As part of a new animal feed trial, veterinary scientists wish to compare the perfor-mance of a new formula with that of an old diet. 50 pigs are assigned to an identicalhandling procedure but half are randomly selected and given the new feed (feed =1) while the other half are given normal food (feed = 0). The weight of the animalsis measured at the beginning (preweight) and at the end of the trial (postweight).Both preweight and postweight are measured in kilograms, and the difference(veightdiff) is calculated using weightdiff = postweight – preweight. Theexperimenters begin their analysis by fitting the following regression model:weightdiff = B + Bifeed + Bapreweight + Bzfeed : preveight +efrom which the following R output is obtained:Coefficients:EstimateStd. Error t value Pr(>|tI)(Intercept)0.4388150.1815722.4170.0175 *feed4.8920930.25043019.535<2e-16 ***preweight0.0481430.001699 28.335<2e-16 ***feed:preweight -0.0292250.002308 -12.664<2e-16 ***(a) Using the R output above, interpret the linear relationship between an animalweightdiff and their preweight under the old diet and explain how thisdiffers when using the new feed. Sketch two straight lines representing the fit.(b) State the assumptions associated with a linear regression model. Using theplots provided at the end of the paper, in addition to any other relevant in-formation, discuss whether the modelling assumptions have been met. Clearlyexplain your reasoning. (c) The scientists wish to obtain a prediction and prediction interval for the likelyweight difference for an animal weighing 121 kilograms. They use the followingR command:predict (weighttrial.lm,newdata = data.frame (preweight = 121, feed = 1),interval = "confidence", level = 0.95)fitlwrupr7.620016 7.550133 7.6899Explain why this calculation is incorrect and explain what the consequence ofthis mistake would be. Interpret the interval that has been calculated.Residuals vs FittedNormal Q-Qo184500400-0204o OcgFitted valuesTheoretical QuantilesFigure 1: Residual plots for the model in question R1(b)Serpise pezpiepuRS8gerpseu

Answered: As part of a new animal feed trial,…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Researchers wanted to test the hypothesis that living in the country is better for your lungs than living in a city. To eliminate the possible variation due to genetic dif- ferences, they located seven pairs of identical twins with one member of each twin living in the country, the other in a city. For each person, they measured the percentage of in- haled tracer particles remaining in the lungs after one hour: the higher the percentage, the less healthy the lungs. They found that for six of the seven twin pairs the one living in the country had healthier lungs. Based on the sample size and distance between the null value and the observed proportion, estimate the strength of evidence: inconclusive, weak but suggestive, moder- ately strong, strong, or overwhelming.
In North America, between 100 million and 1 billion birds die each year by crashing into windows on buildings, more than any other human-related cause. This figure represents up to 5% of all birds in the area. One possible solution is to construct windows angled downward slightly, so that they reflect the ground rather than an image of the sky to a flying bird. An experiment by Klem et al. (2004) compared the number of birds that died as a result of vertical windows, windows angled 20 degrees off vertical, and windows angled 40 degrees off vertical. The angles were randomly assigned with equal probability to six windows and changed daily; assume for this exercise that windows and window locations were identical in every respect except angle. Over the course of the experiment, 30 birds were killed by windows in the vertical orientation, 15 were killed by windows set at 20 degrees off vertical, and 8 were killed by windows set at 40 degrees off vertical. . Clearly state an appropriate…
A survey by Men’s Health magazine stated that 14% of men said they used exercise to reduce stress. Could this be valid, or is it actually less than what the magazine claims? A random sample of 100 men was selected, and 10 said that they used exercise to relieve stress. If a=0.10, test a hypothesis and state your conclusion in context.
Determine your independent and dependent variables, which type of ANOVA would you choose, and how much replication would be needed. Farmer Joe has eight different plots of land. He applies two different nitrogen levels randomly to each of the 8 plots, low and high. In addition, four different strawberry varieties are randomized within the plots. He is measuring mass of fruit for his outcome. I would like to see if there is a difference in birth weight among cat breeds. I will measure 5 different cat breeds.
Answer the whole quesrion
Independent studies show that 26 out of 40 r/s full-time students favor a shorter semester system, while 38 out of 50 r/s part-time students favor the same. Construct a 98% c.i. for the difference between the overall percentage of full-time students favoring a shorter semester system and that of part-time students.
A demographer wants to know what proportion of women in different countries dye their hair. They take a random sample of women from Korea and another random sample of women from China. In Korea, 47% dyed their hair. In China, 56% dyed their hair. (a) What hypotheses would the demographer use to test if women in China are more likely to dye their hair than women in Korea? H0: pChina=pKorea vs. Ha: pChina ≠ pKoreaH0: pChina - pKorea = 0 vs. Ha: pChina - pKorea > 0 H0: μChina-μKorea = 0 vs. Ha: μChina-μKorea ≠ 0H0: pChina-pKorea = 0.09 vs. Ha: pChina-pKorea ≠ 0.09H0: μChina = 0.56 vs. Ha: μKorea = 0.47 (b) Suppose their test yielded a p-value of 0.079. What does this mean in the context of the test? We are 7.9% confident that the results of our test are correct.If the women of two countries have the same rates of coloring their hair, the probability of getting a difference at least this large is 0.079. If we take another sample, the probability of getting the same p̂China and…

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