## A local hotel wants to estimate the average age of its guests that are from out-of-state. Preliminary estimates are that standard deviation

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

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2 weeks 2021-10-08T07:50:54+00:00 2 Answers 0 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: represent the previous estimation for the population deviation represent the confidence level

The margin of error for the true mean is given by this formula: (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: (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 , replacing into formula (b) we got: And if we round up to the nearest integer we got n =139, and the best answer for this case is:

c. 139

c. 139

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So: Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so Now, find the margin of error M as such In which 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 So   Simplifying by 5   We round up,