Question

a newspaper agency wants to estimate the proportion in a population who will buy their new news magazine using a 95 percent confidence interval. information from sale of similar magazines suggest a prior guess of 0.25 for the proportion , how large a random sample would be necessary for the interval to have a margin of error of 0.015 or less?

Answer 1

A random sample of approximately 3192 individuals is needed to estimate the proportion of a population that will buy a new news magazine with a 95 percent confidence interval and a margin of error of 0.015 or less.

Step-by-step explanation:

To estimate the proportion of a population who will buy their new news magazine with a 95 percent confidence interval and a margin of error of 0.015 or less, we can use the formula for the sample size (n) when estimating proportions:

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

Where,

• Z is the Z-value corresponding to the confidence level (1.96 for 95%),
• p is the estimated proportion (0.25), and
• E is the margin of error (0.015).

Plugging in the values, we get:

n = (1.96^2 * 0.25 * (1 - 0.25)) / 0.015^2

n = (3.8416 * 0.25 * 0.75) / 0.000225

n = (0.7182) / 0.000225

n = 3192

Therefore, a random sample of approximately 3192 individuals would be necessary for the interval to have a margin of error of 0.015 or less.