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 sa…

Question

asked Apr 24, 2022 118k views

1 Answer

Abhishek · Answer 1 · 2022-04-28T06:44:05+0000

Answer:

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.