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.

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 , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

The margin of error is:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

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

So

Rounding up

We need a sample size of 564.

The sample size for restaurants that earn less than $50,000 last year is; 564

Where;

z is the critical value at given confidence level.

From tables, the value of z at confidence level of 95% is 1.96.

Now, we are given;

Margin of error; E = 0.04

Sample proportion; p^ = 81/130 = 0.6231

Thus;

n = z²(p^(1 - p^)/E²

Plugging in the relevant values gives;

n = 1.96²(0.6231(1 - 0.6231))/0.04²

n = 563.866

n ≈ 564

Read more about margin of error for proportions at; link

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