A fast food restaurant executive wishes to know how many fast food meals teenagers eat each week. They want to construct a 80% confidence interval with an error of no more than …

Question

asked Oct 18, 2022 185k views

1 Answer

BYZZav · Answer 1 · 2022-10-23T04:11:25+0000

Answer:

The minimum sample size required to create the specified confidence interval is of 565.

Explanation:

We have that to find our
$\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = (1 - 0.8)/(2) = 0.1$

Now, we have to find z in the Ztable as such z has a pvalue of
$1 - \alpha$ .

That is z with a pvalue of
1 - 0.1 = 0.9 , so Z = 1.28.

Now, find the margin of error M as such

$M = z(\sigma)/(√(n))$

In which
$\sigma$ is the standard deviation of the population and n is the size of the sample.

Found the standard deviation to be 1.3.

This means that
$\sigma = 1.3$

Error of no more than 0.07. What is the minimum sample size required to create the specified confidence interval?

This is n for which M = 0.07. So

$M = z(\sigma)/(√(n))$

0.07 = 1.28(1.3)/(√(n))

0.07√(n) = 1.28*1.3

√(n) = (1.28*1.3)/(0.07)

(√(n))^2 = ((1.28*1.3)/(0.07))^2

n = 565.1

Rounding up:

The minimum sample size required to create the specified confidence interval is of 565.