Solutions M203E Practice Quiz 4 for Quiz 4 Friday Nov 18, 2011

Instructions: The quiz is open book (any books) and open notes (any notes or written material).

[1] A farmer who wants to assess the level of pest infestation in her orchard is considering several different sampling methods. For each of (a) through (d), indicate what sampling technique it corresponds to (choose your answers from among convenience sampling, simple random sampling, systematic sampling or stratified sampling).
(a) If a third of the trees are 5 years old, a third are 10 years old and a third are 15 years old, pick any sample of 12 trees as long as 4 trees are from each age range: quota sampling
(b) Pick the first 10 trees near the entrance to the orchard: convenience sampling
(c) Pick a random sample of trees from each of three age ranges: 0 to 3 years old, 3 to 6 years old and 6 or more years old: stratified sampling
(d) Each tree is numbered and a random sample of those numbers is chosen to select the trees to be inspected: simple random sampling

[2] Suppose 24000 fish from a lake were captured and tagged. A second sample of 10300 were later captured and 225 of these fish were found to be tagged. Given this data, how many fish altogether are estimated to be in the lake?

We expect the same fraction of fish to be tagged in the second sample of 10300 as are tagged in the total population N of fish in the lake. Thus 225/10300 = 24000/N so N = 24000*10300/225 or about 1.1 million fish.

[3] Assume a test for a certain disease, which has a prevalence of 10%, has a sensitivity of 70% and a specificity of 90%.  

(a) For people who have the disease, what's the chance the test is negative? Answer: 100% - 70% = 30%

(b) For people who do not have the disease, what's the chance the test is negative? Answer: 90%

(c) If a randomly chosen test subject tests negative, what's the chance the subject nevertheless has the disease? Show how you obtain your answer. Answer: Consider a random sample of 10,000 test subjects:

		Negative Test		Positive Test
Has disease	300 = 1000-700		700 = 70% of 1000	1000 = 10% of 10000

Free of disease	8100 = 90% of 9000	900 = 9000-8100		9000 = 10000 - 1000
								Total=10000
The subject has a chance of 300/(8100+300) = 3.6% of having the disease.

[4] IQ's are normally distributed with mean 100 and standard deviation 15. 

(a) What percentage of the population has an IQ at or above 145? Show how you obtain your answer. Answer: x = 145 is the same as z = 3 (i.e., 3 standard deviations above the mean), so the probability is (100-99.7)/2 = 0.15%.

(b) What percentage of the population has an IQ in the range 70 to 115? Show how you obtain your answer. Answer: This range is z = -2 to z = 1. The range z=0 to z=1 covers 68/2 = 34% of the population, and the range z = -2 to z = 0 covers another 95/2 = 47.5%, so z = -2 to z=1 covers a total of 47.5+34 = 81.5% of the population.

[5] Assume a newspaper runs a poll and finds p^ = 80% with a sample size of n = 900.

(a) What is the margin of error? Show how you obtain your answer. Answer: it is plus or minus 2s, where s is the square root of (p^(1-p^)/n) = (0.80*0.20/900); i.e., 0.0133, so the margin of error is plus or minus  2.7%.

(b) What sample size would be needed to obtain a margin of error of plus or minus  2%? Show how you obtain your answer. Answer: s is half the margin of error, so .01 = s is the square root of (p^(1-p^)/n) = (0.80*0.20/n), hence n = 0.80*0.20/0.01^2 = 1600.