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As the saying goes, any man should accomplish three things in his life: build a house, plant a tree, and raise a child. Let us concentrate on the second part for now. A certain gardening magazine conducted a survey to determine how many people have planted a tree in their life. The results stated that 19% of the respondents have planted at least one tree. For the margin of error to be 3%, how many people should be surveyed? Round up to the nearest whole number, if necessary.

Respuesta :

Using the z-distribution, considering a standard 95% confidence level, it is found that 657 people should be surveyed.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

The margin of error is given by:

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

In which:

  • [tex]\pi[/tex] is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.

The estimate is of [tex]\pi = 0.19[/tex], and we have to solve for n when M = 0.03 to find the desired sample size.

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

[tex]0.03 = 1.96\sqrt{\frac{0.19(0.81)}{n}}[/tex]

[tex]0.03\sqrt{n} = 1.96\sqrt{0.19(0.81)}[/tex]

[tex]\sqrt{n} = \frac{1.96\sqrt{0.19(0.81)}}{0.03}[/tex]

[tex](\sqrt{n}) = \left(\frac{1.96\sqrt{0.19(0.81)}}{0.03}\right)^2[/tex]

n = 656.9.

Rounding up, 657 people should be surveyed.

More can be learned about the z-distribution at https://brainly.com/question/25890103

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