Question

An advertising agency that serves a major radio station wants to estimate the mean amount of time the station’s audience members listen to the radio on a daily basis. From past studies the population standard deviation is known to equal 45 minutes. What sample size is needed if the agency wants to be 90% confident of being within (±) 5 minutes of the true mean amount of time audience members listen to the radio per day?

Answer 1

Answer:

$n=(\frac{1.645(45)}{5})^2 =219.18 \approx 220$

So the answer for this case would be n=220 rounded up to the nearest integer

Step-by-step explanation:

Information given

$\sigma = 45$ represent the population deviation

$ME = 5$ represent the margin of error

The margin of error for the true mean 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 (4) we got:

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

The critical value for 90% of confidence interval now can be founded using the normal distribution. The significance level is $\alpha=1-0.9=0.1$ and the critical value would be $z_{\alpha/2}=1.645$, replacing into formula (b) we got:

$n=(\frac{1.645(45)}{5})^2 =219.18 \approx 220$

So the answer for this case would be n=220 rounded up to the nearest integer

Answer 2

Answer:

The margin of error for the true mean is :

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

$n=(\frac{1.645(45)}{5})^2 \\\\=219.18\\\\ \approx 220$

Therefore, the answer for this case would be n = 220

Step-by-step explanation:

Population standard deviation is 45

Margin of error is 5

|Z(0.05)|=1.645 (check standard normal table)

The margin of error for the true mean is :

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

$n=(\frac{1.645(45)}{5})^2 \\\\=219.18\\\\ \approx 220$

Therefore, the answer for this case would be n = 220