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Exlercise 1 Data Interpretation EXERCISE DATA INTERPRETATION Tho pH scalo Quanba Nvch hovaodic & BI Roneon k The PHICa Qnons Irom 10.milh bor pH beina noit aodi...

Question

Exlercise 1 Data Interpretation EXERCISE DATA INTERPRETATION Tho pH scalo Quanba Nvch hovaodic & BI Roneon k The PHICa Qnons Irom 10.milh bor pH beina noit aodic &nd nalurpn Bling nore bask Aphd7e condeed naubal {nerher acidic nor basic)_ The PH ol Ihe sur- Houn environment can have a grea1 erect Anal Mrooes &8 ae *0 Qiot FuartnatnaQan ITable t chch shotslhe cuniemaola anaeina mum nuf mkroplal colonts 06147.40 904n9 eacn culhure anstnnMa quesnnnsennioyTable PH Versus Number of Microb

Exlercise 1 Data Interpretation EXERCISE DATA INTERPRETATION Tho pH scalo Quanba Nvch hovaodic & BI Roneon k The PHICa Qnons Irom 10.milh bor pH beina noit aodic &nd nalurpn Bling nore bask Aphd7e condeed naubal {nerher acidic nor basic)_ The PH ol Ihe sur- Houn environment can have a grea1 erect Anal Mrooes &8 ae *0 Qiot FuartnatnaQan ITable t chch shotslhe cuniemaola anaeina mum nuf mkroplal colonts 06147.40 904n9 eacn culhure anstnnMa quesnnnsennioy Table PH Versus Number of Microbial Colonies Number Colonles Observed Whal palrers 00 VDJ cosere based ne Inlontaton Taor 4 Davelop hypothests relating the pH level oltthe Q4te 740a 5 "na numdut 0amicrobu colonies Cosonveo each cul Mhal #ould Your orpcnnunlalappionch lcstamport- Mhal enulo tho Indopondonl and dcptnocnt vanabtt? Whal would be your control? Whal type grapn would apbopnae Enstat sn Wi Graph Ule dala tron Tabe Interprctha dallIron Ine grion YoJ made Oeann



Answers

Use the data in KIELMC for this exercise.
(i) The variable dist is the distance from each home to the incinerator site, in feet. Consider the model
log $($price$)=\beta_{0}+\delta_{0} y 8 l+\beta_{1} \log (d i s t)+\delta_{1} y 81 \cdot \log (d i s t)+u$
If building the incinerator reduces the value of homes closer to the site, what is the sign of $\delta_{1} ?$
What does it mean if $\beta_{1}>0 ?$
(ii) Estimate the model from part (i) and report the results in the usual form. Interpret the coefficient on $y 81 \cdot$ log (dist). What do you conclude?
(iii) Add $a g e, a g e^{2},$ rooms, baths, log(intst), log(land), and log(area) to the equation. Now, what do you conclude about the effect of the incinerator on housing values?
(iv) Why is the coefficient on log(dist) positive and statistically significant in part (ii) but not in part (iii)? What does this say about the controls used in part (iii)?

Hello today were given to questions where first we have to find out the difference off the pH value of the acidity between normal going for in the rainfall North niece in the United States that we have to find out the difference in the tolerance off the acid it be being to fish right just of pH values off the ideal moment, officials and for us. Let's look at it. Let me make a new screen right so were given to the pH. Rally off a rain normally is 5.5, and then we're given the pH value for in the northeastern United States. Ph Dash. Let's make it is 4.5, but if we find the difference between them years indifference, that would be one point up. That's a one point of difference, since it's for performance 55 But on the lower the pH is, the higher the acidity is. So since we're given that pH is not going to make weight, so that means an increase or decrease is ah ah. One unit is 10 fold, so an increase off 10 means 10 times less acid, very on decrease of 10 means 10 times more acid e And then please accuse of one I mean right. So since the difference here is one that means that Northeastern United States here is 10 times more acidic than normal estate. That means that the an interest a factor of 10 x unless treated the other question that as the pH only after fresh work. So we're given that the Ph volley off the yellow porch right ph of yellow perch is 4.0, and the pH volley off a common shiner. PH dash is 6.0, so our question basically years. Ah, our question basically is fit. Speech is more likely to survive in a more acidic and warm and bright. And since a shiner can tolerate the values below six, why no, But we know that a yellow porch can easily told her values off four point. Oh, so that means that legal a perch here yellow perch can survive and more in a more acidic environment, and the defense would mean the tolerance of boat is 2.0 p. H units. It's now, since Ph is logarithmic, this would be 10 race were too critical 100 So there's difference being 100 X and comes to Tournament of the PHR. Liza boldface

