Question

In a Princeton Survey of 800 randomly selected teens, 272 of them admitted to texting while driving. Obtain a 95% confidence interval for the proportion of teens who text while driving.

Answer 1

The 95% confidence interval would be given (0.307;0.373).

Explanation:

Data given and notation

n=800 represent the random sample taken

X=272 represent the teens who admit texting while driving

`\hat p=(272)/(800)=0.34` proportion estimated for teens who admit texting while driving

`\alpha=0.05` represent the significance level (no given, but is assumed)

Confidence =0.95 or 95%

p= population proportion of teens who admit texting while driving

The confidence interval for the population proportion would be given by this formula:

`\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}`

For the 95% confidence interval the value of
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`0.34 - 1.96 \sqrt{(0.34(1-0.34))/(800)}=0.307`

`0.34 + 1.96 \sqrt{(0.34(1-0.34))/(800)}=0.373`

And the 95% confidence interval would be given (0.307;0.373).

We are confident (95%) that about 30.7% to 37.3% of the teens are texting while driving