Question

Suppose we wanted to estimate the proportion of houses worth more than $400k. What sample size do we need to get the margin of error for the 95% confidence interval for the proportion to within 0.015? Recall that the z-critical value is 1.96. When we don't have an estimate for the proportion, we use p.

a) 384
b) 567
c) 625

Answer 1

To achieve a margin of error within 0.015 for a 95% confidence interval for the proportion of houses worth more than $400k, you would need a sample size of approximately 4269, which is not among the provided options.

Step-by-step explanation:

To estimate the proportion of houses worth more than $400k with a margin of error within 0.015 for a 95% confidence interval, you can use the following formula for sample size (n):

n = (Z^2 * p * (1-p)) / E^2

Where:
Z = z-critical value (for 95% confidence, Z = 1.96)
p = estimated proportion (since we don't have an estimate, we use 0.5 as a conservative estimate)
E = margin of error (in this case, 0.015)

Plugging in the values gives us:

n = (1.96^2 * 0.5 * (1-0.5)) / 0.015^2

n = (3.8416 * 0.25) / 0.000225

n = 0.9604 / 0.000225

n = 4268.888889

Since the sample size must be a whole number, we round up to the next whole number:

n ≈ 4269

Therefore, none of the options (384, 567, 625) are correct. To achieve a margin of error within 0.015 with 95% confidence, you need a sample size of approximately 4269 houses.