Faculty of Science FINAL EXAMINATION. MATHematics 203 Principles of Statistics I Tuesday, December 6th 9 a.m Noon

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1 Student Name: Student Number: Faculty of Science FINAL EXAMINATION MATHematics 203 Principles of Statistics I Tuesday, December 6th 9 a.m Noon Answer directly on the test (use front and back if necessary). Calculators are allowed. One two-sided sheet of notes is allowed. Language dictionaries are allowed. There are 16 pages to this exam and 3 pages of tables. The total number of marks for the exam is 100. Examiner: Professor Russell Steele Associate Examiner: Professor Masoud Asgharian 1

2 Question 1: (8 points) The amount of lead in a certain type of soil, when released by a standard extraction method, averages 86 parts per million (ppm). A new extraction method is tried on 40 specimens of the soil, yielding a mean of 83 ppm lead and a standard deviation of 10 ppm. Test the hypothesis that the new method yields less lead than the old method (choose any reasonable significance level). (8 points) 2

3 Question 2: (6 points) Assume that Y is normally distributed with mean 0.05 and variance Calculate the following probabilities: (a) Pr(Y > 0.05) = (2 points) (b) Pr( 0.40 < Y < 0.50) = (2 points) (c) Pr(Y = 0.05) = (2 points) 3

4 Question 3: (16 points) A Master s student in Environmental Engineering was studying the effect of two types of treatment on bacterial growth in drinking water. The two treatments were UV40 (ultraviolet light) and PAA+UV20 (phenylacetic acid and ultravoiolet light). Each day the student sampled water from the same source, poured it into two containers and then randomly assigned a treatment to each container. The log bacterial growth was then measured after three hours of being exposed to a light source for each container. The student did this on 33 separate days. Assume the days were far enough apart so that the samples were independent and different for different days. The goal of the analysis is to see if there is a difference between treatments with respect to suppressing bacterial growth. Log Growth by Treatment Difference btw Trts by Day UV40 PAA2+UV UV40 PAA+UV20 Diff by day (UV40 - PAAUV20) Min %ile %ile %ile Max Mean StDev

5 (a) Test the hypothesis that there is no mean bacterial growth after 3 hours for each treatment individually. (8 pts) (b) Is there anything that makes you doubt either of the results you observed in part (b)? (2 pts) 5

6 (c) Test the student s research hypothesis that the PAA+UV20 treatment allows for less bacterial growth than the standard treatment, UV40. (6 pts) 6

7 Question 4: (8 points) A pathologist is doing research on colo-rectal cancer treatments. She measures levels of a particular protein (VEGF) in tumour biopsies of 90 patients and then waits until they undergo a colon resection to determine whether or not they respond to a new treatment. The goal of her analysis is to determine whether one can use VEGF to predict responsiveness to treatment. The protein measurements are categorized into two categories: Low and High. Here is a table of her results: Responded to Treatment VEGF Level No Yes Total Low High Total (a) Do you believe that patient level of VEGF affects patient level of responsiveness to treatment? Test the hypothesis at α =0.05. (6 points) (b) Do you trust the results in part (a)? Why or why not? (2 points) 7

8 Question 5: (10 points) What relationship exists between the standard normal distribution and the boxplot methodology? The answer depends on the true underlying probability distribution of the data. Assume for the remainder of this exercise that the distribution is normal. Remember that the inner fences extend 1.5 IQR from the 25%ile and the 75%ile and the outer fences extend 3.0 IQR from the 25%ile and the 75%ile. (a) Calcluate the values of the standard normal random variable z, call them z L and z U that correspond to the hinges of the boxplot (i.e. the lower and upper quartiles, Q L and Q U ) of the probability distribution. (3 points) (b) Calculate the z values that corespond to the inner fences of the boxplot for a normal probability distribution. (2 points) 8

9 (c) Calculate the z values that correspond to the outer fences of the boxplot for a normal probability distribution. (2 points) (d) What is the probability that an observation lies beyond the inner fences of a normal probability distribution? The outer fences? (3 points) 9

