Respuesta :
Answer:
The necessary sample size is of 5683.
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]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
Past surveys reveal the 30% of tourists going to Las Vegas to gamble spend more than $1,000.
This means that [tex]\pi = 0.3[/tex]
The estimate is to be within 1% of the population proportion. What is the necessary sample size?
A sample size of n is necessary. n is found when [tex]M = 0.01[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.01 = 1.645\sqrt{\frac{0.3*0.7}{n}}[/tex]
[tex]0.01\sqrt{n} = 1.645\sqrt{0.3*0.7}[/tex]
[tex]\sqrt{n} = \frac{1.645\sqrt{0.3*0.7}}{0.01}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.645\sqrt{0.3*0.7}}{0.01})^2[/tex]
[tex]n = 5682.6[/tex]
Rounding up:
The necessary sample size is of 5683.
