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

Respuesta :

Answer:

c. 139

Step-by-step explanation:

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

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

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

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] 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 [tex]M = 5, \sigma = 30[/tex]

So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]5 = 1.96*\frac{30}{\sqrt{n}}[/tex]

[tex]5\sqrt{n} = 1.96*30[/tex]

Simplifying by 5

[tex]\sqrt{n} = 1.96*6[/tex]

[tex](\sqrt{n})^{2} = (1.96*6)^{2}[/tex]

[tex]n = 138.30[/tex]

We round up,

So the correct answer is:

c. 139

Answer:

[tex]n=(\frac{1.960(30)}{5})^2 =138.30 \approx 139[/tex]

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

c. 139

Step-by-step explanation:

For this case we have this previous info:

[tex]\sigma = 30[/tex] represent the previous estimation for the population deviation

[tex]Confidence =0.95[/tex] represent the confidence level

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

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (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:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (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 [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.960(30)}{5})^2 =138.30 \approx 139[/tex]

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

c. 139