Question

Nielsen Media Research wants to estimate the mean amount of time, in minutes, that full-time college students spend texting each weekday.Find the sample size necessary to estimate that mean with a 15 minute margin of error. Assume that a 96% confidence level is desired and that the standard deviation is estimated to be 112.2 minutes.

Answer 1

Answer:

n=237

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=112.2)$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The next step would be find the value of $\z_{\alpha/2}$, $\alpha=1-0.96=0.04$ and $\alpha/2=0.02$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=2.054$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{2.054(112.2)}{15})^2 =236.05$

And if we round up the answer we see that the value of n to ensure the margin of error required $\pm=15 min$ is n=237.