Question

What is the 99% confidence interval for a sample of 52 seat belts that have a mean length of 85.6 inches long and a population standard deviation of 3.8 inches?

Answer 1

We need o find the 99% confidence interval for a sample with:

`\begin{gathered} n=52 \\ \\ \overline{x}=85.6\text{ in} \\ \\ \sigma=3.8\text{ in} \end{gathered}`

A 99% confidence interval has a z-value z = 2.576.

And the confidence interval is given by:

`\overline{x}\pm z*(\sigma)/(√(n))`

Thus, we obtain:

`\begin{gathered} 85.6\text{ in}\pm2.576*\frac{3.8\text{ in}}{√(52)} \\ \\ \cong85.6\text{ }\imaginaryI\text{n}\pm1.4\text{ in} \end{gathered}`

Approximating to the nearest tenth, the 99% confidence interval is:

Answer

`\lbrack84.2\text{ in},87.0\text{ in}\rbrack`