Question

1) A school has 1500 students and they are going to vote as to whether they will completely convert the school completely off fossil fuels. How many students would you have to poll to be 95% confident of the outcome within /- 3% of the vote

Answer 1

Using the z-distribution, it is found that 1068 students have to be polled to be 95% confident of the outcome within +/- 3% of the vote.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

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

We have no estimate for the true proportion, hence we use $\pi = 0.5$, and solve for n when M = 0.03. Then:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{(0.5)(0.5)}{n}}$

$0.03\sqrt{n} = 1.96(0.5)$

$\sqrt{n} = \frac{1.96(0.5)}{0.03}$

$(\sqrt{n})^2 = \left(\frac{1.96(0.5)}{0.03}\right)^2$

n = 1067.11.

Rounding up, 1068 students have to be polled to be 95% confident of the outcome within +/- 3% of the vote.

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

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