We are studying the proportion of cities that have private garbage collector. We want a maximum error of 0.10 and a 95% confidence level, and tentatively estimate the proportion at 0.50. What is the minimum number of cities we need to contact?

Respuesta :

Answer:

The minimum number of cities we need to contact is 96.

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 interval [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 interval

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 = 0.5, M = 0.1[/tex]

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.1 = 1.96*\sqrt{\frac{0.5*0.5}{n}}[/tex]

[tex]0.1\sqrt{n} = 0.98[/tex]

[tex]\sqrt{n} = 9.8[/tex]

[tex](\sqrt{n})^{2} = (9.8)^{2}[/tex]

[tex]n = 96[/tex]

The minimum number of cities we need to contact is 96.

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