Question

An insurance company checks police records on 582 accidents selected at random and notes that teenagers were at the wheel 15.6% of the time. The standard deviation is 1.2%. Determine the confidence interval that is likely to contain the percent of all teenagers at the wheel when an accident occurred.

Answer 1

At 95% confidence level, the confidence interval that is likely to contain the percent,
`\hat p`, of all teenagers at the wheel when an accident occurred is given as follows;

12.7% <
`\hat p` < 18.6%

Explanation:

The parameters given are as follows;

Sample size, n = 582

Proportion of accidents where teenagers are at the wheel,
`\hat p` = 15.6%

x = 0.156*582 ≈ 91

The formula for the confidence interval is as follows;

`CI=\hat{p}\pm z* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

Where:

z = z value at the confidence level

We therefore have;

`CI=0.156\pm z* \sqrt{(0.156(1-0.156))/(582)}`

`CI=0.156\pm z* 1.504 * 10^(-2)`

At a confidence level of 95%, z = 1.96

The confidence interval at 95% confidence level is therefore;

12.7% <
`\hat p` < 18.6%.