Question

A certain forum reported that in a survey of 2005 american adults, 27% said that they believe in astrology.
(a) Calculate a confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology. (Round your answers to three decimal places.) Interpret the resulting interval.
i. We are 99% confident that the true population mean lies below this interval.
ii. We are 99% confident that the true population mean lies above this interval.
iii. We are 99% confident that this interval contains the true population mean.
iv. We are 99% confident that this interval does not contain the true population mean.
(b) what sample size would be required for the width of a 99% CI to be at most 0.05 irrespective of the value of p? Round your answer up to the nearest integer.)

Answer 1

Answer:

               

Step-by-step explanation:

                 

Answer 2

Using confidence interval concepts, it is found that:

a)

The 99% confidence interval for the proportion of all adult Americans who believe in astrology is (0.244, 0.296).

The interpretation is:

iii. We are 99% confident that this interval contains the true population mean.

b)

A sample size of 664 would be required.

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

$\pi \pm M$

The margin of error is:

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

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

The interpretation of a x% confidence interval between a and b is that we are x% confidence that the true population mean lies in this interval, thus, option iii is correct.

In this problem:

• Survey of 2005 adults, thus $n = 2005$.

• 27% believe in astrology, thus, $p = 0.27$.

• 99% confidence interval, thus z has a p-value of $\frac{1 + 0.99}{2} = 0.995$, thus z = 2.575.

Then:

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

$M = 2.575\sqrt{\frac{0.27(0.73)}{2005}}$

$M = 0.026$

The interval is:

$M - \pi = 0.27 - 0.026 = 0.244$

$M + \pi = 0.27 + 0.026 = 0.296$

The 99% confidence interval for the proportion of all adult Americans who believe in astrology is (0.244, 0.296).

Item b:

• Irrespective of the value of p, thus $\pi = 0.5$.

• Margin of error of $M = 0.05$, and we have to find n:

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

$0.05 = 2.575\sqrt{\frac{0.5(0.5)}{n}}$

$0.05\sqrt{n} = 2.575(0.5)$

Dividing both sides by 0.05.

$\sqrt{n} = 2.575(10)$

$(\sqrt{n})^2 = [2.575(10)]^2$

$n = 663.1$

Rounding up:

A sample size of 664 would be required.

A similar problem is given at https://brainly.com/question/16807970