Question

Seriously help I’m about to submit this: A researcher wishes to estimate the percentage of adults who support abolishing the penny what size sample should be obtained if he wishes the estimate to be within three percentage points with a 95% confidence if he:
A: uses a previous estimate of 26%?
B: he does not use any prior estimates?

Answer 1

Using the z-distribution, the sample sizes are given as follows:

a) 822.

b) 1068.

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.

Item a:

The estimate is of $\pi = 0.26$, hence we solve for n when M = 0.03.

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

$0.03 = 1.96\sqrt{\frac{0.26(0.74)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.26(0.74)}$

$\sqrt{n} = \frac{1.96\sqrt{0.26(0.74)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.26(0.74)}}{0.03}\right)^2$

n = 821.2

Rounding up, a sample of 822 is needed.

Item b:

No prior estimate, hence $\pi = 0.5$.

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

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

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

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

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

n = 1067.11

Rounding up, a sample of 1068 is needed.

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

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