A large on-demand, video streaming company is designing a large-scale survey to determine the mean amount of time corporate executives watch on-demand television. A small pilot …

Question

asked Nov 3, 2022 124k views

1 Answer

Gert Arnold · Answer 1 · 2022-11-07T16:42:45+0000

Answer:

554 executives should be surveyed.

Explanation:

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

$\alpha = (1 - 0.95)/(2) = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of
$1 - \alpha$ .

That is z with a pvalue of
1 - 0.025 = 0.975 , so Z = 1.96.

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.

Standard deviation of 3 hours.

This means that
$\sigma = 3$

The 95% level of confidence is to be used. How many executives should be surveyed?

n executives should be surveyed, and n is found with
M = 0.25 . So

$M = z(\sigma)/(√(n))$

0.25 = 1.96(3)/(√(n))

0.25√(n) = 1.96*3

√(n) = (1.96*3)/(0.25)

(√(n))^2 = ((1.96*3)/(0.25))^2

n = 553.2

Rounding up:

554 executives should be surveyed.