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The following iS tht mcasured radiation rales (WAg) of a random sample of cell phones: V 1,A 149 JA 445 074 0 89 14 L5 At 138 Sketch Ihe Boxplot; Lah-ling thc S-num...

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The following iS tht mcasured radiation rales (WAg) of a random sample of cell phones: V 1,A 149 JA 445 074 0 89 14 L5 At 138 Sketch Ihe Boxplot; Lah-ling thc S-number summary on the plot0y Pet d4 - # H6 List > t<a 72 Ll Stak 1-Var'51 LL (3 List Iha mncan and tnc stand_naviation - Dv De9 /o9 Ws/te(2 pts! ATe an1 ofthz Viluzs considered outliers (use the 5 IQR methodi? Show the valucs Ihat @c uscd I0 dclerina the outliers Ukes 51 I+y eignfiartly louxr Skowr M2 Wit 46 L5 IQr mchsdrcceni

The following iS tht mcasured radiation rales (WAg) of a random sample of cell phones: V 1,A 149 JA 445 074 0 89 14 L5 At 138 Sketch Ihe Boxplot; Lah-ling thc S-number summary on the plot 0y Pet d4 - # H6 List > t<a 72 Ll Stak 1-Var '51 LL (3 List Iha mncan and tnc stand_naviation - Dv De9 /o9 Ws/te (2 pts! ATe an1 ofthz Viluzs considered outliers (use the 5 IQR methodi? Show the valucs Ihat @c uscd I0 dclerina the outliers Ukes 51 I+y eignfiartly louxr Skowr M2 Wit 46 L5 IQr mchsd rcceni Cn_ thc mcan ACT composile scure WaS 21, and the standard deviation Was 5.1. Find the Z-scure Ior &n ACT scor 0[ 28, 28 - 26.r5,/ = 2 -=cele student oblaintd Zcur 01 2 $ on & Iest she took, The Mlan ECOIL 75 ; the standard devialion Was points Whal Was hct score on thc exam? 75+ 5+2.5 82.,5 X # Celculale 2-Su6res an4 how Revien and UAl Yaem



Answers

Find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as "minutes") in your results. (The same data were used in Section 3-I, where we found measures of center. Here we find measures of variation.) Then answer the given questions. Listed below are the measured radiation absorption rates (in W/kg) corresponding to these cell phones: iPhone $5 \mathrm{S}$, BlackBerry $\mathrm{Z} 30$, Sanyo Vero, Optimus V, Droid Razr, Nokia N97, Samsung Vibrant, Sony Z750a, Kyocera Kona, LG G2, and Virgin Mobile Supreme. The data are from the Federal Communications Commission. If one of each model of cell phone is measured for radiation and the results are used to find the measures of variation, are the results typical of the population of cell phones that are in use? $$\begin{array}{cccccccccc} 1.18 & 1.41 & 1.49 & 1.04 & 1.45 & 0.74 & 0.89 & 1.42 & 1.45 & 0.51 & 1.38 \end{array}$$

All right. So in today's example, we're gonna be looking at the high and low values to get our range as well as a measure for our variance and standard deviation for cell phone radiation emitted in watts per kilogram So are high value is going to be 1.55 and our low value is going to be 0.38 together. When we subtract them, we get a range of 1.17 watts or kilogram. Next up we can calculate the variance using this formula right here, what's unique about this is that we can use the average, which is denoted as X with a bar on top of it. And what we do to find that average is simply add up all the values we have and then divide by the number of observations right here And when we do that, we get an average of 0.9382 or a mean, the book refers to it. So what we do is we subtract each of our terms from the main and then square it. Once we add up all those differences with his summation, simple here, we divide by one less than our sample size. So put this into a couple of math numbers here, All we're gonna do is track each one of our terms. So the first one would be 0.38 -0.9382. We're gonna square that and add up the other 10 differences. And then once we divide that by 11 -1, we will end up with a standard deviation or sorry, variants, uh 0.18. And we are swearing this. So that means we have lots per kilogram squared. Not the most useful unit ever, but to make this easier to understand, we can convert this into a standard deviation by simply taking the square root of the variance. So are variants. Our standard deviation Is going to be equal to the square roots of zero. Plant 18. Yeah, And this is going to be equal to 0.423. Yeah. And the last question asked if we select one cell phone, is that going to be representative of the whole population of our cell phones that are in use? And the answer to this would simply be no. Um not only because we have a one sample of a population, but that not all cell phones will have the same types going uh making the same radiation. Mhm Yeah.

