Past surveys reveal the 30% of tourists going to Las Vegas to gamble spend more than $1,000. The Visitor's Bureau of Las Vegas wants to update this percentage. The new study is to use the 90% confidence level. The estimate is to be within 1% of the population proportion. What is the necessary sample size?

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.

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