Question

In a certain rural county, a public health researcher spoke with 238 residents 65-years or older, and 186 of them had obtained a flu shot. The researcher wants to calculate a 95% confidence interval for the percent of the 65-plus population that were getting the flu shot.

Do not use StatCrunch. Show all formulas used, work and steps. Be sure to define your variables.

As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 1.96 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval

Give your answer in interval notation rounded to 3 decimal places. Then interpret your answer in a complete sentence within the context of the problem.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 186 / 238 = 0.782

1 - $\hat p$ = 1 - 0.782 = 0.218

Z$\alpha$/2 = Z0.025 = 1.96

Margin of error = E = Z$\alpha$ / 2 * (($\hat p$ * (1 - $\hat p$ ))$\sqrt$ / n)

= 1.96 ($\sqrt$((0.782 * 0.218) / 238)

= 0.052

A 95% confidence interval for population proportion p is ,

$\hat p$ ± E

= 0.782  ± 0.052

= ( 0.730, 0.834 )

We are 95% confident that the true proportion that were getting the flu shot between 0.730 and 0.834.