Question

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 survey of 10 executives indicated that the mean time per week is 12 hours, with a standard deviation of 3 hours. The estimate of the mean viewing time should be within 0.25 hour. The 95% level of confidence is to be used. How many executives should be surveyed? (Use z Distribution Table.)

How many executives should be surveyed? (Round the final answer to the next whole number.)

Answer 1

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.