All right. So in number 13 are not 13. 19. Sorry. In 3.1, we're talking about pH and water in tap versus bottled water. So notice that the tap numbers are a little bit higher than the bottle numbers for the pH. So we're trying to determine the me meeting a mode of pH for each type of water in part A. So I already have the numbers listed in order from smallest to largest. That makes it easier to find the median later on. And so we have the mean is going to be adding up all of those numbers. So when you add up 7.1 all the way up to the 7.69 that should add up to 90 and there are 12 numbers in that list, so 90/12 is going to be 7.5. The median is going to be smacked up in the middle between those six between the six number. So 123456 So halfway between the 7.47 to 7.5, So you're gonna do 7.47 plus 7.5 over to that's going to give you a median of seven 0.485 Then mode is the number that shows up the most, and it turns out that 7.47 shows up three times, so the mode is 7.47 So now to continue in part A, we are going to look at the bottled numbers doing the same thing. We're going to get the mean median and mode. So when I add up all 12 numbers of the bottle data 5.2 all the way up to the 5.2 8/12, I add up all those numbers and you should get 62 0.33 And when you divide that by 12, you get 5.19 Now the median for this one is the one that shows up the most. 5.26 Oh, no. So that was the mode. So we'll go ahead and say mode real quick. I apologize. Mode is 5.26 It showed up the most, um, but then the median is halfway between the sixth and the seventh items. So when I go 123456 at the 5.23 and 5.24 Double check that. 123456 5.23 And the 5 to 4. I think I might have missed a number here. 5.25 point 95.13515 5.21 Well, there's 5.2 here. Okay, Apologize. There's a 5.2 here. So that means this 5.21 and 5.23 or the middle numbers. I apologize. Okay, so we add up here, we're going to add up the 5.21 plus 5.23 over to, and that's going to be 5.22 Mhm. All right. And then it says be supposed the pH of 7.1 in tap water was incorrectly recorded and it was recorded as 1.7. What will that do to the mean the median? And, um, what property will that median illustrate? And so, for part B. Um, because here's how you can do this without having to redo all the audition. You take 90. The total of the mean that you had before 90 minus 7.1, which is a difference of 7.1. And then you add the 1.7. So you take away 90 minus 7.1 because that was the bad one. And add 1.7, which is the error that will get you a new top number. So if you do 90 minus 7.1, add 1.7. You now have 84 point 6/12, which will get you seven point oh five. And so that's gonna be your new mean. The media is going to still be the same. Mhm. Yeah. So the median is unchanged and that term that they're trying to ask for is resistant. The median is resistant because it is unaffected by the extreme value.

For the match pairs given on the right. We want to conduct a sign tested matched pairs testing P does not equal 0.5 at alpha equals 0.5 significance. This question is testing our understanding of non parametric tests in particular how to conduct a scientist of matched pairs. We proceed there steps A through D below to solve. So first, in a we stayed alpha hypotheses, this gives alpha equals 0.5 H. And r P does not equal five. H. A P does not equal five. So the no hypothesis P is equal to have, the alternative is not A and B. We compute the test out. So are signs for these matched pairs are as follows and equals 12 is the total number of matched pairs. So X equals the number of plus or the total number plus and minus which is 4/12. Thus we have seen it equals x minus 5/45 over and equals negative 1.155 Thus the p value is for a normal distribution, P equals two PZ greater than zero equals 248 Thus we conclude that we fail to reject asian off because he is greater than alpha, which means that we lack evidence to support the claim P does not equal five.

Mm In this problem we want to get a 90% confidence interval for the difference between the phosphorus content of the two regions. Once you've verified our summary statistics for the two regions, the mean the standard deviation, then we can use our formula for confidence interval. It is I've already plugged in, our means are standard deviations in our sample sizes into this formula, the only thing I have left out is the T. Star, which we're going to get from the Table six of Appendix two in the back of your book, The smaller sample sizes 12. So our degrees of freedom is going to be one less than that at 11 And according to the chart at 90% T star is 1.796. Mhm Yeah, So now it's a matter just plugging using our calculator here is when we find the difference in the means that comes out to be 8.6 and then when we multiply our T star by this big square root, that's going to give us a margin of error, which is 1.796 times 72.33 which is 1 29.9. Yeah. And then when we add 8.6 and subtract when we take 8.6 and add 1 29.9 and 8.6 minus 1 29.9, we're going to get a confidence interval of negative 1 21.3 up to positive 1 38.5. So this tells us we are 90% confident the average difference and phosphorus content in the two regions is between those two numbers negative 1 21 3 and 1 38.5. Yeah. And notice since this interval is going from negative to positive, we're having different signs, negatives and positives in our interval. So this would tell us there's a good chance that there is zero difference between the two regions, so no region is is of any more interest than the other. And the last question we use the T. Start here, because this is quantitative and we don't know the population standard deviations.


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