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16 The datathe following table gives the nurnbers of uudienee lnDrCSsOnE (in hundreds of millions) listening SOnES Td the corresponding numbers of CD sold (in hundr...

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16 The datathe following table gives the nurnbers of uudienee lnDrCSsOnE (in hundreds of millions) listening SOnES Td the corresponding numbers of CD sold (in hundreds of thousands). The number of Judience impressions is count of the number of times people have hcard the JOm The dala fromrandomly selected CD Audience Impressions;CD' $ Sold,Assume that the populations of both variables normally distributed The sample correlation coeflicient for the above data is =0.796 Sketch scatter plot of

16 The data the following table gives the nurnbers of uudienee lnDrCSsOnE (in hundreds of millions) listening SOnES Td the corresponding numbers of CD sold (in hundreds of thousands). The number of Judience impressions is count of the number of times people have hcard the JOm The dala from randomly selected CD Audience Impressions; CD' $ Sold, Assume that the populations of both variables normally distributed The sample correlation coeflicient for the above data is =0.796 Sketch scatter plot of the dala: points b. Describe the information that provides in the context of this bivariate data set: points Test; using the 4-step procedure; whether the linear population correlation coeflicient between audience impressions and CD s sold is positive points



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In $11-17 :$ a. Draw a scatter plot. b. Does the data set show strong positive linear correlation, moderate positive linear correlation, no linear correlation, moderate negative linear correlation, or strong negative linear correlation? c. If there is strong or moderate correlation, write the equation of the regression line that approximates the data.
A sociologist is interested in the relationship between body weight and performance on the SAT. A random sample of 10 high school students from across the country provided the following information:
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So we're dealing with people with low and high iron levels and or excuse me, lead levels. And we're going to assume that the people with low uh levels have an equal like to those that have a high level for the lab. And alternately that the low level of uh the chemical that the I. Q. Is greater if you have a low lad greater than the high lab. Kind of an interesting study. So we will be assuming again that the difference between these two and I'll take the low minus the high is zero and our test statistic, we have a smaller sample sizes 21. So we'll say that we have 20 degrees of freedom instead of using that formula. And we'll take the 92.88462 minus the 15. Excuse me, 86 0.90476 So the difference between the means and then divided by the square root up. And we're going to take the standard deviation of the first population square divided by the sample size, which was 78. So we do notice that the sample sizes are different. Vote. That's the nice thing about this study is we can still use this test even though the sample sizes are different and this sample size was 21. And when I put that in my calculator, I got that, that test statistic comes out to be 2.282 So that's that value. So we know that we want to find what is the likelihood if we mark that test statistic on this chart, that's about here, where that T value is 2.282 and we want to know what that p value is, and if it's less than 5% then we can reject the null. So what's the likelihood of getting this type of a sample with 20 degrees of freedom and having it be 2.282 So what's the likelihood of getting a number like this or higher? If the main difference is actually zero? So I used my TCD Epp and to find that value and when I did I got the p value to come out to be 0.16 About eight. So that's the likelihood of that happening. This type of difference or greater if the mean is actually zero and that a significance level of 5% we have evidence to reject them all. So we reject the null. And we can conclude that it does appear as though the I. Q. Of the low lad does seem to be higher. Then the high lad folks kind of interesting. Now in part b we want to find a confidence interval and the appropriate for this test and since we're using 5% for one tail, test 5% at one tail and then we would need 5% at the other tail. We would want to find a 90% confidence interval. So we need to find what the T. Star value is for. And again I will use 20 degrees of freedom and we'll look up either 5% in the upper tail or we'll look up the number that corresponds with the 90% confidence interval and I'm looking that up in my table right now and it looks to me that that is 1.7 to 5. So we're going to take that difference and why don't we just subtract those two? Let's find out what we get there, that 92.8846 to minus the 86.905 476 That difference comes out to be 5.97986 plus or minus. And then we're gonna use that T. Star value and then we'll use that whopping large. Standard deviation are standard air we have here, so that 15.34451 square divided by the sample size of 78. And then that 8.988352 They sure did give us a lot of dozen places. Didn't they divided by 21? And now let's put this in. I hadn't done that ahead of time. And let's find let's find what the Let me write this part down first. Let's find the margin of error. So 1.725 times the square root of 15.34451 square divided by 78 plus that 8.988 352 square divided by the 21. And I'm getting that margin of air to come out to be 4.5 to 0 actually. And there are lots more displaced. But I'm going to actually store that value is X. And let's find out what those numbers are. So when we take that five 5.97 986 which I should have been smart and should have stored that in my calculator to begin with, shouldn't I? And subtract away X. When I do that, I get about 1.46 and then change that subtraction signed to an addition sign. And we're going to have that being 10 point 50 So we are 90% confident that the actual difference is somewhere in here and notice it does not include zero. So not including zero. Mhm. Which means we have evidence to reject them all now. So what do we think is true? In part, it does appear as though the I. Q. Uh does seem to be higher for the low lead people as opposed to the high lead people and why that might be don't know. And uh I think it also would be nice if we had maybe a little bit bigger sample sizes to make a better conclusion.

The following is a solution to number 48 9. And this just looks at three different data sets that are very, very similar. Just the sample sizes are a little bit different and this is taken from population where the mean is 100 the population standard deviations 15. So we're gonna do a few Z intervals. But the first part of this is to find the sample means for each data set. And I did technology use technology for this, I use a. T. I. T. Four because it, you know, really cuts down especially as the days get bigger, but you can use any sort of technology, Excel works great mini tab. Or you can do the formula, it might take a little longer but that's fine too. So stat and edit. And you see I have these three datasets here and I'm just basically gonna do the data set X. Bar for one, L two and L three. These are my three different data sets of stat. And then air over to cal can it's the one of our stats and I'll just go and change that to L one and this is my first X bar. So X one bar 99.125 So X1 bar is 99.125. Okay, so I do the same thing, go to stat, cowpoke one of our stats, but this time it's going to be L two. So the medium data set and I get about 99.1 99.1 for X bar. And then very last go to stat, cowpoke One of our stats and then I'm going to change this to L three And that's about 90.03, repeating. So we'll say 90, I'm Sorry, So, you see these sample means they're roughly the same. There may be slightly off by, you know, a couple hundreds, but it's not huge. Right? So that you can essentially think, okay, these are about the same um sample means. So now we're gonna do the 95% confidence interval for each one of these and then we're gonna compare. So let's take a look if we go to stat and then tests now, since we know the population standard deviation, we can use the Z interval and we have data Um since we're given a list of data, the sigma is 15 and I'm just gonna go in order. So my list is a one first, the frequency is always one and then the sea level, the confidence levels, 95% of .95. And then I go in and calculating this top one here, That's my confidence interval. So it's a pretty big confidence interval. And that's 20 Ish, you know, maybe a little more than 20 Units apart. So let's go and write that down. 88.731 88.731 All the way up to 109.52. Okay, so that's whenever the data set was quite small, it was about eight data values I think. So we're gonna do the same thing if we go to stat and then tests and it's that seventh option again. But this time we're going to go to L two. So the medium data set size and let's see what happens here. So 92.5-6 all the way up to 105.67. So you see that's already getting a little bit smaller, it's about 13 units apart as opposed to 20. Alright, so 92 .5-6 All the way up to 105.67. Yeah. And then finally the last data set So that's whenever an equals 20 so stat tests and it's still the seventh option. But this time it's going to be the third column and let's see what happens here. So this is where the sample size was 30 93.666 to 1 oh 4.4. So you see that's even smaller, it's only about 11 10 10.8 I guess units apart. So let's go and write that one down. So 93 .666 to one oh 4.4. So these are three data sets and were asked to compare or you know, look at as the sample size increases. You know what happens? And as we've already talked about it as in increases that margin of air is going to decrease. Which makes sense because you're dividing by a bigger number. If you look back at that format sigma over squared of in your divided by a larger number. So therefore that's a horrible. Therefore I guess it's not really shorthand whenever it's that bad. So therefore that confidence interval width decreases. Okay, so as in increases, the confidence interval will get narrower and you can see that as we increase the sample size, It gets narrow, narrow goes from 22, 13 To like 10 and change. Okay, so then these last two questions kind of deal with the same thing. So let's say you you miss um put something, answer that first. I think it's the first data value was 106. Well let's say you put that in a 016 so let's see what happens with the new confidence intervals. So let's go ahead and make that error. So stat edit. So yeah, it's that first one, so 016 which is 16, so I'm just gonna change it to 16, 16 and 16, So let's see what happens here. So stat tests And it is the easy and also seven, and then let's just kind of keep the same order. I could do L three first, I guess, but let's just keep it to the same order and see what happens. So 77.481 to 98.269 77.481 to 98 269. And this is kind of the last part, but you see that this actually already doesn't it's been uh Affected so much by that outlier of 16, that this has pulled it down so much that it doesn't actually contain the population mean. So you can see the smaller that sample size, the more it's affected. So let's go and take a look at the second one. See what happens here, stat tests seventh option and list is L. two this time. So it is significantly bigger, 88-101, So 88 points 0 to 6 All the way up to 101.17. So that one barely contains The mean 100 but it's definitely been shifted down. So I'm also comparing two up here, you know, this first one's been shifted down like 11 units right from 88 to 77. Well this one's only been shifted down about 4.5 units. So you can see it's not nearly as affected by that outlier. And then the very last one stat Tests 7th option. And then this time we're gonna change this to L. three and 90.66-1 on 1.4 9666 To 1014. So you can see this one also contains the population mean? And it's been shifted down looks like three units instead of four, you know? So it's even less affected because the sample size was 30 instead of 20 and that's kind of what part E talks about Which ones are still good um indicators or which one is still capture the population mean? Are are good estimates and the last two seem to still be good estimates. So data set golden scroll down. So if you look back this dataset one that's the only one that doesn't contain 100. So data sets two and 3. Data Sets two and 3 still capture New Equals 100. And so what does this show? This shows. Okay that as in increases the more accurate the confidence interval becomes because outliers become less influential. Okay, so the greater the sample size, the less influential those outliers become.

