Question

you want to estimate the mean time college students spend watching online videos each day. the estimate must be within 1 minute of the population mean. determine the required sample size to construct a 99% confidence interval for the population mean. assume the population standard deviation is 2.4 minutes.

Answer 1

The required sample size is 39.

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 - 0.99)/(2) = 0.005`

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 - 0.005 = 0.995`, so Z = 2.575.

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.

Minimum sample size, within 1 minute of the population mean, population standar deviation of 2.4 minutes.

This minimum sample size is given by n, which is found when
`M = 1`, and we have that
`\sigma = 2.4`. So

`M = z(\sigma)/(√(n))`

`1 = 2.575(2.4)/(√(n))`

`√(n) = 2.575*2.4`

`(√(n))^2 = (2.575*2.4)^2`

`n = 38.2`

Rounding up

The required sample size is 39.