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

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

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 length of the sample.

In this problem, we have that:

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

`M = z*(\sigma)/(√(n))`

`60 = 1.96*(400)/(√(n))`

`60√(n) = 784`

`√(n) = 13.07`

`n = 170.7`

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