European number 23. Uh, it's, you know, they're not have processes. That new one is one of them repeal than YouTube. Alternative I busted in that new one is large in their music. Eso the critical value in table four, corresponding toe probability open 95. So the critical very equal to 1.645 So that the statistic is that equal toe X one bar, which is the mean off the first temple, minus the mean on the second sample minus new one minus you two over square, hold off. Think minus squared over and one lost signal to squared over. A two, which is a 2.6 and 2.62 is larger than 1.645 So we reject well, hypotheses, so there is sufficient evidence to support this claim.

This problem covers constructing normal probability plots and determining information from them. So, to construct a normal probability plots from this data and this data is on south Carolina's Cisco abstract um about the daily charges in 15 hotels, so and equals 15. So this data, there is 15 different terms corresponding to this end. We want to find table three with the number 15 because we want to construct a normal um probability plot. So these are the normalized values. We want to first organize our data into increasing order. So what I would do is I would and put these values in a calculator and sort list. So you get the lowest values and the highest values right in increasing order and the corresponding um table three lowest values and the highest values right here. And it looks like the lowest value here is 47.72 and the highest value is 130 0.17 So 47 72 and correspondingly 3.17 and all the in between also had the same pattern. So once you have that, you can graph it, you can graph the X and Y. Of these things. And once you grab the X and Y, the plot will look something like this. And you can generally see the trend follows this line right here, right? It follows this line. We can determine if there is any outliers outsiders. So an outlier is when is points that don't follow a trend. So at this point, that is just true, for example, would not follow the trend because it doesn't fall into this line generally. But in this case there is no outliers because all the points fall into the line, so there's no outliers. And finally, if this trend is linear, right? If it goes into a line, then we can say that it is normal. Normally distributed. Okay, Only shredded pot. Okay, Yeah, right. If it wasn't um linear, then it wouldn't be normally distributed.


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