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.)

Answers

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 [tex]\pi[/tex], and a confidence level of [tex]\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm M[/tex]

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

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

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 [tex]n = 2005[/tex].27% believe in astrology, thus, [tex]p = 0.27[/tex].99% confidence interval, thus z has a p-value of [tex]\frac{1 + 0.99}{2} = 0.995[/tex], thus z = 2.575.

Then:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]M = 2.575\sqrt{\frac{0.27(0.73)}{2005}}[/tex]

[tex]M = 0.026[/tex]

The interval is:

[tex]M - \pi = 0.27 - 0.026 = 0.244[/tex]

[tex]M + \pi = 0.27 + 0.026 = 0.296[/tex]

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 [tex]\pi = 0.5[/tex].Margin of error of [tex]M = 0.05[/tex], and we have to find n:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.05 = 2.575\sqrt{\frac{0.5(0.5)}{n}}[/tex]

[tex]0.05\sqrt{n} = 2.575(0.5)[/tex]

Dividing both sides by 0.05.

[tex]\sqrt{n} = 2.575(10)[/tex]

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

[tex]n = 663.1[/tex]

Rounding up:

A sample size of 664 would be required.

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


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