A weighted (unfair) die that was purchased at a magic shop was rolled 30 times. The die landed on five dots 25 times. Based on the results, what is the margin of error for the true proportion of times this die would land on five dots? Base your margin of error on a 95% confidence level.

Respuesta :

Using the z-distribution, it is found that the margin of error for the true proportion of times this die would land on five dots is of 0.1338.

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.

The margin of error is given by:

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

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 and the sample size are given as follows:

[tex]\pi = \frac{25}{30} = 0.8333, n = 30[/tex].

Hence the margin of error is given by:

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

[tex]M = 1.96\sqrt{\frac{0.8333(0.1667)}{30}}[/tex]

M = 0.1338.

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

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