Question

assume that a sample is used to estimate a population proportion p. Find the margin of errorS that corresponds to the following: a 98% confidence ; the sample size is 800, of which 40% are successes

Answer 1

Explanation:

To find the margin of error for a 98% confidence level, we need to use the z-score that corresponds to this confidence level, which is 2.33 (found in a z-table or using a calculator).

The formula for the margin of error (E) is:

E = z*(sqrt(p*q/n))

Where:

z is the z-score for the desired confidence level (2.33 for 98% confidence)

p is the sample proportion (0.4, or 40%)

q is the complement of the sample proportion (1 - 0.4 = 0.6)

n is the sample size (800)

Substituting these values into the formula, we get:

E = 2.33*(sqrt(0.4*0.6/800)) = 0.045

So the margin of error for a 98% confidence level, with a sample size of 800 and a sample proportion of 40%, is 0.045 or approximately 4.5%. This means that we can be 98% confident that the true population proportion is within 4.5% of the sample proportion.