate the numberFacebock friends for this individual7, The number of Facebook friends for Subject =541 . Complete the following sentence:The residual for Subject >This mieans the number of Facebook friends for Subject _the same as lower than higher than the number Facebook friends predicted by the regression modeiAnother individ

An increase of 1.0 in grey matter density corresponds to alan number of Facebook friends: Subject has recorded grey matter density of-0.6_ Use the regression cquation estimate the number Facebock friends for this individual 7, The number of Facebook friends for Subject = 541 . Complete the following sentence: The residual for Subject > This mieans the number of Facebook friends for Subject _ the same as lower than higher than the number Facebook friends predicted by the regression modei Another individual (not included in this data set} has grey matter density of = number of Facebook friends for this individual? Would be appropriate use tnis Iinear model to predict the No, because 2.5 beyond the range of the data used to build the regression model; Yes, because 2.5 is reasonable grey matter density for humans. C. No; bocause 575.1928 number of Facebook friends too largo to bo reasonable number of Faccbock friends, oven for this point. What are the conclusions of this study? Choose all that apply: Grey matter density correlates counts of Facebook friends B. Large grey matter density cause individuals to have smaller counts of Facebook friends C. Increasing the number of Facebook friends an individual has will increase the grey matter density they have, D: Large grey matter density cause individuals have larger counts of Facebook friends E: Increasing the grey matter density an individual has will increase the number Facebook friends they have F Grey matter density do not correlate to counts of Facebook friends Help Entering Answers Preview My Answers Submit Answers



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This is a technique to break down the variation of a random variable into useful components (called stratum) in order to decrease experimental variation and increase accuracy of results. It has been found that a more accurate estimate of population mean $\mu$ can often be obtained by taking measurements from naturally occurring subpopulations and combining the results using weighted averages. For example, suppose an accurate estimate of the mean weight of sixth grade students is desired for a large school system. Suppose (for cost reasons) we can only take a random sample of $m=100$ students, Instead of taking a simple random sample of 100 students from the entire population of all sixth grade students, we use stratified sampling as follows. The school system under study consists three large schools. School A has $N_{1}=310$ sixth-grade students, School B has $N_{2}=420$ sixth-grade students, and School $C$ has $N_{3}=516$ sixth-grade students. This is a total population of 1246 sixth-grade students in our study and we have strata consisting of the 3 schools. A preliminary study in each school with relatively small sample size has given estimates for the sample standard deviation s of sixth-grade student weights in each school. These are shown in the following table: School A $N_{1}=310$ $s_{1}=3 \mathrm{lb}$ School B $N_{2}=420$ $s_{2}=12 \mathrm{lb}$ School C $N_{3}=516$ $8_{3}=6 \mathrm{lb}$ How many students should we randomly choose from each school for a best estimate $\mu$ for the population mean weight? A lot of mathematics goes into the answer. Fortunately. Bill Williams of Bell Laboratories wrote a book called $A$ Sampler on Sampling (John Wiley and Sons, publisher). which provides an answer. Let $n_{1}$ be the number of students randomly chosen from School A. $n_{2}$ be the number chosen from School $\mathrm{B}$, and $n_{3}$ be the number chosen from School $\mathrm{C}$. This means our total sample size will be $m=n_{1}+n_{2}+n_{3} .$ What is the formula for $n_{i} ?$ A popular and widely used technique is the following. $$n_{i}=\left[\frac{N_{i} s_{i}}{N_{1} s_{1}+N_{2} s_{2}+N_{3} s_{3}}\right] m$$ The $n_{i}$ are usually not whole numbers, so we need to round to the nearest whole number. This formula allocates more students to schools that have a larger population of sixth graders and/or have larger sample standard deviations. Remember, this is a popular and widely used technique for stratified sampling. It is not an absolute rule. There are other methods of stratified sampling also in use. In general practice, according to Bill Williams, the use of naturally occurring strata seems to reduce overall variability in measurements by about $20 \%$ compared to simple random samples taken from the entire (unstratified) population. Now suppose you have taken a random sample size $n_{i}$ from each appropriate school and you got a sample mean weight $\bar{x}_{i}$ from each school. How do you get the best estimate for population mean weight $\mu$ of the all 1246 students? The answer is that we use a weighted average. \begin{equation}\mu \approx \frac{n_{1}}{m} \overline{x_{1}}+\frac{n_{2}}{m} \overline{x_{2}}+\frac{n_{3}}{m} \overline{x_{3}}\end{equation}

