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EnteredAnswer PreviewResultMessage0.02720.0272incorrectThe conditions for large-sample are not met0.39380.3938incorrectThe conditions for large-sample are not metAt...

Question

EnteredAnswer PreviewResultMessage0.02720.0272incorrectThe conditions for large-sample are not met0.39380.3938incorrectThe conditions for large-sample are not metAt least one of the answers above is NOT correct:point) Some shrubs have the useful ability to resprout from their roots after their tops are destroyed Fire is particular threat to shrubs in dry climates _ as it can injure the roots as well as destroy the aboveground material: One study of resprouting took place in dry area of Mexico_ T

Entered Answer Preview Result Message 0.0272 0.0272 incorrect The conditions for large-sample are not met 0.3938 0.3938 incorrect The conditions for large-sample are not met At least one of the answers above is NOT correct: point) Some shrubs have the useful ability to resprout from their roots after their tops are destroyed Fire is particular threat to shrubs in dry climates _ as it can injure the roots as well as destroy the aboveground material: One study of resprouting took place in dry area of Mexico_ The investigation clipped the tops of samples of several species of shrubs_ In some cases, they also applied propane torch t0 the stumps tO simulate fire. Of 19 specimens of particular species resprouted after fire. Estimate with 95% confidence the proportion of all shrubs of this species that will resprout after fire. Interval: 0272 3938



Answers

Shrubs and fire Fire is a serious threat to shrubs in dry climates. Some shrubs can respraout from their roots after their taps are destroyed. One study of resprouting tack place in a dry area of Mexico. "The investigators randomly assigned shrubs to treatment and control groups. Ibey clipped the taps of all Har shrubs. They then applied a propane torch to the stumps of the treatment group to simulate a fire All 12 of the shrubs in the treatment group resprouted. Only 8 of the 12 shrubs in the control gratip resprouted.

Coach in mine. The normal requirements required that the number off failures, uh, in the number off successes input champions is greater than 10. So we know that there are eight successes in the second sample. Eso the crew is a control who so the normal requirements is not satisfying.

For this question, we have three different types of Vienna types. We have a big ass for single. They use a big D. For double. And they use SD for a super double. So the key to this is looking at the first cross they conduct were they are crossing a pure S. And a pure two. So the offspring from this cross results in all S organisms. So they're all going to have that single flower. This indicates typically some sort of Hamas, I guess a dominant and Hamas. Itis recess of nature. And since S. Occurs, you can figure that your pure S. Is going to be a dominant to the D. Trade, which is going to be recessive. So with this in mind I'm going to use a big S. For the single genotype. And I'll use little S. Is for the recessive or double genotype. So if we move down the line for across two, we have F. One cross F. One. So this is going to be this Os organism that came out of this first cross. So for this pure as we can imagine, this would be our homeless Vegas, dominant and pure D. Would be a homeless Vegas recessive. So the offspring we should get from this should all be hetero Ziegenfuss. So if we do a cross of two hetero sickest individuals, big S. Little s. Cross, big s little S. We should get about a 3 to 1 ratio of dominant to recessive. Because about 25% will be homeless. I guess dominant 50% will be hetero Z. Ghous but will display the dominant phenotype and about 25% will be homeless I guess recessive. And this matches up to the numbers that were given of about 78 2 27 which is about a 3 to 1 ratio. So this fits that are single and double finna types have this dominant and recessive relationship. So from here we have to look at the effects of this super double population. So for cross three we have a pure D. So little as little as cross this super double. And the results we get out of this is we're going to have about a 1 to 1 ratio of the super double to the de population. So how this works is I'm going to use an X. For this super double population. So if we imagine we have a large X. And a little X. And that only this large X. Phenotype is going to be displayed. The cross result of this. Third cross would be a big X. Little s about 50% of the time and about little X. Little s 50% of the time. And if we expect only the big X. To display this super double and the little X. To not, we would have this proper 1 to 1 ratio for four. We are doing a pure S cross this super double again, so Big S. Big S. Cross, Big X. Little X. And again we get about a 1 to 1 ratio of super double to the S. For genotype. So here you can imagine that you would have big X. Big S. Cross, Little X. Big S. So this super double is only going to appear as a phenotype if this large X trait is passed on with it. So here there's sort of a hierarchy to how these are determined first if a super double is present and is dominant, this is going to be displayed in the phenotype. First from there we're going to move on to the single flower genotype. So this is going to be a dominant expression. And then lastly we have the recessive expression of the double. So we move on to part or the cross five to see how this works. We have a pure D. Cross so little s. Little X. Across the result of Cross four of the SD project, which would be this big X. Little S. So here if we imagine that we carry over one trait and cross this to each, we would have about a 50 50 ratio. Or I went to one ratio of big X. Little S. And big S. Little S. And of course this hetero zika straight is going to display the dominant genotype whereas this other one will display the super double because it has the large dominant X. Trade. So here you should get 50% super double and 50% showing the big S. Genotype for the single. And that matches as well for the last cross six we are crossing a pure D. Organism with the four S. So the little X. Big S. So we have little S. Little S. X. Little X. Big S. So here you get again a 1 to 1 ratio or a 50 50 split of big yes little ass as well as little X. Little S. So the little X. Of the super double will not be displayed. And you should get a double phenotype for the big S. Little S. It's hetero zegas and it will display the dominant trait of single. So here you will get a 50 50 or a 1 to 1 ratio of big S. Two. Little S. So how this works is we have those three different prototypes. We have the super double which displays in an all or nothing display of the phenotype and is dominant to both the other singles or doubles. From there, we have the dominant trait so the big as big as for our single. And then lastly, we have the display of the recessive trait or the little s little s. Or double. So how this could occur is a super double organism. Could have a missing alil or it could have some part of the gene that's missing, so it's not going to display these same type of flower when it's passed on to these next organisms. However, these typically have to have a hetero zegas display or if they're both dominant, it's going to be sterile and not result in any fertile offspring. So here it has to do with some sort of activation or deactivation to display the single or double trait. Or it's going to present that single or the super double itself. So the origin for this is that we have this hierarchy of genetic display or the phenotype that the super double is greater than the single is greater than the double. What this does not explain, however, to keep in mind is that for some reason only the females are sterile. So although this hierarchy works, it does not explain the sterility of only the females. The other thing that is not well explained is that are super double has to originate from our double flower organisms or our little s little s. So the connection between these two gina types and Fiona types is not clear and is also not explained by this hierarchy. However, this is the hierarchy that works best to explain the genotype and phenotype relation between each of these three multi A lilic displays

