Biometrics 301 Fall 2000 Exam II(total # points = 55)Question 1 (4 points) An avian community that utilizes a pe15% robins,5% catbirds,7% chickadees,6% starlings,10% crows, and57% other less abundant species.Answer the following questions:MinutesQuestion 8 Recall the experiment involving singing times ofMaximum 1920.0Median 1410.0Minimum 1180.0Mean 1444.62Std Dev 183.13N 39Variance 33536.032CV 12.677Biometrics 301 Fall 2000 Exam II NAME: SSN: Section: (total # points = 55) Question 1 (4 points) An avian community that utilizes a pear orchard in West Virginia consists of the following species in the amounts indicated: 15% robins, 5% catbirds, 7% chickadees, 6% starlings, 10% crows, and 57% other less abundant species. Answer the following questions: 1a. In a single selection, what is the probability of randomly selecting a crow? a. 0.15 b. 0.05 c. 0.10 d. 0.57 1b. In two selections, what is probability of randomly capturing two crows? a. 0.20 b. 0.01 c. 0.10 d. 1.00 1c. In a single selection, what is the probability of randomly capturing either a starling or a chickadee? a. 0.16 b. 0.0042 c. 0.42 d. 0.13 1d. In a single selection what is the probability of capturing a bird other than a robin or a catbird? a. 0.20 b. 0.0075 c. 0.9925 d. 0.80 Questions 2, 3 and 4 Consider an experiment in which the researcher is observing the behavior of ghost crabs found on beaches in Maryland. Ghost crabs live in the upper reaches of sandy beaches in areas which are inundated with tidal water only infrequently. They burrow into the sand but leave the burrow periodically to feed. Of interest is the number of times a crab leaves it burrow in a six hour period and the length of time a crab forages when it leaves the burrow (time from leaving to returning). 2. (2 points) State the variables being studied and their type (categorical? discrete? continuous?). number of times a crab leaves its burrow - discrete length of foraging time - continuous3. (3 points) Suppose the probability distribution for X = foraging time (minutes) is given by the following density curve: 1/25 Density 0 25 Minutes What is the probability that a randomly selected crab will forage for more than 10 minutes? Pr (X>10) = area under the curve between 10 and 25 minutes = (25-10)*(1/25) = 15/25 = 3/5 = 0.60 4 (6 points). Suppose a random sample of 50 ghost crabs is taken from randomly selected beaches where crabs are found. From such a sample we can calculate the sample mean foraging time. What is the sampling distribution of the sample mean foraging time? Explain your answer including any assumptions you made. (You may assume that the population of foraging times for ghost crabs has a mean of 12.5 minutes and a standard deviation of σ=7.22 minutes). n=50 The sample mean will have a normal distribution with mean = 12.5 minutes and standard deviation = 7.22 / sqrt(50) = 7.22 / 7.07 = 1.021 The assumptions are 1) sampling is random and independent, 2) the sample size is small relative to the population size (< 5%) , and 3) the sample size is large enough to invoke the central limit theorem (n >30).Questions 5 and 6. Water samples are regularly collected from a monitoring station in the Choptank River and dissolved oxygen (DO in mg/l) is measured. The population of DO values for this site in May are normally distributed with a mean of 8.0 mg/l and a standard deviation of 2.6. 5 (5 points). What is the probability that a single sample of water in May will have a DO level between 7 and 12 mg/l? Show all work please. Pr (7 < X < 12) = Pr (X < 12) – Pr (X < 7) = = Pr (Z < (12-8) / 2.6) – Pr(Z < (7-8) / 2.6) = Pr (Z < 1.54) – Pr (Z < -0.38) = 0.9382 – 0.3520 = 0.5862 6 (5 points) Suppose a random sample of three bottles of water are taken and the DO level in each is measured. What is the probability that the mean of the three samples will fall below 4.8 mg/l? Pr(Xbar < 4.8) = Pr(Z < (4.8-8) / (2.6/sqrt(3)) = Pr(Z < -3.2 / 1.501) = Pr(Z < -2.13) = 0.0166Question 7. A new antibiotic drug (call it Drug A) for a certain type of bacterial infection is under study. 7a (5 points). In the original clinical trial to determine the efficacy of the drug, a random sample of 100 patients was given the drug. Of those, 80 were cured of the bacterial infection. Using these data, construct a 90% confidence interval for the true probability that a patient would be cured of the infection. Interpret your interval for the managers at the company. 90% C.I. = p ± z-score • √ {(p(1-p)/n} p = 80/100 = 0.80 z for 90% confidence = 1.645 So, a 90% interval is 0.80 ± 1.645 √ {(.8(1-.8)/100} = 0.80 ± 1.645 (0.04) = 0.80 ± 0.0658 based on the data, we are 90% confident that the true proportion of patients cured of their bacterial infection when taking Drug A lies between 0.735 and 0.865. 7b (6 points). Assuming that the true population proportion of successes (patients cured) to be 76%, what is the sampling distribution of the sample proportion of patients cured by the drug in a clinical trial involving 100 patients? Explain your answer. The sample proportion for a sample of 100 patients given the drug has a a normal distribution with mean = 0.76 and standard deviation = √ {(.76(1-.76)/100} = 0.0427 The assumptions are 1) sampling is random and independent, 2) the sample size is small relative to the population size (< 5%) , and 3) the sample size is large enough to invoke the central limit theorem (nπ and n(1-π) are both 10 or larger). Here 100*0.76 = 76 and 100*(.24) = 24 so the central limit theorem can be applied.7c (4 points). Another drug company has developed an antibiotic (Drug B) that they believe to be as good as or better at curing the bacterial infection than Drug A. They intend to run a clinical trial to demonstrate the success rate of Drug B. If they wish to estimate the population proportion of patients cured when given Drug B using a 95% confidence interval with no more than a 2% margin of error, how many patients should they use in their clinical trial? 95% confidence gives z = 1.96 ME = 0.02 Guesstimate of π for Drug B is 0.76 since they feel confident that it is as good or better
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