Question

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 necessary to estimate that mean with a 15-minute sampling error. Assume 96% confidence level is desired. Also assume that a pilot study showed that the standard deviation is estimated to be 112.2 minutes.

Answer 1

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.