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4Adog breeder has asked you for advice: She would like to breed Alaskan huskies that run faster for sled dog racing events The table below shows the running speeds ...

Question

4Adog breeder has asked you for advice: She would like to breed Alaskan huskies that run faster for sled dog racing events The table below shows the running speeds (m/s) of 15 familles of dogs ! in the breeder'$ kennel:Note: Midparent values average trait values of fatherand mother; Midoffspring values average trait values of the ofispring:Running speed of Family Midparent (m[s)_ 113 45 13.8 12.6 29 8.8Running speed of Midoffspring (mIs) 10.9145 12.3 6,593145 9.6 29 13.6 11.4 12.5 48 14.953

4Adog breeder has asked you for advice: She would like to breed Alaskan huskies that run faster for sled dog racing events The table below shows the running speeds (m/s) of 15 familles of dogs ! in the breeder'$ kennel: Note: Midparent values average trait values of fatherand mother; Midoffspring values average trait values of the ofispring: Running speed of Family Midparent (m[s)_ 113 45 13.8 12.6 29 8.8 Running speed of Midoffspring (mIs) 10.9 145 12.3 6,5 93 145 9.6 29 13.6 11.4 12.5 48 14.9 53 9.1 {55 14.2 10.6 0 a) Use Excel Lo construct scatterplot of- midoffspring running speeds versus midparent running speeds (3 pts} Plot a regression line to show the correlation of the midoffspring values with midparent values (1 pt) Include the equation of the regression line on the scatterplot (1 pt) See lecture slides for examples ofthe scatterplot b) From the scatterplot in Q4(a} estimate the heritability (h2) ofrunning speed Round numbers t0 decimal places: (1 pt) c) Five pairs of parents with the highest running speeds were selected as breeders. Their mean ngspereds were: 14.9 m/s,145m/s,13,8 m/s, 13,6 m/s,and 12.5 m/s Assuming the Roand unning speed ofthe population is 10.71 m/s Calculate the selection differentiai (S) Round numbers to 2 decimal places: (2 pts)



Answers

A dog breeder has asked you for advice. The breeder keeps Alaskan huskies, which she races in sledding events. She would like to breed huskies that run faster. The table below gives data on the running speeds $(\mathrm{m} / \mathrm{s})$ of 15 families of dogs in the breeder's kennel.
$$\begin{array}{ccc} \text { Family } & \text { Midparent } & \text { Midoffspring } \\ 1 & 12.7 & 10.8 \\ 2 & 7.6 & 8.0 \\ 3 & 14.4 & 8.0 \\ 4 & 4.3 & 9.7 \\ 5 & 11.3 & 6.6 \\ 6 & 12.5 & 6.2 \\ 7 & 8.9 & 12.5 \\ 8 & 8.2 & 7.4 \\ 9 & 6.3 & 3.4 \\ 10 & 12.7 & 6.7 \\ 11 & 13.9 & 7.9 \\ 12 & 7.3 & 13.6 \\ 13 & 5.9 & 7.4 \\ 14 & 12.8 & 12.1 \\ 15 & 12.5 & 11.3 \end{array}$$ a. Use a piece of graph paper to prepare a scatterplot of midoffspring values versus midparent values. Approximately what is the heritability of running speed in the breeder's dog population? b. If the breeder selectively breeds her dogs, will the next generation run substantially faster than the dogs she has now? c. $W$ hat else would you suggest the breeder should try if she wants to win more races?

So in this problem, we are asked to make a dot plot for the data and problem. 15 regarding the finish, time and number of hours for the sled dog race. So looking at that data, um, and then knowing that a dot plot is another display technique that's similar to a history, Graham. But it's the data. Values are displayed along the horizontal axis, and the needle is plotted over each data value in the data set. So going up to the data and problems 15, we can see that it is already arranged from, So we're just going to make our part. This is finished time. So how to make a dot plot? You would just do each you can do ranges so we can look for our lowest value, which we see would be about 2. 50. so let's go out and start our lowest value at 200 2, 50 300 3, 50 and 400 because in the dot plot you want to put over where it is, but it doesn't have to be exact. You don't have to have a mark for each one, so now we'll just go down the list and start putting in our values. We have to 61 2, 71 2, 36. Yeah. 2, 44 2 79 2 96 2 84 2. 99 28. Another one at 2 88. 2. 47 2. 56. So I'm just for all of these data points. And as you can see, it starts to build kind of a mound where the peak of the distribution would be. So you just want to go ahead and put dots where all the values would follow your line, and that's how you create a dot thought. And this is just kind of a different way to visualize. It still shows you the frequency in the same way that a history Graeme Wood. But instead of lumping all values into a A bar, um, you would see each individual data points, so this could be nice if you care about visualizing each individual data point rather than just lumping them into groups

