Question

A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States that has sleep deprivation. How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 4%?

Answer 1

The size of a confidence interval depends on several factors, including the margin of error and the level of confidence.


In this case, the researcher has set a level of confidence of 90%. This means that they are aiming for an interval that includes 90% of the values that would be observed if the experiment was conducted many times.


The researcher also wants to ensure that their confidence interval includes the true proportion of the US population that is sleep-deprived—within a margin of error of 4%.

The exact value of the sample size is 1722 (1/.04 × .9 × 100).

Answer 2

We can use the formula for the margin of error:

Margin of error = z* * sqrt(p*(1-p)/n)

where z* is the z-score for the desired level of confidence, p is the estimate of the proportion (we'll assume 0.5 since we don't have any prior information), and n is the sample size.

We want the margin of error to be no more than 0.04, and we want to be 90% confident, so the z-score is 1.645 (from a standard normal distribution table).

0.04 = 1.645 * sqrt(0.5*(1-0.5)/n)

Solving for n, we get:

n = (1.645/0.04)^2 * 0.5*(1-0.5) = 601.25

So we need a sample size of at least 602 to be 90% confident that the sample proportion will not differ from the true proportion by more than 4%.