1) Answer the following questions with one or two short sentences.

Transcription

1 1) Answer the following questions with one or two short sentences. a) What is power and how can you increase it? (2 marks) Power is the probability of rejecting a false null hypothesis. It may be increased by increasing sample size or using equal sample sizes (one or the other of these is fine). b) A husband and wife decide to have 6 children. What is the probability that they will have 3 children of each gender? (2 marks) Pr(X) = n!/{x!(n-x)!} p X (1-p) n-x For this question, we want X = 3 females (or males), and n=6 and p=0.5 So just plug those into equation for binomial distribution. Probability of 3 children of each gender is c) What term(s) best describes the distribution shown below? (1 mark) Skewed Left

2 2. Given a normally distributed population with mean μ = 8 cm and σ 2 = 16, what is the probability of randomly sampling an individual having a length x > 4 cm? (2 marks) You need the shaded area below. Also z = (x - μ)/ σ σ = 4, μ = 8. To get the shaded area convert to standard normal: z = (4-8)/4 = -1. To get the answer we need 1 pr[z > 1] = = So probability of sampling an individual > 4 is Given a normally distributed population with a mean length μ = 6 cm and σ = 2, what is the probability of randomly sampling an individual that falls into the range 3 < x < 7? (3 marks) z = (x - μ)/ σ μ = 6 cm and σ = 2 need pr[x<3] = pr [z>1.5] = need pr[x>7] = pr [z>0.5] = now subtract these from one. pr[3 < x < 7] = = Given a normally distributed population with μ = 6 and σ = 8, what is the probability of obtaining a random sample of n = 16 individuals with a mean x < 10? (3 marks) Since we are asked a question about a mean based on n =16, we need to determine the standard error of the relevant distribution which is given by σ x = σ/ n, so σ x = 8/ 16, σ x = 2.0 z = (x - μ)/σ x z = (10-6)/2 = 2 so we need pr[x > 10] = pr[z > 2] = but prob we want is 1 minus that so pr[x < 10]= =

3 5) You wish to determine whether the crown-of-thorns starfish (Acanthaster planci) is randomly distributed along a coral reef. You obtain set of photographs taken of randomly selected 5x5 meter sections of a reef, and you count the number of starfish in each (7 marks) Results: 150 photos had 0 starfish; 30 photos had 1 starfish; 20 photos had 2 starfish Use the most appropriate hypothesis test to address this question. Ho: starfish are randomly distributed along reef Ha: starfish are NOT randomly distributed along reef Include calculations in here, use a goodness of fit test to the Poisson distribution Pr (x) = e -μ μ x /x!, we must estimate the mean, x, and the standard deviation, s. n = 200 x = (0 x x x 20 )/200, x = 0.35 starfish in each photo x 2 = (0 2 x x x 20) = 110 So s 2 = ( /200) /199, or s 2 = 0.43 Num Obs num Poisson probs Expect number starfish e /0!= e /1!= = Get last one by subtracting from 1 Get last one by subtraction X 2 calc = (obs-exp) 2 /exp = ( ) 2 / ( ) 2 /9.73 X 2 calc = 18.98, df = =1 critical value of with 1 degrees of freedom, χ 1 2 = 3.84 Reject null hypothesis since calculated X 2 calc= is greater than the critical value χ 1 2 = 3.84 so the starfish are not randomly distributed The fact that the variance s 2 = 0.43 is greater than the mean, x = 0.35, indicates that the starfish have a clumped or aggregated or contagious distribution.

4 6) A geneticist collects a random sample to test whether the colour of mice follows the ratio of: 9/16 Agouti : 3/16 Black : 3/16 Yellow : 1/16 White. The geneticist observes the following numbers in the random sample: 40 Agouti, 20 Black, 8 Yellow, 7 White Conduct the most appropriate hypothesis test. (5 marks) Ho: proportion of coat colours is 9/16 Agouti : 3/16 Black : 3/16 Yellow : 1/16 White Ha: proportion of coat colours is NOT 9/16 Agouti : 3/16 Black : 3/16 Yellow : 1/16 White Include calculations in here Agouti Black Yellow White Total observed n = 75 expect 9/16x75 = /16x75 = /16x75= /16x75=4.69 Since there are too many expected numbers less than 5, I ll pool the white and yellow categories to Conduct the goodness of fit test Agouti Black Yellow&White Total observed n = 75 expect X 2 calc = ( ) 2 / ( ) 2 / ( ) 2 /18.75 X 2 calc = 3.37 df = 3 categories 1 = 2 critical value of χ 2 2 = 5.99 Since X 2 calc < χ 1 2 = 5.99, don t reject the null hypothesis. There is no evidence that the mouse colours deviate from the expected

5 7) Banana slugs (Ariolimax variabilis) show variation in colour, where some individuals are yellow and others are brown. A biologist sets out to test whether the two colour forms are equally frequent. In a random sample they obtain: 9 yellow slugs and 2 brown slugs. (5 marks) Conduct the most appropriate statistical test. Ho: proportion of brown slugs is 0.5 Ha: Conduct proportion a statistical of brown test slugs of the is not hypothesis 0.5 using the spaces provide below Include calculations in here Given that there are two categories and n is small use binomial test n = 11 Pr (x) = n!/(x!(n-x)!) p x (1-p) n-x, where p = 0.5, x = number of brown slugs So here we want the probability of obtaining a result as extreme or more extreme than 2 brown. Pr (x=0) = 11!/(0!11!) = Pr (x=1) = 11!/(1!10!) = Pr (x=2) = 11!/(2!9!) = Since this is a two-sided test, sum and multiply these by 2, to get the P-value P-value = 2 x P-value = We do not reject that null Hypothesis because our P-value is greater than We have no evidence that the proportion of brown vs yellow slugs differ

6 8) You wish to know if different species of bee are more likely to become contaminated with a pesticide. You randomly sample a number of bees of each of three species and perform a chemical test to determine whether each individual is contaminated with a neonicotinoid pesticide, or not contaminated. Your counts of the numbers of bees contaminated or not contaminated for each of the three species are tabulated below. Conduct the most appropriate statistical test (5 marks) Contaminated Apis melifera Bombus vagans Megachile pluto Yes No Ho: Pesticide contamination is independent of bees species Ha: Pesticide contamination is dependent on bees species Include calculations in here Total number of bees n = 200 Expected table Contaminated Apis melifera Bombus vagans Megachile pluto Yes No X 2 calc = (obs-exp) 2 /exp = ( ) 2 / ( ) 2 /23.75 X 2 calc =13.53 df = (nrows-1)(ncols-1) = (3-1)(2-1) df = 2 critical value with 2 degrees of freedom, χ 2 2 = 5.99 Since X 2 calc is greater than the critical value, we reject the null hypothesis. There is an association between pesticide contamination and bees species. Bombus vagans shows many fewer contaminated bees than expected (while the other two species have more contaminated then expected).

7 9) Write all the SAS programming statements (including the data) necessary to carry out the statistical test for the experiment in question 8 above. (5 marks). DATA BEECONTAM; INPUT SPECIES $ CONTAM $ COUNTS; DATALINES; APIS YES 60 APIS NO 40 BOMBUS YES 15 BOMBUS NO 35 MEGACHILE YES 30 MEGACHILE NO 20 ; proc freq; tables SPECIES*CONTAM /chisq; weight counts; run;