Respuesta :
Answer:
a) At least 842 adults must be surveyed.
b) At least 1068 adults must be surveyed.
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].
a) 73% of adults used the Internet.
At least n adults must be surveyed.
n is found when [tex]M = 0.03, \pi = 0.73[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.73*0.27}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.73*0.27}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.73*0.27}}{0.03}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.73*0.27}}{0.03})^{2}[/tex]
[tex]n = 841.3[/tex]
Rounding up
At least 842 adults must be surveyed.
b. No known possible value of the proportion.
Same as above, the difference as the since we do not know the value of the proportion, we use [tex]\pi = 0.5[/tex], which is when the largest sample size is going to be needed.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.5*0.5}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.5*0.5}}{0.03}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.5*0.5}}{0.03})^{2}[/tex]
[tex]n = 1067.11[/tex]
Rounding up
At least 1068 adults must be surveyed.
