Question

Franchise Business Review stated over 50% of all food franchises earn a profit of less than $50,000 a year. In a sample of 130 casual dining restaurants, 81 earned a profit of less than $50,000 last year. Based upon a 95% confidence interval with a desired margin of error of .04, determine a sample size for restaurants that earn less than $50,000 last year.

Answer 1

Answer:

We need a sample size of 564.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$\pi = \frac{81}{130} = 0.6231$

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Based upon a 95% confidence interval with a desired margin of error of .04, determine a sample size for restaurants that earn less than $50,000 last year.

We need a sample size of n

n is found when $M = 0.04$

So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.6231*0.3769}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.6231*0.3769}$

$\sqrt{n} = \frac{1.96\sqrt{0.6231*0.3769}}{0.04}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.6231*0.3769}}{0.04})^{2}$

$n = 563.8$

Rounding up

We need a sample size of 564.