So to get the mean, all we have to do is that up all of 11 of our values. And once we do that, we're gonna get 10.34. And what do we do with this? 10, 34? Well, we just divided by the number of values we have, which is 11 And that will get us our mean, which is going to be 0.94. And our units are going to be watts are ws berger kilogram Yeah, this makes sense. Since we're talking about batteries and then we can find the median. So that is going to be the 6th value in order from least to greatest. So that will be our sixth. It's going to be in the middle. So that's six value is a 0.92. And we go for the sixth value because that's the one in the middle. Uh it would be five away from the first and five away from the 11th value. So that's how we get that median. And for the mode, which one appears the most often we don't have one. So that does not exist. And finally, our mid range, we just have to add our smallest value, which is 0.38 to our largest value of 1.55, divide that by two. And we got a mid range of 0.965. And what can we say about this? We're looking for the least amount of radiation emissions. So we're gonna go with the lowest value that we found. Our average was 0.94 and the meeting was 0.92. So we'll go with that, Yeah.

A question number 30 five number. Somebody order the that. The values from smallest to largest we have the smallest value is 0.51 the largest value is 1.49. So the minimum we have, it's 4.51. Number of the data values is hot. We can say that the median is the middle value off. The sort that I set is AM equal. Q two equals 1.38. The first quarter is the meeting of state of values, with those the median or at 25% cute name Cuban equal to oh, point 89. The third quoteit is the median of the data values above the median at 75% Q three go to 1.35. The maximum we have is 1.49 moving to the next. We have his book spot those cars from the box book at the minimum and maximum values the box start at the first court I at the circle tile and has a vertical line at the medium. The versatile is that 25% of assorted at a list. The meeting is at 50% and the third at 75%. So we have the vertical lines had the minimum value which is four point 51 the maximum values at 1.49. The million was at and the million is at 1.38 and the circle tile is at, and the third call tile is at one point or to try. And the first school tire is at four point you have in mind. We connect with the lines so you get and that's the point. That answer. Thank you.

So we're going to assume that for these non smokers, that those people who are exposed to second hand smoke, I have going to have equal lovers of Kolkata Ning, uh uh as those who are not exposed to secondhand smoke, alternately, we think that that exposed group is going to have a higher level. And We know that we need to find our test statistic. And since our sample size is 40, for each, will use the conservative measure. and we have the difference between our two groups, which we look at these numbers and they look different. However, we look at the samples standard deviations, we find they are very large, so 138.08 sq divided by the sample size, And then the 62.53 squared Divided by the sample size. And when we do that calculation, we find that that test statistic comes out to be 1.8455. And so we're assuming that the difference between the two groups Is equal to zero. But we're getting a test statistic that's up here at like 1.8455. And again, this is the test statistic. The actual difference is like 40 something, but we want to find this which will be our p value. So what is the likelihood if these two are equal or their differences? 0? What's the likelihood of getting a test statistic that is greater than or equal to this value. And I used my T C D E. F. To find that and I got a p value of 0.36 and that is definitely less than 5%. So a five significance level we would have evidence to reject to reject the null and claim that the mean of the experimental does seem to be higher is higher than that uh group of non I said, experimental. I mean, the exposed group than the non experimental group that they don't end up having exposure. Now let's find a confidence interval, the appropriate confidence interval. Because we're using five significance for a one tailed test, five down here for a confidence interval. We would also want five in the higher level. So we really want to find a 90 confidence interval. And again, I'm going to use this degrees of freedom of 39. Now, my table does not have that. I was using, it has degrees of freedom for 30 and 40. And probably we could very safely use this one for 40 if the degrees of freedom is 39. But I'm actually going to look up and use my inverse nor inverse T Button. So if I go to inverse T and go to second and distribution and I go to inverse T And I'm going to plug in an area of 25. Let me type the Button in the area .5. And then I'm going to type in my degrees of freedom of 39 and then find out what that gives me. It tells me that that lower T value is For 39° of freedom is negative 1.68. And we would round that to five. And I believe if you look up this when it's 1.684, so notice that these are very very close. So when we find that uh that difference and let's just find what this difference is. This difference if we subtract eight and then five becomes a three A three and the five minus two, Three becomes a two. And then these two have a difference of 44. So that difference between those is 44.23 plus or minus. And then we put our test statistic, R r Rt star value here and then we'll use that times the square root of and we have that first standard deviation was again very large squared over 40 Plus the 62.53 squared over 40. And let's find this margin of air first. Okay. And so I have that 1.685 times the square root of now enter this in 1 38.8 squared divided by 40 Plus that 62.53 squared divided by that 40. And I get that margin of air to be 40 0.38 And so let's find the two values and I'm going to just store that value as X. And so I have 44.23- the x value. And that gives me 3.85. And then I can go back and second entry and just change that subtraction sign to an addition sign And I get 84.61. And so we're 90 confident that the actual confident that the actual difference between those two is somewhere in here and notice it does not it doesn't include zero. Which means we definitely do not think that they're equal, means we think they are different. So in part C. It says, can we conclude we definitely look at the group that had exposed proof that they definitely have seemed to have a higher level of that uh that narcotic or the drug the nicotine offset of them, the non exposed group.


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