Mathematics, 24.04.2020 21:55 uchapman6286

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?

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Answer from: samiam61

The first interval, with 9000 viewers, is 3 times as narrower as the second interval, with 1000 viewers.

Given data:

The percentage of confidence level interval is, 90%.

Number of viewers in the simple random sample is, 9000.

Let the number of people surveyed be n. So, in a sample of n surveyed people, the probability of success is $\pi$ . And the confidence level is, $(1-\alpha)$ .

Then, the confidence interval of proportion is,

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

Here,

z is the zscore having the pvalue of $(1- \dfrac{\alpha}{2})$ .

Then, the width of the interval is,

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

Now, in the given sample:

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

For interval A of 9000 viewers, n = 9000.

Then,

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

And for interval B of 1000 viewers, n = 1000.

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

Take the ratio of equation (1) and (2) to obtain the relation between the width at each interval.

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

Thus we can conclude that the first interval, with 9000 viewers, is 3 times as narrower as the second interval, with 1000 viewers.

Learn more about the confidence interval here:

link

Answer from: sin16

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:

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