Question

A company manufacturing CDs is working on a new technology. A random sample of 705 Internet users were​ asked: "As you may​ know, some CDs are being manufactured so that you can only make one copy of the CD after you purchase it. Would you buy a CD with this​ technology, or would you refuse to buy it even if it was one you would normally​ buy?" Of these​ users, 64​% responded that they would buy the CD. ​

a) Create a 99​% confidence interval for this percentage. ​
b) If the company wants to cut the margin of error in​ half, how many users must they​ survey?

Answer 1

a) The 99​% confidence interval for this percentage is (0.5934, 0.6866).

b) They should survey 2820 users

Explanation:

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

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

A random sample of 705 Internet users. This means that
`n = 705`.

Of these​ users, 64​% responded that they would buy the CD. ​This means that
`\pi = 0.64`.

a) Create a 99​% confidence interval for this percentage. ​

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`z = 2.575`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.64 - 2.575\sqrt{(0.64*0.36)/(705)} = 0.5934`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.64 + 2.575\sqrt{(0.64*0.36)/(705)} = 0.6866`

The 99​% confidence interval for this percentage is (0.5934, 0.6866).

b) If the company wants to cut the margin of error in​ half, how many users must they​ survey?

The margin of error is the subtraction of the upper limit of the confidence interval by the lower limit. This subtraction is

`M = 2z(\pi(1-\pi))/(n)`.

This means that the margin of error is inverse proportional to the square root of the sample size. This means that to cut the margin of error in half, we must increase the sample size 4 times.

So, they should survey 4*705 = 2820 users