So we got a sample size of 51 and when we plug it in to make a stem and leaf plot, we got a data that looks a little bit like this little spike right here and then we go down a little and then we see our normal curve robert up here. So we got to say that this is this is approximately normal and very good to do the T procedure on. So the next thing you can do is uh find in terms of that 90% confidence interval which is gonna go a little something like this, which is going to be given by this formula right here, your average plus or minus the significance level, half the significance level of the T statistic time to standard deviation divided by the square root event. So these values, you'll be in love fines from the calculator you use and then the square event it's gonna be square to 51 tea with a significant level of half that is going to be at while. If you're taking a alpha of 0.1, that means south to is gonna be 55%. So then you go to the back of the book yeah. To find that beautiful uh test statistic, make sure you're going to the tease. So it's gonna be table four and the degrees of freedom arguing given by it's like here. Mhm. No wait in minus one sums can be 21 at the 5% significance level. Here we go, 1.7 to 1. Mhm. I'll go and write that down. Okay. It was whatever the standard deviation is, the square to 51 X. Bar was ar minus at And then we're gonna come back the book here for the answer or the competence interval. We see that as 14 point 4636 to 16 or 15.6384 Okay. Yeah.

All right. Consider to buy new experiments. The 1st and 112 with observed successes are one equal 69 the second and two equals 1 40. With article 26. We want to construct a confidence interval for the different difference in population proportions. P one minus P two for which we need to proceed to the following three steps to complete. Step is to determine whether or not it's appropriate to the normal distribution to approximate P hat one minus P had to. The answer is yes. P hot one is our one over N one P had to is our to ever end to. And with RP hat one, he had to queue hat one equals one minus P hat one and so on. We see that in one piece at one and one that one and two P had two and then two. If you had to are all greater than five, which means the necessary conditions are met. Use the normal. Next for part B. Let's constructed 99% confidence interval given by the expression here on the right. We see that we have to find the margin of error. E is just given by the formula left, which is the critical Z score times square expression for 99% confidence RZ is 2.33 from a Z table on the floor and textbook plugging in RP hats to hats and Zeke is equals 0.132 And plugging into our interval gives point to 98 is less than 21 minus P two is less than 20.56 To finally, we interpret this to mean that since our confidence interval only contains positive values, we conclude at 99% confidence that we expect P one to be greater than P two.


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