Respuesta :
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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]\pi = \frac{81}{130} = 0.6231[/tex]
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
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 [tex]M = 0.04[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.6231*0.3769}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96\sqrt{0.6231*0.3769}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.6231*0.3769}}{0.04}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.6231*0.3769}}{0.04})^{2}[/tex]
[tex]n = 563.8[/tex]
Rounding up
We need a sample size of 564.
The sample size for restaurants that earn less than $50,000 last year is; 564
- The formula for margin of error of proportions is; E = z√(p^(1 - p^)/n)
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
- To determine a sample size for restaurants that earn less than $50,000 last year, we just make sample size (n) the subject of the formula in the formula for margin of error to get and then solve with the relevant values.
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; https://brainly.com/question/7522552
