Question

A 90 percent confidence interval is to be created to estimate the proportion of television viewers in a certain area who favor moving the broadcast of the late weeknight news to an hour earlier than it is currently. Initially, the confidence interval will be created using a simple random sample of 9,000 viewers in the area. Assuming that the sample proportion does not change, what would be the relationship between the width of the original confidence interval and the width of a second 90 percent confidence interval that is created based on a sample of only 1,000 viewers in the area?

Answer 1

Answer:

This means that the first interval, with 9000 viewers, is 3 times as narrower as the second interval, with 1000 viewers.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The width of the interval is:

$W = 2z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this sample:

Two 90% intervals, with different lenghts. So both have the same values for z an $\pi$

Interval A:

9000 viewers.

So the width is

$W_{A} = 2z\sqrt{\frac{\pi(1-\pi)}{9000}}$

Interval B:

100 viewers

So the width is

$W_{B} = 2z\sqrt{\frac{\pi(1-\pi)}{1000}}$

Relationship between the widths:

$R = \frac{W_{A}}{W_{B}} = \frac{2z\sqrt{\frac{\pi(1-\pi)}{9000}}}{2z\sqrt{\frac{\pi(1-\pi)}{1000}}} = \frac{\sqrt{1000}}{\sqrt{9000}} = \frac{\sqrt{1}}{\sqrt{9}} = \frac{1}{3}$

This means that the first interval, with 9000 viewers, is 3 times as narrower as the second interval, with 1000 viewers.