Question

Suppose that a marketing research firm wants to conduct a survey to estimate the meanμof the distribution of the amount spent on entertainment by each adult who visits a certain popularresort. The firm would like to estimate the mean of this distribution to within $60 with 95% confidence.From data regarding past operations at the resort, it has been estimated that the standard deviation ofthe entertainment expenditures is no more than $400. How large does the firm’s sample size need to be?

Answer 1

Answer:

The firm's sample size must be of at least 171 adults.

Step-by-step explanation:

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

$\alpha = \frac{1-0.95}{2} = 0.025$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the length of the sample.

In this problem, we have that:

$M = 60, \sigma = 400$. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$60 = 1.96*\frac{400}{\sqrt{n}}$

$60\sqrt{n} = 784$

$\sqrt{n} = 13.07$

$n = 170.7$

The firm's sample size must be of at least 171 adults.