Question

Suppose the U.S. president wants to estimate the proportion of the population that supports his current policy toward revisions in the health care system. The president wants the estimate to be within 0.03 of the true proportion. Assume a 98% level of confidence. The president’s political advisors found a similar survey from two years ago that reported that 63% of people supported health care revisions.

How large of a sample is required? (Round intermediate values to 3 decimal points. Round up your answer to the next whole number.)

Answer 1

We have the information for the proportions. The president wants the estimate to be within 0.03 of the true proportion.This means if confidence interval is calculated then the margin of error should be 0.03. So the true population proportion can be either greater by upto 3% or can be less upto 3% of the sample proportion.

MOE = 3%

The president’s political advisors found a similar survey from two years ago that reported that 63% of people supported health care revisions. This is from a survey so this is sample proportion

= 63%

For the binomial proportion  98% level of confidence means 2% (1-0.98) is the level of significance

= 2%

Margin of error =

=

= 2.3264 * .............using normal distribution tables

3% = 2.3264 *

squaring both sides

n =

n = 1401.681

= 1402