Question

A federal report stated that 88% of children under 18 were covered by health insurance in 2000. How large a sample is needed to estimate the true proportion of covered children with 90% confidence with a confidence interval of .05 wide?

Answer 1

`n = 115`

Explanation:

Do you mean with a margin of error of `5%`?

The formula for a confidence interval for a proportion is given by `hat p +- ME`, where `ME = z`* `* SE (hat p)`.

• `hat p` is the sample proportion
• `z`* is the critical value of `z`, which you can obtain from a graphing calculator or a table
• `SE(hat p)` is the standard error of the sample proportion, which can be found using `sqrt ((hat p hat q)/n)`, where `hat q = 1 - hat p` and `n` is the sample size

We know that the margin of error should be `0.05`. With a `90%` confidence interval, `z`* `~~ 1.64`.

`ME = z`* `* SE (hat p)`

`0.05 = 1.64 * sqrt ((0.88 * 0.12)/n)`

We can now solve for `n` algebraically. We get `n ~~ 114.2`, which we round up to `115` because a sample size of `114` would be too small.

We need at least `115` children to estimate the true proportion of children who are covered by health insurance with `90%` confidence and a margin of error of `5%`.

Answer 2

458

Explanation: