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

The margin of error for the true mean is :

`ME=z_(\alpha/2)(\sigma)/(√(n))`

`n=((1.645(45))/(5))^2 \\\\=219.18\\\\ \approx 220`

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

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)(\sigma)/(√(n))`

`n=((1.645(45))/(5))^2 \\\\=219.18\\\\ \approx 220`

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

Answer 2

`n=((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

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)(\sigma)/(√(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=((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=((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