Tuesday, 11/12
𝑟
➢ Point estimates for p and q: p̂ = 𝑛, q̂ = 1-p̂
Finding Confidence Interval for p
➢ In a binomial experiment with n trials, p is equal to the population probability of success,
q is equal to 1-p(probability of failure), r is equal to the number of successes, and n is
equal to the number of trials
𝑝𝑞 𝑝(1−𝑝)
➢ Confidence Interval: (p̂-E)<p<(p̂+E), where E ≈ zc√ 𝑛 = zc√ 𝑛
➢ np̂ and nq̂ must both be greater than 5
➢ Use Table 5(b) Appendix II to find frequently used values
Example
➢ 800 random students are chosen from a student body of 20,000 and are given flu shots.
600 of the 800 students did not get the flu. Let p represent the probability that a shot will
be successful, and q the probability that a shot will not be successful.
➢ What are the values of n and r?
→ n=800, r=600
➢ What are the point estimates for p and q?
→ p̂ = 600/800 = 0.75; q̂ = 1-0.75 = 0.25
➢ Is the number of trials large enough to justify a normal approximation to binomial?
→ n≥30 and np̂ and nq̂ both >5, so yes.
➢ Find the 99% confidence interval for p.
0.75(0.25) 0.1875
→ E = z0.99√ = 2.58√ = 2.58(0.15) = 0.0395
800 800
→ p̂-0.0395<p<p̂+0.0395 = (0.75-0.0395)<p<(0.75+0.0395) = 0.71<p<0.79
Video
➢ To find z score given a confidence level(90%, 95%, etc.) use formula to find area then
find z score corresponding to value
➢ AL = area under the curve to the left
➢ Formula: AL = (1+CL)/2
(1+0.95) 1.95
➢ For 95% confidence level: AL = = = 0.975
2 2
→ Using table, we find value corresponding to 0.975=1.96
𝑟
➢ Point estimates for p and q: p̂ = 𝑛, q̂ = 1-p̂
Finding Confidence Interval for p
➢ In a binomial experiment with n trials, p is equal to the population probability of success,
q is equal to 1-p(probability of failure), r is equal to the number of successes, and n is
equal to the number of trials
𝑝𝑞 𝑝(1−𝑝)
➢ Confidence Interval: (p̂-E)<p<(p̂+E), where E ≈ zc√ 𝑛 = zc√ 𝑛
➢ np̂ and nq̂ must both be greater than 5
➢ Use Table 5(b) Appendix II to find frequently used values
Example
➢ 800 random students are chosen from a student body of 20,000 and are given flu shots.
600 of the 800 students did not get the flu. Let p represent the probability that a shot will
be successful, and q the probability that a shot will not be successful.
➢ What are the values of n and r?
→ n=800, r=600
➢ What are the point estimates for p and q?
→ p̂ = 600/800 = 0.75; q̂ = 1-0.75 = 0.25
➢ Is the number of trials large enough to justify a normal approximation to binomial?
→ n≥30 and np̂ and nq̂ both >5, so yes.
➢ Find the 99% confidence interval for p.
0.75(0.25) 0.1875
→ E = z0.99√ = 2.58√ = 2.58(0.15) = 0.0395
800 800
→ p̂-0.0395<p<p̂+0.0395 = (0.75-0.0395)<p<(0.75+0.0395) = 0.71<p<0.79
Video
➢ To find z score given a confidence level(90%, 95%, etc.) use formula to find area then
find z score corresponding to value
➢ AL = area under the curve to the left
➢ Formula: AL = (1+CL)/2
(1+0.95) 1.95
➢ For 95% confidence level: AL = = = 0.975
2 2
→ Using table, we find value corresponding to 0.975=1.96
