Modeling data distributions #7 - Questions and Answers

Question 1 a) To calculate: The least squares regression line for the data points using the table given below. Fertilizer x 100 150 200 250 Yield y 35 44 50 56 b) To calculate: The approximate yield when 175 pounds of fertizers were used per acre of land. Answer Calculation: The regression quation can be obtained using Maple. Maple Command: restart: with(plots): with(CurveFitting): with(Statistic): a = [100, 150, 200, 250]: b = [35, 44, 50, 56]: q = evalf (Least Squares(a, b, x)), 22.10000000 + 0.1380000000 2 This implies that the regression equation is y = 0.138x + 22.1 b) The regression equation is y = 0.138x + 22.1 Substitute for x in the model to obtain the desired value. y = 0.138x + 22.1 = 0.138(175) + 22.1 = 24.15 + 22.1 = 46.25 Question 2 How do frequency tables, relative frequencies, and histograms showing relative frequencies help us understand sampling distributions? They show each data point in the sampling distribution and how it relates to the overall distribution. They provide insight into the overall behavior of the population, therefore helping us understand the sampling distribution. They help us visualize the sampling distribution by using tables and graphs that approximately represent the sampling distribution. They help us visualize the sampling distribution by using tables and graphs that approximately represent the population distribution. Answer When visualizing data, the graphical representation is a very helpful phenomenon. The magnitude and spread of the data, as well as any distribution patterns, are shown in the frequency and relative frequency tables and the graphical representation. Therefore, the correct statement is, They help us visualize the sampling distribution by using tables and graphs that approximately represent the sampling distribution. Question 3 In each of the following situations, the sampling frame does not match the population, resulting in undercoverage. Give example of population members that might have been omitted. a) The population consists of all 250 students in your laege statistics class. You plan to obtain a sample random sample of 30 students by using the sample frame of students present next Monday. b) The population consists of all 15-year-olds living in the attendance district of a local high school. You plna to obtain a simple random sample of 200 such residents by using the student roster of the high school as the sampling frame. Answer (a) Nonresponse bias is the results of not having data for everybody in the sample. A simple random sample of 30 students was drawn from the students present in the statistics class next Monday. However, some students do not come to class or some might have another class at the same time (to which they choose to go to) or some might be il, and thus all students in the class do not have the same chance of being selected. This then means that the sample suffers from undercoverage. (b) Nonresponse bias is the results of not having data for everybody in the sample. A simple random sample of 200 residents was drawn was drawn from the student roster of the local high school, where we are only interested in 15-year-old. However, some 15-year-olds will no go to that particular local high school, because they might go to some private school or are homeschooling. This then means that the sample suffers from undercoverage. Question 4 √2π This exercise requires the use of a graphing calculator or computer programmed to do numerical integration. The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the function p(x) = e-(2-4)²/(202) where π = 3.14159265... and sigma and mu are constants called the standard deviation and the mean, respectively. Its graph(for o = 1 and μ = 2)is shown in the figure. With σ = 5andµ = 0, approximate foto p(x) dx. Answer Here, XN (M, 52) +00 -X-1/2/282 top(x)dx, Where P(x) = 12 1/9/252 To find fo XER. 0 P(x) is the PDF of x +00 => P(x)dx = P [x>0] x~N (M. 82) XM ~N/0.2) 0 For M=0, 5=5M중~~ (0,1) e Then, P [x>0] = P[570] = P[Y>0] =1-P[Y≤0]=1-(0) << 1-0.5=0.5