Question

Chapter 6/7 Assignment 1. Of the 67,410 surgeries tracked in astudy in the U.K, 2832 were followed by surgical site infections (Coello et al. 2005). a. Assuming the data are a random sample, determine a a 95% confidence interval for the true percent of surgeries resulting in a surgical site infection. Report your answer as a percentage rounded to 2 decimal places. For partial credit, make sure to indicate all the formulas and values you're using in your calculations. (2 pts) -=67410 = 2832 = =2832 67410 = 0.042 b. If the study had only followed 647 surgeries from the same population, would the confidence interval have been wider or narrower? Why?(1 pt) 2. If 48% of a sea turtle population are female, what is the probability I'd find 156 or more male sea turtles in a sample of 312 sea turtles? For partial credit, make sure to indicate all the formulas and values you're using in your calculations. (2 pts) 3. What is the most likely number of females I should find in this sample? Explain your reasoning? (1 pt)

Answer 1

Q1:

a)

n = 67410, x = 2832

p̅ = x/n = 0.042

95% Confidence interval :

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.042 - 1.96 *√(0.042*0.958/67410) = 0.0405

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.042 + 1.96 *√(0.042*0.958/67410) = 0.0435

4.05% < p < 4.35%

b) The confidence interval will be narrower as the proportion has decreased.

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2. n = 312, p = 0.48

Mean, µ = n*p = 312 * 0.48 = 149.76

Standard deviation, σ = √(n*p*(1-p)) = √(312 * 0.48 * 0.52) = 8.8247

P(X ≥ 156)

= P((X - µ)/σ ≥ (156 - 149.76)/8.8247)

= P(z ≥ 0.7071)

= 1 - P(z < 0.7071)

Using excel function:

= 1 - NORM.S.DIST(0.7071, 1)

= 0.2398