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

Answer:

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

Step-by-step 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.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 150, \pi = \frac{120}{150} = 0.8$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.576$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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%