Solved Problems on Statistics - Final Exam | MTH 243, Exams of Statistics

Download Solved Problems on Statistics - Final Exam | MTH 243 and more Exams Statistics in PDF only on Docsity! Math 243 Answers to Final Exam Practice Problems 1. Carbon dioxide baited traps are typically used by entomologists to monitor populations. An article in the Journal of the American Mosquito Control Association investigated whether temperature influences the number of mosquitoes caught in a trap. Six mosquito samples were collected on each of nine consecutive days. For each day two variables were measured: average temperature (degrees Centigrade) and mosquito catch ratio (the number of mosquitoes caught in each sample divided by the largest sample caught). a. Find the least squares regression line using Excel or your calculator. y = 0.0638x - 0.6432 b. What fraction of the variation of the catch ratio can be explained by the least squares regression line using the average temperature? Use Excel or your calculator to find the required value. R2 = 0.5861 c. If the average temperature is 18°C what is the predicted catch ratio? 0.638(18)-0.6432= 0.5052 The predicted catch ratio is for a temperature of 18°C is 0.51. d. [4] The biologist use the regression line to predict catch ratios. Use this scenario to explain what is meant by extrapolation. Average Temperature Catch Ratio 16.8 .66 15 .30 16.5 .46 17.7 .44 20.6 .67 22.6 .99 23.3 .75 18.2 .24 18.6 .51 Extrapolation involves attempting to predict a catch ratio for a temperature that is either lower that 15° Celsius, or for a temperature that exceeds 23.3° Celsius. e. Find the corresponding residual values for the temperatures 15°C and 18.6°C. 0.30 - (0.0638(15) - 0.6432) = -0.0138. The corresponding residual for 15°C is -0.0138. 0.51 - (0.0638(18.6) - 0.6432) = -0.0335. The corresponding residual for 18.6°C is -0.0335. 2. A 95% confidence interval about a sample mean indicates the probability of the procedure used to create the interval of actually containing the population mean. a. Two individuals both take a SRS from the same population, both working independently from each other. What is the probability that both 95% confidence intervals contain the population mean µ? Let the random variable X count the number of confidence intervals containing µ. P(X = 2) = 0 .952 = 0.9025 2b. Eight individuals take SRS from the same population, all working independently from each other. What is the probability that 6 of the eight 95% confidence intervals contain the population mean? P(X = 6) = 0.05146. I recognize this as a binomial situation. I want to compute the probability of getting exactly 6 successes out of eight attempts. I arrived at my answer by using binomdist(6,8,.95,false) I could also have used the actual formula ( ) ( ) ( ) 6 28!P(X = 6) = 0.95 0.05 6! 8 6 !− 3. It is believed that adult human heights follow an approximately normal distribution. Suppose that the mean of this distribution is 68 inches with a standard deviation of 5 inches. a. Tim Duncan the basketball player for the San Antonio Spurs is 7 feet tall. Approximately what percentage of people are taller than 7 feet tall? Let the random variable X measure the height of an individual. 84 68P(X > 84) =P 5 P(Z > 3.2) = 0.000687 Z − >    = b. If we randomly chose 4 people from the population what is the probability that average height is less than 72 inches? P( X < 72) = 72 685 4 P Z    − <       a. = P(Z < 1.6) = 0.945201 4. You are given that P(A) = .34, P(B) = .45, and P(A or B) = 0.68. Do not assume that events A and B are independent. Calculate P(A and B) = .34 + .45 – (.68) = 0.11 b. Are events A and event B independent? You must show evidence for your conclusion in order to receive credit. P(A)P(B) = (.34)(.45)= 0.153 Which does not 0.11 the probability of A and B. So the two events are not independent. 4c. [4] P(A| B) = 0.11 0.45 = 0.244444.