Respuesta :
Answer:
We need a sample of at least 1797 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage.
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:
[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].
In this problem, we have that:
[tex]p = 0.55[/tex]
How large a sample is needed if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage?
We have to find n for which [tex]M = 0.023[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.023 = 1.96\sqrt{\frac{0.55*0.45}{n}}[/tex]
[tex]0.023\sqrt{n} = 0.9751[/tex]
[tex]\sqrt{n} = \frac{0.9751}{0.0023}[/tex]
[tex]\sqrt{n} = 423.95[/tex]
[tex]\sqrt{n}^{2} = (42.395)^{2}[/tex]
[tex]n = 1797[/tex]
We need a sample of at least 1797 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage.
