Respuesta :
Answer:
A sample of 11352 is needed.
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].
The margin of error is given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
For this problem, we have that:
400 cards, 120 cases were correct. So
[tex]p = \frac{120}{400} = 0.3[/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].
How large a sample n would you need to estimate p with margin of error 0.01 and 95% confidence
We would need a sample of size n, and n is found when [tex]M = 0.01[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.01 = 2.325\sqrt{\frac{0.3*0.7}{n}}[/tex]
[tex]0.01\sqrt{n} = 2.325\sqrt{0.3*0.7}[/tex]
[tex]\sqrt{n} = \frac{2.325\sqrt{0.3*0.7}}{0.01}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.325\sqrt{0.3*0.7}}{0.01})^2[/tex]
[tex]n = 11351.8[/tex]
Rounding up
A sample of 11352 is needed.
