Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If we want to estimate this proportion to within 4% at the 95% confidence level, how many randomly selected teenagers must we survey?
To estimate the proportion of teenagers who are lactose intolerant with a desired level of precision, we can use the formula for sample size estimation. In this case, we want to estimate the proportion within 4% at the 95% confidence level.
The formula for sample size estimation is:
n = (Z^2 * p * (1-p)) / E^2,
where:
- n is the required sample size,
- Z is the Z-score corresponding to the desired confidence level,
- p is the estimated proportion of teenagers who are lactose intolerant,
- E is the desired margin of error.
Given that we want a 95% confidence level, the Z-score is approximately 1.96. The margin of error is 4%, which can be written as 0.04.
To estimate p, we can use a known value or assume a conservative estimate. Let's assume that 50% of teenagers are lactose intolerant since we do not have a specific value.
Plugging in the values into the formula, we have:
n = (1.96^2 * 0.5 * (1-0.5)) / (0.04^2).
Simplifying this equation gives us:
n = (3.8416 * 0.25) / 0.0016,
n = 96.04 / 0.0016,
n ≈ 60,025.
Therefore, to estimate the proportion of lactose-intolerant teenagers with a margin of error of 4% at the 95% confidence level, we would need to survey approximately 60,025 randomly selected teenagers.
To estimate the proportion of teenagers who are lactose intolerant with a certain level of confidence and margin of error, we can use the formula for sample size calculation for proportions.
The formula is:
n = (Z^2 * p * (1-p)) / E^2
where:
- n is the required sample size
- Z is the z-value corresponding to the desired level of confidence (in this case, 95% confidence level)
- p is the estimated proportion of teenagers who are lactose intolerant (we'll estimate this later)
- E is the desired margin of error (in this case, 4%)
To determine the estimated proportion of teenagers who are lactose intolerant, we can look up previous research or studies in the field. Let's assume that the estimated proportion is 30%, which is a common estimate for lactose intolerance prevalence. Therefore, p = 0.3.
Now, to find the z-value corresponding to a 95% confidence level, we can refer to a standard normal distribution table or use a calculator. The z-value for a 95% confidence level is approximately 1.96.
Plugging in the given values, we have:
n = (1.96^2 * 0.3 * (1-0.3)) / 0.04^2
Simplifying the equation, we get:
n = (3.8416 * 0.21) / 0.0016
n = 80.349
Rounding up to the nearest whole number, we find that we need a sample size of approximately 81 teenagers to estimate the proportion of lactose intolerance among teenagers aged 13-18 with a 4% margin of error at a 95% confidence level.