Biostatistics
1 BIOL 2300 Name____________________ Elementary Biostatistics, Homework 2, page 1 Due: Thursday, February 20, 2020 (first 10 min of class!) Directions: Write a report of your answers on separate sheets of paper, taking care to prepare a legible and organized report. Put your name on your report. STAPLE ALL PAGES TOGETHER. Answer questions in order and underline or draw a box around any final numerical answers. Show your work for all answers. Full credit will not be given to answers without work, and partial credit may be given when work is shown. Write any written answers or explanations clearly, and be as neat as possible. 1. You will need the same dataset used for problem 3 in homework 1 (the dataset obtained from the yahoo website with the company you selected). Use Excel to calculate the average and standard deviation of the close data column. Assume that these two numbers represent the population (=parametric) mean and population standard deviation, respectively, for the variable length (in cm) in a population of a species of fish. Attach a printout of the data to your homework and write down the ticker code on it. a. Calculate the probability of sampling at random a fish that is smaller in size than the value you would obtain by subtracting half the standard deviation from the average [x will be equal to: μ – (σ/2)] b. Calculate the probability of sampling at random a fish that is greater in size than the value you would obtain by adding half the standard deviation from the average [x = μ + (σ/2)] c. Calculate the probability of sampling at random a fish that has a size between the two values [x = μ – (σ/2), x = μ + (σ/2)] used in parts “a” and “b,” respectively d. Calculate the 25th and 75th percentiles of fish size for the population using the normal distribution table. e. Imagine that 5 individuals are sampled at random from this fish population. Calculate the probability that the average calculated will be less than the value: μ – (σ/3) NOTE: Assume the variable is normally distributed and use bell-shaped curve diagrams to shade the areas that correspond with the answers to questions “a” through “d” 2. Write the last two digits of your 1000 student ID number with a decimal point in between these two digits:____ Assume that this number is the average number of squirrels you come across when you walk from the Life Science Building (LS) to the Engineering Research Building (ERB). Tomorrow you plan to walk from LS to ERB and you wonder what it the probability of observing at most three squirrels. Assume that squirrels are randomly distributed and that each squirrel is an independent observation. NOTE: if your last two digits happen to be both 0, use 4.2 as the average number of squirrels encountered. 3. Write the last digit of your 1000 student ID number: __ You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a random sample of 9 individuals from this population: a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites. b. Calculate the probability that at least two individuals carry intestinal parasites. NOTE: you can still calculate “a” if your last digit is “0.” 1. The frequency distribution table for the provided data is given below: X Frequency 0 6 1 9 2 5 3 4 4 4 5 0 6 1 7 1 The frequency distribution for these data is given below: From the above graph, the distribution is unimodal. The distribution is not symmetric but skewed to right. There is no apparent outlier. 2. The average of the provided data is: Average Variance Standard Deviation = sqrt (Variance) Standard Deviation, Standard Error = Standard Deviation / Sqrt (n) Standard Error, 3. I choose the Tata Steel Limited (TATASTEEL.NS) to perform our data collection and summary preparation. TATA Steel Limited (TATASTEEL.NS) Date Open High Low Close Volume Adj Close 12/31/2015 260 260.8 256.95 259.8 4446600 254.0752 12/30/2015 256.4 261.8 255.1 258.7 6812700 252.9994 12/29/2015 255.8 257.15 253.35 255.3 3800500 249.6743 12/28/2015 263.95 264.3 253 254.1 5317600 248.5008 12/25/2015 263.35 263.35 263.35 263.35 0 257.5469 12/24/2015 266 266 260.6 263.35 3843700 257.5469 12/23/2015 262.25 266.4 262.25 264.45 7702900 258.6227 12/22/2015 259 261.8 256.55 257.4 4544200 251.728 12/21/2015 255.5 261.4 254.65 258.45 6195200 252.7549 12/18/2015 256.5 257.95 254 255.65 6944200 250.0166 12/17/2015 246.9 257.95 246.05 257.1 12112200 251.4347 12/16/2015 246.55 248.6 242.1 244.7 5759800 239.3079 12/15/2015 247 247.25 241.25 244.35 5218800 238.9656 12/14/2015 239 248.9 238.45 245 11314400 239.6013 12/11/2015 235.1 243.9 235.1 240.9 11922400 235.5916 12/10/2015 227.25 234.75 226.5 233 4804800 227.8657 12/9/2015 234 235.85 224.9 226.85 4727600 221.8512 12/8/2015 242.05 243.05 233.5 234.75 4545900 229.5772 12/7/2015 244.8 247.5 242.45 243.55 4936400 238.1832 12/4/2015 238.05 244.25 237.1 240.25 5538100 234.956 12/3/2015 240.15 243.5 236.8 240.15 7107800 234.8582 12/2/2015 239.95 245.9 238.1 243.85 8858300 238.4766 12/1/2015 230.8 238.5 230.3 237.55 6586500 232.3155 The values of the Close data are as shown: Date Close 12/1/2015 237.55 12/2/2015 243.85 12/3/2015 240.15 12/4/2015 240.25 12/7/2015 243.55 12/8/2015 234.75 12/9/2015 226.85 12/10/2015 233 12/11/2015 240.9 12/14/2015 245 12/15/2015 244.35 12/16/2015 244.7 12/17/2015 257.1 12/18/2015 255.65 12/21/2015 258.45 12/22/2015 257.4 12/23/2015 264.45 12/24/2015 263.35 12/25/2015 263.35 12/28/2015 254.1 12/29/2015 255.3 12/30/2015 258.7 12/31/2015 259.8 The five number summaries of the data can be calculated by arranging the data in increasing order as: S.No. Close 1 226.85 2 233 3 234.75 4 237.55 5 240.15 6 240.25 7 240.9 8 243.55 9 243.85 10 244.35 11 244.7 12 245 13 254.1 14 255.3 15 255.65 16 257.1 17 257.4 18 258.45 19 258.7 20 259.8 21 263.35 22 263.35 23 264.45 The maximum value of data = M = 264.45 The minimum value of the data = m = 226.85 There are 23 observations in the data, so the median would be the 12th observation from the beginning. So, Median = 12th term = 245 The first quartile is the center of the first half of the data. As there are 11 terms in each half, so the first quartile would be the 6th term from beginning and similarly, the 3rd quartile would be the 6th term from end (i.e. 18th term from beginning). So, first quartile = 6th term = 240.25 And, third quartile = 18th term = 258.45 So, the five point summary is - {226.85, 240.25, 245, 258.45, 264.45} The box plot for the given summary data can be directly plotted as shown below: 4. The given data is: Fatal Not Fatal No Seat Belt 1,563 161,220 Seat Belt 564 412,482 a) Probability that a seat belt is not worn b) Probability that a fatal injury occurs c) Probability that both a fatal injury occurs and a seat belt is not worn P (Fatal injury ∩ No Seat Belt) d) Probability of a fatal injury or that the seat belt was not worn or both. P (Fatal injury U No Seat Belt) e) Conditional probability that a fatal injury occurs given that a seat belt was not worn, f) P (Fatal injury ∩ Not wearing seat belt) = P (F) * P(Not wearing) Therefore, the events are not independent. Histogram Frequency69544011Frequency