Respuesta :
Answer:
We need a sample size of at least 657.
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]
45 percent of its claims have errors.
So [tex]\pi = 0.45[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
What sample size is needed if they wish to be within 5 percent of the actual
This is a sample size of at least n, in which n is found when M = 0.05.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 2.575\sqrt{\frac{0.45*0.55}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.28[/tex]
[tex]\sqrt{n} = \frac{1.28}{0.05}[/tex]
[tex]\sqrt{n} = 25.62[/tex]
[tex]\sqrt{n}^{2} = (25.62)^{2}[/tex]
[tex]n = 656.4[/tex]
We need a sample size of at least 657.
