Nielson Media Research wants to estimate the mean amount of time (in minutes) that full-time college students spend watching television each weekday. Find the sample size necess…

Question

asked Apr 2, 2022 140k views

1 Answer

Miro Solanka · Answer 1 · 2022-04-06T11:00:58+0000

Answer:

The sample size necessary is 237.

Explanation:

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

$\alpha = (1 - 0.96)/(2) = 0.02$

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 - 0.02 = 0.98 , so Z = 2.056.

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.

Find the sample size necessary to estimate that mean with a 15-minute sampling error.

A sample size of n is necessary.

n is found when
M = 15

We have that
$\sigma = 112.2$

So

$M = z(\sigma)/(√(n))$

15 = 2.056(112.2)/(√(n))

15√(n) = 2.056*112.2

√(n) = (2.056*112.2)/(15)

(√(n))^2 = ((2.056*112.2)/(15))^2

n = 236.5

Rounding up,

The sample size necessary is 237.