An epidemiologist is planning a study on the prevalence of oral contracep- tive use in a certain population. She plans to choose a random sample of n women and to sue the sample proportion of oral contraceptive users (pˆ) as an estimate of the population proportion (p). Suppose that in fact p = 0.12. Use the normal approximation (with continuity correction) to determine the probability that pˆ will be within ±0.03 of p if
(a) n = 100; (b) n = 200.
To determine the probability that pˆ will be within ±0.03 of p, we can use the normal approximation with continuity correction. The formula for the continuity-corrected approximation is:
P(p - 0.03 < pˆ < p + 0.03) = P((p - 0.5/n) - 0.03 < pˆ < (p - 0.5/n) + 0.03)
Let's calculate the probability for both scenarios:
(a) When n = 100:
Using the formula, we have:
P(p - 0.03 < pˆ < p + 0.03) = P(0.12 - 0.5/100 - 0.03 < pˆ < 0.12 - 0.5/100 + 0.03)
P(0.09 < pˆ < 0.15)
To find this probability, we need to determine the z-scores for 0.09 and 0.15, and then use a standard normal distribution table or calculator to find the area between these z-scores.
(b) When n = 200:
Similarly, using the formula, we have:
P(p - 0.03 < pˆ < p + 0.03) = P(0.12 - 0.5/200 - 0.03 < pˆ < 0.12 - 0.5/200 + 0.03)
P(0.08 < pˆ < 0.16)
Again, we need to find the z-scores for 0.08 and 0.16, and then use a standard normal distribution table or calculator to determine the probability between these z-scores.
Please note that we require the actual z-scores or a standard normal distribution table or calculator to find the probabilities.
To solve this problem, we can use the formula for the standard error of a proportion:
SE(p̂) = sqrt[(p * (1 - p)) / n]
(a) For n = 100:
p = 0.12
n = 100
SE(p̂) = sqrt[(0.12 * (1 - 0.12)) / 100]
= sqrt[(0.12 * 0.88) / 100]
= sqrt(0.01056)
≈ 0.1027
Now, we can use the normal distribution to find the probability that p̂ is within ±0.03 of p. We need to calculate the z-scores for both sides:
Z1 = (0.03 - 0) / 0.1027
≈ 0.2919
Z2 = (-0.03 - 0) / 0.1027
≈ -0.2919
Next, we find the cumulative probabilities for both z-scores:
P(X < Z1) = P(X < 0.2919) ≈ 0.6159
P(X > Z2) = P(X > -0.2919) ≈ 0.6187
Since we want the probability that p̂ is within ±0.03 of p, we need to find the area between Z2 and Z1:
P(Z2 < X < Z1) = P(Z2 < X < Z1) ≈ P(X < Z1) - P(X > Z2)
≈ 0.6159 - 0.6187
≈ -0.0028
Therefore, the probability that p̂ will be within ±0.03 of p when n = 100 is approximately -0.0028.
(b) For n = 200:
p = 0.12
n = 200
SE(p̂) = sqrt[(0.12 * (1 - 0.12)) / 200]
= sqrt[(0.12 * 0.88) / 200]
= sqrt(0.00528)
≈ 0.0727
Now, we can calculate the z-scores:
Z1 = (0.03 - 0) / 0.0727
≈ 0.4124
Z2 = (-0.03 - 0) / 0.0727
≈ -0.4124
Find the cumulative probabilities for both z-scores:
P(X < Z1) = P(X < 0.4124) ≈ 0.6601
P(X > Z2) = P(X > -0.4124) ≈ 0.6601
Calculate the area between Z2 and Z1:
P(Z2 < X < Z1) = P(Z2 < X < Z1) ≈ P(X < Z1) - P(X > Z2)
≈ 0.6601 - 0.6601
≈ 0
Therefore, the probability that p̂ will be within ±0.03 of p when n = 200 is approximately 0.