All right. We have some data and we have the the MRI values is the explanatory variable. And the I. Q. Is response variable and we want to find the least squares regression line and I use this uh spreadsheet program to do that for me and how I did that. Just highlight the data. And then you go to insert chart which is right in here, insert chart and then you get some options and then you basically see this and you can select what do you want scatter diagram, some other lines, whatever you want there. And then to get the line you go down to series, you have to select trend line and then you might need to go down here where it says label. You have to click use equation. And that is the trend line that's the least squares regression line. And what do you notice about the value of the slope. Look at that thing, it's super close to 2.86 times 10 to the negative five. That's uh for all intents and purposes for most humans that's gonna be zero. But it's not exactly zero. So it's technically not zero. But it's it's very very close to zero. Which means that this is actually rather horizontal line very close to the 20. And this seems to fit. Because if we're looking at this data it's it's kind of linear, but at the same time it's very flat. It's like it means it as one variable goes up, the other variable doesn't go up by much. It looks like it because we've scaled in. Um But then if you look at the correlation coefficient are here and I mean it's equals corral for the correlation coefficient here, it's like the data. Yeah. And we see this value .54. So that's the linear correlation coefficient. So kind of makes sense that it's it's kind of it's it goes in a line, but at the same time there's still quite a bit of scatter. And so when we're looking at the scatter plot and we see the linear correlation coefficient, this seems like a reasonable result that this data doesn't really show or that that the correlation between these two variables is not uh not that much pretty low, not correlated at all for all intensive purposes. Uh And so when there is no relation between the explanatory response variables, we just use the mean of the white make prediction. So Yeah. Yeah. I mean, which in this spreadsheet program is average? Was average, get your data And it's 136.4. And so what that means is we're going to take the average and that's what the MRI that's the of anybody. So the average IQ predicted like an individual whose recounts a million and whose MRI candace 830,000? Well 136.4. Which 136.4 is going to be roughly here. And if you can imagine drawn the line here straight across, which would actually kind of fit with the data, is just very mhm horizontal, looking at the horizon. If we zoom out, so there you go.

Soon. Number 28. What's your name? The level of significance is mentioned is all for equal 0.1 on our hypothesis, which state that the population mean new is equal to the value mentioned in the claim, so it would two new equal 4.55 g. The alternative hypothesis stated the opposite off the novel hypothesis according to the clean, thus using lesson. So each one to mu is less than 4.55. Dreams is alternative hype with his old listen. Then the test is left. 15 is the alternative. Hypotheses uses Biggers in, so the desk is right field. The alternative hypothesis uses non equal, so the test is toe field, so the answer will be left field question. Number be given export equal 3.75 being equal 4.55 Zita equal 0.7 and unequal six. The sampling distribution off the sample mean export is normal because the population distribution X is assumed to be normal. The sampling distribution off the sample mean has mean new and stirred and standard deviation. Zita over square root and does it? Value is a sample mean decreased boy. The population mean divided by the standard division so that equal explore minus new over Zita over square root in equal 3.75 minus 4.55. Mhm 0.7 over square root six equal. Minus 2.8. Question number C Riddle Sport A and B. It snowed too. Mu equal 4.55 g each one to mu is less than 4.55 g and did equal minus 2.8. The B value is the population off obtaining a value more extreme or equal to the standard test static Zet that remind the probability using tepidly. So, probability equal probability off export is less than 3.75 Equal probability off. That is less than minus 2.8. Equal 0.26 Question number We? Yeah, given Alpha giving Goma equal 0.1 Result port. See, P equal points here is you 26 If the value is smaller than significance level Alfa, then then I'll Hypothesis is rejected. Mm. Is listening 0.1 So I reject h No. If we rejected them on hypothesis the data is the data easy, statically significant at level Alfa Question number e result, 40 each node to new equal, 4.55 g and each one to mu is listening four point 55 graham result for the project. Each node. There is sufficient evidence to support the claim that the mean, the mean weight off these periods in three sports off the Grand Canyon is less than 4.55 g.


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