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

Answer:

A sample size of 385 should be used.

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 $1 - \alpha$.

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

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

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 $\sigma = 50$

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

$5 = 1.96\frac{50}{\sqrt{n}}$

$5\sqrt{n} = 1.96*50$

Simplifying by 10

$\sqrt{n} = 1.96*10$

$\sqrt{n} = 19.6$

$(\sqrt{n})^2 = (19.6)^2$

$n = 384.16$

Rounding up

A sample size of 385 should be used.