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For a 99 % confidence level, find where
\begin{equation*} F(z_{ \frac{0.01}{2}}) = P( z \lt z_{ \frac{0.01}{2}}) = 0.995 = 1 - \frac{0.01}{2} . \end{equation*}
The calculators InvNorm(0.995) gives
\(z_{ 0.005} \approx 2.576\text{.}\)
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