Hello. We're talking about squirrels and how they're all spring are affected by in index off pine cones. You can think of the schools as you Why? And the index of Punk O's as yours show. Given the information that is given, we can write that a white hat we're trying to estimate the offspring of the squirrels is going to be equal to the Y in Jersey, which in this case is one point for 146 And then this number would be how many offspring of squirrels you would have given? Uh, there are in any index of pine cones you into that. We're gonna add 0.4399 which this is going to be the soap off our line. And now the X is going to be the index. A pint goes no. The red is the fostering well, squares. And the 1.41 for six is your intercept does number. Here is the slope in this X is the index. Oh, plan comes now. We won't want to be, and in be they give us a graph. Let's say it looks like this. We have Blanco's, and then we have the residuals here, and we have a line with Slope zero. And they were just gonna have thoughts just random all around. So as you can tell, there is no pattern in dots show. That's a good thing. Um, and because of that lack of pattern, we could do least squares regression. So this graph tells us how good the data fits the model so we can say that data fit's the lean year model. Well, okay, now we move on to part C, an import. See, we want to know what s is so what is s as it's also known as the standard error. And this tells us how far apart predictions are from our regression line. And the number given is your 0.600309 and are square. This is a coefficient of the termination and what this tells us. ISS The variance explained by the model divided by the total there. So how much of various is explained and it's ah percentage the number is 57 going to percent

Yeah. In this problem, we're comparing the weights of wolves from two different regions. Once you've verified the summary statistics and we can discuss, we're going to be using a T distribution for this problem because the data is quantitative and we don't know the population standard deviations. All we have are two simple random samples from two separate populations. So we're gonna be using this formula for quantitative confidence intervals for the difference of And I've already plugged in our summary statistics. The one thing that I have included in here is for an 85% confidence level are critical T score with the degrees of freedom of nine is 1.574. Now, when we go to do our calculations here, the difference in the weights of the two groups is 8.97. You just, And then for our standard DVD for our margin of error, it's going to be 5.289. So when I add and subtract the margin of error from our middle, we're going to get 3.681 on the low side and 14.259 on the high side. And what this means is we are 85% confident that the Chihuahua Wolves Weigh on average between £3.68 more and £14.259 more than the wolves from the Durango Region, you think? Mhm.

Then Question eight, given the mean is 27.2. Standard deviation is equal 4.3. So it's as if the score is equal to 30 minus 27.2 over 4.3, which approximately 0.65 But please, the value in the intervals corresponding to, say, the value and the place extras that is smaller than open 65 Ah, uh, you know, 27.2. But here signal 4.3 so that this court is for 19 minus 27.2 over 4.3, which approximately negative. 1.91. So negative 0.91 smaller than which in C Escort is 30 to minus 27 point 2/4 27.0.3. Which folks, um field but 1.12 Justin score for the other fairly 55 minus 27 point 2/4 27.0.3, which approximately equals one point 81 So I'm going to want to his lives between 1.81 and 1.12 For questions, D X is equal to 27.2, minus 2.17 off the black 4.3. There's already is 17 point 869 So 17.869 is smaller than X do. To me. The X is equal to 27.2 plus 1.28 multiplying by 4.3, which is equal to 52.744 So X is smaller than 52.704 with chamber um access equal to 27.2, minus 1.99 times 4.3, which is able to 18.643 and the other, very for the equality is 27.2 plus 1.44 times 4.3, which is 33.392 So exploits between the two right you on. This score here is equal to 14, minus 27.2 over some division 4.3 is approximately 3.47 So the animal is unusual. Small if that school isn't the negative, too. So within two Standard Division, and thus this phone is unusual, sport. Uh, it's, um, new is 27.2. Standard Division is 4.3, so the animal is unusual. Large existing school is more than two standard giving. Um So the score off a score off three correspondent was unusual, Lord. So while other score off here and negative food on so three is unusual Lord, your a negative, Toto.


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