52% of 1000 registered voters intend to vote for Steven Collins for mayor. What is the 95% confidence interval to describe the total percentage of registered voters who intend to vote for Steven Collins?


A. (50.1%, 53.9%)

B. (48%, 56%)

C. (49.9%, 54.1%)

D. (48.9%, 55.1%)

Respuesta :

Using the z-distribution, the 95% confidence interval to describe the total percentage of registered voters who intend to vote for Steven Collins is:

D. (48.9%, 55.1%)

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]

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 other parameters are given as follows:

[tex]\pi = 0.52, n = 1000[/tex]

Hence the bounds of the interval are given by:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 - 1.96\sqrt{\frac{0.52(0.48)}{1000}} = 0.489[/tex]

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 + 1.96\sqrt{\frac{0.52(0.48)}{1000}} = 0.551[/tex]

As a percentage, option D is correct.

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

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