Question

A survey of 1060 randomly selected US teens ages 13 to 17 found that 605 of them say they have made a new friend online

Answer 1

Answer:

The 90% confidence interval for the proportion of all US teens who have made a new friend online is (0.55, 0.60).

Step-by-step explanation:

The complete question is:

A survey of 1060 randomly selected US teens ages 13-17 found that 605 of them say they have made a new friend online.  Find and interpret a 90% confidence interval for the proportion, p, of all US teens who have made a new friend online.

Solution:

In a sample of 1060 randomly selected US teens ages 13 to 17, 605 said they have made a new friend online.

Compute the sample proportion of US teens ages 13 to 17 who said they have made a new friend online as follows:

$\hat p=\frac{605}{1060}=0.571$

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The critical value of z for 90% confidence interval is:

$z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645$

*use a z-table.

Compute the 90% confidence interval for the proportion of all US teens who have made a new friend online as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

     $=0.571\pm 1.645\sqrt{\frac{0.571(1-0.571)}{1060}}\\=0.571\pm 0.025\\=(0.546, 0.596)\\\approx (0.55, 0.60)$

Thus, the 90% confidence interval for the proportion of all US teens who have made a new friend online is (0.55, 0.60).