Question

An advertising agency that manages a major radio station wants to estimate the mean amount of time that the station’s audience spends listening to the radio on a daily basis. From past studies, the standard deviation is assumed to be 20 minutes. What sample size is needed if the agency wants to have a 95% confidence interval with a margin of error equal to 5 minutes?

Answer 1

Answer:

$n=(\frac{1.960(20)}{5})^2 =61.46 \approx 62$

So the answer for this case would be n=62 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=20$ represent the population standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$, replacing into formula (b) we got:

$n=(\frac{1.960(20)}{5})^2 =61.46 \approx 62$

So the answer for this case would be n=62 rounded up to the nearest integer