Respuesta :
Answer:
A sample of 997 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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
A previous study indicates that the proportion of left-handed golfers is 8%.
This means that [tex]\pi = 0.08[/tex]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].
How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?
This is n for which M = 0.02. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 2.327\sqrt{\frac{0.08*0.92}{n}}[/tex]
[tex]0.02\sqrt{n} = 2.327\sqrt{0.08*0.92}[/tex]
[tex]\sqrt{n} = \frac{2.327\sqrt{0.08*0.92}}{0.02}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.327\sqrt{0.08*0.92}}{0.02})^2[/tex]
[tex]n = 996.3[/tex]
Rounding up:
A sample of 997 is needed.
