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Construct the confidence interval for $p_{1}-p_{2}$ for the level of confidence and the data given. (The samples are sufficiently large.) a. $80 \%$ confidence, $$ \begin{array}{l} \mathbf{n}_{1}-300, \hat{p}_{1}-0.255 \\ \approx_{2}-400, \hat{p}_{2}-0.1 Q 3 \end{array} $$ b. $95 \%$ confidence, $$ \mathrm{w}_{1}-3500, \hat{p}_{1}-0.147 $$ $$ \mathrm{w}_{2}-3750, \hat{p}_{2}-0.131 $$

In this video, we're going to look at how would you determine whether a group of data is a bootstrap sample of a given data set? So there are a couple important things to remember with bootstrapping one is that they have to come from the original data set and Tuesday they have to be the same size as the original data set. And so up here we have the random sample of taxi out times from JFK Airport in minutes and below we have a series of data sets and we're going to see whether each of them are or are not a bootstrap sample. So for the first one we have 12 1913, 43 15. We see this is the same as the original data set and more importantly, it's the same size, it's five values and all the numbers are taken from the original data. So this is a bootstrap sample. And the second one we have 12, 19 and 15. Now this one is only three values and we need five values for a bootstrap of this data. So this one is not a bootstrap sample. And below that we have 12, 12, 12 43 and 43. And this is a bootstrap sample because it's five values just like our original dataset and all of the numbers come from the data set. This is also a bootstrap sample. Then we have 14 2012, 19 and 15. And this is not a bootstrap sample because it includes data points that were not from the original dataset. 14 and 20 are not in our original data. This one is not a bootstrap SAm And then finally we have 12, 13, 13, 12 43 15 and 19. And all of these numbers are from the original data set, but we see the instead of five values, we have seven. So this one is not a bootstrap sample and that's a quick rundown of how to determine whether or not a given set of data is a bootstrap sample of given data set. You look at whether the numbers are from the original dataset and whether it's the same size as the original dataset.

The following is a solution for # nine. And we have a uh hypothesis tests where the saying the difference of two proportions is zero. And then the alternative being the difference of the two proportions is less than zero. We're testing it at the 5.5% level of significance. And we're actually supposed to do the P value method this way. Okay so whenever you can I like to do a little shortcut here. So if you take 1100 times.22 you get 242. So I'll call that X one and X two. If you take 1300 times 13000.27 that gives you 351. So this is enough information for me to use the calculator. So I need to find the Z. And the P value. I don't need to find a critical value. Whenever I'm doing a P. Value method and I'm going to compare the P value with my alpha value that's gonna tell me whether to reject or not reject. So fortunately for us I can just go to tests and to prop Z. Test. The X one, remember was to 40 to the N. One was 1100. So if you just divide those that should give you the P. Hat and then the X two is 3 51. And that um sample size was 1300. And we're looking at less than so the alternative is that is less than and whenever we calculate It gives us a test statistic of negative 2.82 Uh negative 2.83 and AP value of .0023. Okay so let's write those down and then we'll analyze them. So this is negative 2.83. And we have a P value of 0.002. And we compare that to the alpha and .002 is less than .005. So whenever this happens we reject H not so we're rejecting the null hypothesis saying the two proportions are equal. Okay so the 2nd 1 We're not as lucky. So you get a decimal whenever you multiply here so we can't do it the easy way. So we gotta do you use a formula which is okay. So if we take .35 -11 and then zero. So I'm not gonna write that Divided by Big Square root. And then the standard area here is .35 so P one and then one minus p one which is 10.65 Divided by that 6:50. And then plus .41 Times 1 -1159 Divided by 6 15. And that equals negative 2- 3 3. Okay so the way we find this p Valium is we find The probability that we get a test statistic less than negative 2.233 Or greater than positive 2.233. So what we're gonna do is just find the probability that's less than the negative and then just multiplied by two because we have symmetry here, so negative infinity. And then this one will be negative 2, 2, 3, 3. And then we go ahead and paste and then we need to account for that other tales. So we multiply by two. So R. P value is 20.255 point 02 55, 0.0255. And then we compare that to the alpha value That's greater than the .1. So whenever the P values greater than alpha, we fail to reject H not so we are not rejecting the null hypothesis.

Yes. Yeah. Okay. So in this question they have given us um a set of data that represents the weight in grams of Tylenol capsules and they ask us to create a box plot and describe the shape. And so the first thing that you need to realize is that this set of data is not ordered from least to greatest. So you know, you can see this is 608 This is 601 Here's 6010.598 So the very first thing that you have to do, however you choose to do it is you have to be sure that your data is ordered from least to greatest. And if you have technology that's exactly what I would use and I would actually use the technology to find Our five Number summary. Because when we're creating our box plot we need to know our five number summary. So minimum Q one median Q three and max easy enough to find our minimum. Once we have the data ordered. Yeah, Easy enough to find the max. Um They had told us that there were 25 capsules. So if I want to find the median it's gonna be um 25 plus one divided by two. It's going to be the 13th term Will be the median. So if I just count 12345 678910111213. That gives me a median of .608. Mhm. And then that would leave me with. If this is the 13th term then I know that between here and here is 12 terms. So 12 divided by two is 6123456. This divides it in half. So I need to find the average of these two which is going to be .604. That's gonna be my Q. One and then I can count six. The other direction 123456 halfway between these two terms, It's gonna be .610. So there's my five number summary. Um Just make that go away. We want to find the I. Q. R. Which is Q. three minus Q. One. It's gonna be .006. We're gonna use that to help us determine if we have outliers. So to check our upper fence we take Q. three And we add 15 times the IQ are. And that is gonna give us Mhm. Yeah. Um .619. So my highest values .612. So there are no high up outliers and I'm going to check for lower outliers so I'm gonna take Q one -15 times the IQ are. Uh huh. Sure. Yeah because it's going to give me point 595. Now I don't have anything lower than .595, my lowest .598. So we have no outliers. So let's go ahead and think about making our box plot. We need to go from .598 up. 2.61 two. Yeah. Community. Okay. So um we're going to use our Middle three numbers to draw our box. So Q1 is at .604 vertical line. There, Medians .608 vertical line there. Q three is .610 vertical line there. So that forms the box. My max value is .612 here. That's my high whisker. .598 is low and that's my low whisker so we can see as far as shape. Um It is elongated longer to the left than it is to the right. So I'm looking at the median, the distance from the median to the minimum is much larger than from median to max. So this is skewed left.


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