Question

Researchers are interested in estimating the percentage of Americans who will get a flu shot this year. How many Americans should be surveyed to be 80% confident that the sample proportion of Americans who will get a flu shot this year is within 0.100 of the population proportion Assume we have a prior estimate of 45%. Round your z critical value to four decimal places.

Answer 1

Answer:

The sample size bedded is
`n =41`

Explanation:

From the question we are told that

The margin of error is
`E = 0.100`

The sample proportion is
`\^ p = 0.45`

From the question we are told the confidence level is 80% , hence the level of significance is

`\alpha = (100 - 80) \%`

=>
`\alpha = 0.20`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.282`

Generally the sample size is mathematically represented as

`n = [\frac{Z_{(\alpha )/(2) }}{E} ]^2 * \^ p (1 - \^ p )`

=>
`n = [(1.282)/( 0.100) ]^2 *0.45 (1 - 0.45 )`

=>
`n =41`