## Essentials of Statistics (5th Edition)

$\mu$ is between -6.59 and 7.39. The result suggests that the method is not too effective because the interval contains 0.
$\alpha=1-0.98=0.02.$ $\sigma$ is 21, hence we use the z-distribution with $df=sample \ size-1=49-1=48$ in the table. $z_{\alpha/2}=z_{0.01}=2.33.$ Margin of error:$z_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=2.33\cdot\frac{21}{\sqrt{49}}=6.99.$ Hence the confidence interval:$\mu$ is between 0.4-6.99=-6.59 and 0.4+6.99=7.39. The result suggests that the method is not too effective because the interval contains 0.