Question

A local hotel wants to estimate the average age of its guests that are from out-of-state. Preliminary estimates are that standard deviation of population of guests from out-of-state is 30. How large a sample should be taken to estimate the average age of out-of-state guests with a margin of error no larger than 5 and with a 95% level of confidence? a. 12 b. 11 c. 139 d. 138

Answer 1

`n=((1.960(30))/(5))^2 =138.30 \approx 139`

And if we round up to the nearest integer we got n =139, and the best answer for this case is:

c. 139

Explanation:

For this case we have this previous info:

`\sigma = 30` represent the previous estimation for the population deviation

`Confidence =0.95` represent the confidence level

The margin of error for the true mean is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

The desired margin of error is ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

The critical value for a 95% of confidence interval given now can be founded using the normal distribution. For this case the critical value would be given by
`z_(\alpha/2)=1.960`, replacing into formula (b) we got:

`n=((1.960(30))/(5))^2 =138.30 \approx 139`

And if we round up to the nearest integer we got n =139, and the best answer for this case is:

c. 139

Answer 2

c. 139

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

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.

How large a sample should be taken to estimate the average age of out-of-state guests with a margin of error no larger than 5 and with a 95% level of confidence?

We need a sample size of n.

n is found when
`M = 5, \sigma = 30`

So

`M = z*(\sigma)/(√(n))`

`5 = 1.96*(30)/(√(n))`

`5√(n) = 1.96*30`

Simplifying by 5

`√(n) = 1.96*6`

`(√(n))^(2) = (1.96*6)^(2)`

`n = 138.30`

We round up,

So the correct answer is:

c. 139