10 Question 6: (10 points) An association of Christmas tree growers in Indiana sponsored a sample survey of Indiana households to help improve the marketing of Christmas trees. A simple random sample of 500 housholds was contacted by telephone and asked several questions in a 2-minute interview. One question was Did you have a Christmas tree this year? Of the 500 respondents, 421 answered Yes. (a) Give a 95% confidence interval for the population proportion of people who will have a Christmas tree. (3 points) (b) Give a 90% confidence interval for the population proportion of people who will have a Christmas tree. (3 points) (c) Determine the sample size needed to obtain a 99% confidence interval of 1% length TOTAL for this problem. (4 points) 10

11 Question 7: (10 points) Physical fitness may be related to personality characteristics. In one study of this relationship, middle-aged college faculty who had volunteered for a fitness program were divided into lowfitness and high-fitness groups based on a physical examination. The subjects then took the Cattell Sixteen Personality Factor Questionnaire. Here are the results for the ego strength personality factor: Fitness n x s Low High (a) Test the hypothesis that the personality score depends on fitness. (5 points) Now, someone else did a different study and got somewhat similar results with a different sample of people: Fitness n x s Low High (b) Test the hypothesis that the personality score depends on fitness for this population (you do not have to worry about multiple testing errors for this problem). (5 points) 11

12 Question 8: (8 points) Many assume that IQ is distributed normally. One common IQ test (the Wechsler Adult Intelligence Scale) is approximately normal with mean 100 and standard deviation 15. Assume also that an IQ of 140 or higher means that a person will definitely love statistics and possibly consider a career in the field. (a) What is the probability that a randomly selected person will love statistics (just based on their IQ score, not the quality of their professor)? (4 points) (b) How many people in a sample of 10,000 will love statistics (again, just based on their IQ score, not the quality of their professor)? (4 points) 12

13 Question 9: (12 points) A bored statistician wants to do a test on his name in the phone book. He wants to test his research hypothesis that less than 5% of the population has a name he considers to be cooler than his own. He does this by opening a Montreal phone book to 30 random pages and taking a random name from each page. (a) What is the null hypothesis of interest (2 points)? (b) Why shouldn t the statistician feel comfortable using a standard test of proportions to test his hypothesis? (3 points) 13

14 (c) The statistician creates a different type of test to get around this problem. He will reject his null hypothesis if and only if he observes NO people with a name cooler than his in his sample of 30. What is the probability that he makes a Type I error with this type of a test? (4 points) (d) The statistician, just to make sure that he knows how cool his name is, also grabs phone books from Toronto, Ottawa, Vancouver, and Quebec City and repeats his experiment in those 4 cities too (assume all 5 cities are independent). What is the probability that he makes at least one Type I error with his 5 phone books? (3 points) 14

15 Question 10: (12 points) Experimental design can be a very sticky subject for researchers from an ethical standpoint. It is not very ethical to allow for patients to continue to receive a treatment for a serious disease when you know that treatment has been unsuccessful in a large number of patients. Therefore, often researchers will design experiments that will allow patients to switch to the other treatment if they are not responding to the treatment that they were initially assigned to. For this particular experiment, let s assume the following: Patients are randomized to Treatment A and Treatment B at the beginning (probability of receiving Treatment A is 0.5 and the probability of receiving Treatment B is 0.5) Patients are observed at two timepoints in their treatment At the first timepoint, if a patient has NOT responded to their treatment, they are switched to the other. If they have responded, then they are NOT switched to the other treatment Once patients have responded to a treatment at one timepoint, they will ALWAYS be responders (i.e. they can never relapse ) Assume that patients respond to treatment according to the following table (at each timepoint, the probability of responding to a treatment is the same): Treatment Pr(Respond) A 0.10 B 0.30 (a) What is the probability that a patient will receive Treatment A at Time 1 and Treatment B at Time 2? 15

16 (b) What proportion of patients will respond to the treatments overall? (4 points) (c) Given a randomly selected patient who responded, what is the probability that they responded to Treatment A, i.e. that Treatment A cured them? (4 points) 16