## Essentials of Statistics (5th Edition)

$\mu$ is between 22.858 and 25.042.
$\alpha=1-0.99=0.01.$ $\sigma$ is known(2.55), hence we use the z-distribution with $df=sample \ size-1=40-1=39$ in the table. $z_{\alpha/2}=t_{0.005}=2.708.$ Margin of error:$z_{\alpha/2}\cdot\frac{s}{\sqrt {n}}=2.708\cdot\frac{2.55}{\sqrt{40}}=1.092.$ Hence the confidence interval:$\mu$ is between 23.95-1.092=22.858 and 23.95+1.092=25.042. We are 99% confident that the mean falls within the confidence interval of chocolate chips.