Question

A market researcher analyzing the fast-food industry noticed the following: The historical average amount spent at an upscale restaurant was $150.30, with a standard deviation of $50. The researcher wishes to have a sampling error of $5 or less and be 95 percent confident of an estimate to be made about average amount spent at an upscale restaurant from a survey. What sample size should be used (round the number up)

Answer 1

A sample size of 385 should be used.

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`.

That 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.

What sample size should be used (round the number up)

A sample of n should be used.

n is found when M = 5.

We have that
`\sigma = 50`

`M = z(\sigma)/(√(n))`

`5 = 1.96(50)/(√(n))`

`5√(n) = 1.96*50`

Simplifying by 10

`√(n) = 1.96*10`

`√(n) = 19.6`

`(√(n))^2 = (19.6)^2`

`n = 384.16`

Rounding up

A sample size of 385 should be used.