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Statistical Analysis of Tick Experiment: Hypothesis Testing and Confidence Intervals, Study notes of Statistics

An example of a statistical analysis of an experiment involving ticks and a deer gland substance. The analysis includes hypothesis testing using a binomial distribution and calculating confidence intervals for the proportion of ticks going to the treated tube. The document also discusses the challenges of working with the binomial distribution and the importance of choosing the correct rejection region.

Typology: Study notes

Pre 2010

Uploaded on 09/02/2009

koofers-user-ye6
koofers-user-ye6 🇺🇸

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Download Statistical Analysis of Tick Experiment: Hypothesis Testing and Confidence Intervals and more Study notes Statistics in PDF only on Docsity! Example [Carroll, J Med Entomol 38:114–117, 2001] Place tick on clay island surrounded by water, with two capillary tubes: one treated with deer-gland-substance; one untreated. Does the tick go to the treated or the untreated tube? Tick sex Leg Deer sex treated untreated male fore female 24 5 female fore female 18 5 male fore male 23 4 female fore male 20 4 male hind female 17 8 female hind female 25 3 male hind male 21 6 female hind male 25 2 Is the tick more likely to go to the treated tube? 1 Test for a proportion Suppose X ∼ binomial(n, p). Test H0 : p = 12 vs Ha : p 6= 1 2 Reject H0 if X ≥ H or X ≤ L Choose H and L such that Pr(X ≥ H | p = 12) ≤ α/2 and Pr(X ≤ L | p = 1 2) ≤ α/2 Thus Pr(Reject H0 | H0 is true) ≤ α. The difficulty: The binomial distribution is hard to work with. Because of its discrete nature, you can’t get exactly your desired significance level (α). 2 Rejection region Consider X ∼ binomial(n=29, p) Test of H0 : p = 12 vs Ha : p 6= 1 2 at significance level α = 0.05 Lower critical value: qbinom(0.025, 29, 0.5) = 9 Pr(X ≤ 9) = pbinom(9, 29, 0.5) = 0.031→ L = 8 Upper critical value: qbinom(0.975, 29, 0.5) = 20 Pr(X ≥ 20) = 1-pbinom(20,29,0.5) = 0.031→ H = 21 Reject H0 if X ≤ 8 or X ≥ 21. (For testing H0 : p = 12, H = n – L.) 3 0 5 10 15 20 25 Binomial(n=29, p=1/2) 1.2%1.2% 4 Example X ∼ binomial(n=29, p); observe X = 24 Lower bound of 95% confidence interval: Largest p0 such that Pr(X ≥ 24 | p = p0) ≤ 0.025 Upper bound of 95% confidence interval: Smallest p0 such that Pr(X ≤ 24 | p = p0) ≤ 0.025 In R: binom.test(24,29) 95% CI for p: (0.642, 0.942) Note: p̂ = 24/29 = 0.83 is not the midpoint of the CI 9 0 5 10 15 20 25 Binomial(n=29, p=0.64) 2.5% 0 5 10 15 20 25 Binomial(n=29, p=0.94) 2.5% 10 Example 2 X ∼ binomial(n=25, p); observe X = 17 Lower bound of 95% confidence interval: pL such that 17 is the 97.5 percentile of binomial(n=25, pL) Upper bound of 95% confidence interval: pH such that 17 is the 2.5 percentile of binomial(n=25, pH) In R: binom.test(17,25) 95% CI for p: (0.465, 0.851) Again, p̂ = 17/25 = 0.68 is not the midpoint of the CI 11 0 5 10 15 20 25 Binomial(n=25, p=0.46) 2.5% 0 5 10 15 20 25 Binomial(n=25, p=0.85) 2.5% 12 The case X = 0 Suppose X ∼ binomial(n, p) and we observe X = 0. Lower limit of 95% confidence interval for p: 0 Upper limit of 95% confidence interval for p: pH such that Pr(X ≤ 0 | p = pH) = 0.025 =⇒ Pr(X = 0 | p = pH) = 0.025 =⇒ (1− pH)n = 0.025 =⇒ 1− pH = n √ 0.025 =⇒ pH = 1− n √ 0.025 In the case n = 10 and X = 0, the 95% CI for p is (0, 0.31) 13 A mad cow example New York Times, Feb 3, 2004: The department [of Agriculture] has not changed last year’s plans to test 40,000 cows nationwide this year, out of 30 million slaughtered. Janet Riley, a spokeswoman for the American Meat Institute, which represents slaughterhouses, called that “plenty sufficient from a statistical standpoint.” Suppose that the 40,000 cows tested are chosen at random from the population of 30 million cows, and suppose that 0 (or 1, or 2) are found to be infected. How many of the 30 million total cows would we estimate to be infected? What is the 95% confidence interval for the total number of infected cows? No. infected Obs’d Est’d 95% CI 0 0 0 – 2763 1 750 19 – 4173 2 1500 181 – 5411 14
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