Question

Evelyn wants to estimate the percentage of people who own a tablet computer she surveys 150 indvidals and finds that 120 own a tablet computer. Identify the values needed to calculate a confidence interval at the 99% confidence level. Then find the confidence interval.

0.10 0.05 0.025 0.01 0.005
1.282 1.645 1.960 2.326 2 576

Answer 1

The 99% confidence interval for the percentage of people who own a tablet computer is between 71.59% and 88.41%

Explanation:

Confidence interval for the proportion of people who own a tablet:

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{(\pi(1-\pi))/(n)}`

In which

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

For this problem, we have that:

`n = 150, \pi = (120)/(150) = 0.8`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.576`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.8 - 2.575\sqrt{(0.8*0.2)/(150)} = 0.7159`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.8 + 2.575\sqrt{(0.8*0.2)/(150)} = 0.8841`

Percentage:

Multiply the proportion by 100.

0.7159*100 = 71.59%

0.8841*100 = 88.41%

The 99% confidence interval for the percentage of people who own a tablet computer is between 71.59% and 88.41%